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Page No 23.17:

Question 1:

If P, Q and R are three collinear points such that PQ =a  and QR =b . Find the vector PR .

Answer:

Given: P, Q and R are collinear such that PQ = a and QR = b. Then,
PQ+ QR = PR           
 PR = a + b

Page No 23.17:

Question 2:

Give a condition that three vectors a, b and c form the three sides of a triangle. What are the other possibilities?

Answer:

Let ABC be a triangle such that BC = aAB = c and CA = b. Then,
   a+b+c= BC+CA+AB
a+b+c = BA+AB                          [∵ BC+CA = BA]
a+b+c = BB                                   [ Using triangle law]
 a+b+c = 0                                    [ By definition of null vector]

Other possibilities are
i  c+ a  = b
ii  a + b = c
iii b+ c = a

Page No 23.17:

Question 3:

If a and b are two non-collinear vectors having the same initial point. What are the vectors represented by a + b and ab.

Answer:



Given: a , b are two non-collinear vectors having same initial points. Complete the parallelogram ABCD such that AB= a and BC= b. 
In ABC,
      AB + BC =AC a + b = AC
In ABD,
     AD + DB = AB b + DB = a DB = a - b

Therefore, AC and DB are the diagonals of a parallelogram whose adjacent sides are a and b respectively.

Page No 23.17:

Question 4:

If a is a vector and m is a scalar such that m a = 0, then what are the alternatives for m and a?

Answer:

Given: a is a vector and m is a scalar such that, ma = 0
Then either m= 0  or,  a = 0

Page No 23.17:

Question 5:

If a, b  are two vectors, then write the truth value of the following statements:
(i) a =-b  a = b 

(ii)  a = b a =±b 

(iii)  a = b a =b 

Answer:

(i) True.
   a = -b
 Taking modulus on both sides of the equation, we get,
   a = -b
a = b                            [ âˆµ -b = b ]

(ii) False.
We cannot say  a = b a =±b 
Consider an example,
a=i+3j and b=2i+2ja=12+32 =2 and b=22+22=2Thus, a= b but a±b

(iii) False.
We cannot say  a = b a =b 
Consider an example,
a=i+3j and b=2i+2ja=12+32 =2 and b=22+22=2Thus, a= b but ab

Page No 23.17:

Question 6:

ABCD is a quadrilateral. Find the sum the vectors BA, BC, CD and DA.

Answer:

Given: ABCD is a quadrilateral. 
We need to find the sum of BA + BC + CD + DA.
Consider,
      BA + BC + CD + DA= BA +DA + BC + CD 
 = BD + 2 DA + BD                            [∵ BD + DA = BA and BC + CD = BD]
 = 2 BD + DA= 2 BA

Page No 23.17:

Question 7:

ABCDE is a pentagon, prove that
(i) AB+BC+CD+DE+EA=0

(ii) AB+AE+BC+DC+ED+AC=3 AC

Answer:

Given: ABCDE is a pentagon.
(i) To Prove: AB + BC + CD + DE + EA = 0.
    Proof: We have,
       LHS = AB + BC + CD + DE + EA         
              = AC + CD + DA                              [ âˆµ AB + BC = AC  and  DE + EA = AD]
              = AD + DA                                         [ âˆµ AC + CD = AD]
              = 0= RHS
(ii) To Prove: AB + AE + BC + DC + ED + AC = 3 AC.
    Proof: We have,
    LHS = AB + AE + BC + DC + ED + AC
          = AB + BC + AE + ED + DC + AC      ​ 
          = AC +AD + DC + AC     [∵​ AB + BC = AC  and  AE + ED = AD]
             =  AC + AD + AC - AD + AC                     [∵  AD + DC= AC  DC = AC - AD ]
            = 3 AC= RHS
Hence proved.
           

Page No 23.17:

Question 8:

Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.

Answer:

Given: A regular octagon of eight sides with centre O.
To show: OA + OB + OC + OD + OE + OF + OG + OH = 0.
Proof: We know centre of the regular octagon bisects all the diagonals passing through it.
OA =-OE ,  OB =-OF , OD =-OH and OC =-OG.
 OA + OE = 0 , OB + OF = 0 , OD + OH = 0 and  OC + OG = 0.    ............(i)
Now, 
      OA + OB + OC + OD + OE + OF +OG+ OH=OA + OE + OB + OF + OC + OG+ OD + OH= 0 + 0 + 0 + 0=0
Hence proved.
  

Page No 23.17:

Question 9:

If P is a point and ABCD is a quadrilateral and AP+PB+PD=PC, show that ABCD is a parallelogram.

Answer:

Given: ABCD is a quadrilateral such that AP + PB + PD = PC.
To show: ABCD is a parallelogram.
Proof: Consider,  
                          AP + PB + PD = PC AP + PB = PC  - PD
                     AB = DC                         [ âˆµ AP + PB = AB   and    PD + DC = PC]
 Again,
                AP + PB + PD = PC AP + PD = PC - PB
          AD = BC                                      [ âˆµ  AP + PD = AD    and   PB + BC = PC]
       
Since, opposite sides of the quadrilateral are equal and parallel.
Hence, ABCD is a parallelogram.

Page No 23.17:

Question 10:

Five forces AB, AC, AD, AE  and AF  act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is
6 AO, where O is the centre of hexagon.

Answer:



 AB+AC+ AD+AE+AF

Consider ∆ADE,

AD+DE+EA=0AD+DE=AE2AO+AB=AE                AD=2AOand ED ||ABDE=ABAE+AB=2AO                 .....(1)

Now, consider ∆ADC

AC+CD+DA=0AC+CD=AD                  CD=AFAC+AF=2AO                  .....2

Using (1) and (2),

AB+AE+AC+AF+AD2AO+2AO+2AO=6AO
 



Page No 23.23:

Question 1:

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors OP=2a+b and OQ=a-2b, respectively in the ratio 1 : 2 internally and externally.                                                                  [NCERT EXEMPLAR]

Answer:


It is given that P and Q are two points with position vectors OP=2a+b and OQ=a-2b, respectively.

When R divides PQ internally in the ratio 1 : 2, then

Position vector of R = 1×a-2b+2×2a+b1+2=5a3

When R divides PQ externally in the ratio 1 : 2, then

Position vector of R = 1×a-2b-2×2a+b1-2=-3a-4b-1=3a+4b



Page No 23.24:

Question 2:

Let a, b, c, d  be the position vectors of the four distinct points A, B, C, D. If b -a =c -d , then show that ABCD is a parallelogram.

Answer:

Given: a, b, c and d are the position vectors of the four distinct points A, B, C and D.
Also, we have, b - a =c  - d.     
  AB = DC
Again,
   b - a = c - d b - c = a - d CB = DA
Consequently, AB  DC , CB  DA and AB = DC ,  CB = DA. Thus two of its opposite sides are equal and parallel.
Hence, ABCD is a parallelogram.

Page No 23.24:

Question 3:

If a, b are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3 AB and that a point D in BA produced such that BD = 2BA.

Answer:


Let the position vectors of C and D are c and d respectively. We have,
     AC=3 AB. AC = 3 (AC-BC).2AC =3BC.
 ACBC = 32.
So C divides AB in the ratio of 3:2 externally.
c = 2a- 3b 2-3 = 3b- 2a.

Position vector of point C is 3b-2a
Moreover,
    BD = 2 BA.BD = 2(BD-AD). BD = 2AD.
 BDAD = 21.
∴ d = b- 2a1-2= 2a - b.
Position vector of point D is 2a-b 

Page No 23.24:

Question 4:

Show that the four points A, B, C, D with position vectors a, b,  c,  d respectively such that 3a - 2b + 5c - 6d =0, are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

Answer:

Let AC and BD intersects at a point P
We have,
    3a-2b+5c-6d = 0.3a+5c =2b+6dSince sum of coefficients on both sides of the above equation is 8.so we divide the equation on both sides by 8.3a+5c8 = 2b+6d83a+5c3+5 = 2b+6d2+6

Therefore, P divides AC in the ratio of  3:5 and P divides BD in the ratio of 2:6.
Therefore, position vector of the point of intersection of AC and BD will be
3a+ 5c8=2b+6d8
 

Page No 23.24:

Question 5:

Show that the four points P, Q, R, S with position vectors p, q, r, s respectively such that 5p−2q+6r−9s=0, are coplanar. Also, find the position vector of the point of intersection of the line segments PR and QS.

Answer:

Let the point of intersection of the line segments PR and QS is A. Then
    5p-2q+6r-9s =0.5p+6r=2q + 9sthe sum of the coefficients on both the sides of the above equation is 11.So, we divide the given equation with 11.5p+6r11 = 2q+9s11

5p+6r5+6=2q+9s2+9



Therefore, A divides PR in the ratio of 5:6 and QS in the ratio of 2:9.
The position vector of the point of intersection of the line segment is 5p+6r11, 2q+9s11.

Page No 23.24:

Question 6:

The vertices A, B, C of triangle ABC have respectively position vectors a , b , c  with respect to a given origin O. Show that the point D where the bisector of ∠A meets BC has position vector d =βb +γc β+γ, where β= c -a  and, γ= a-b .
Hence, deduce that the incentre I has position vector αa +βb +γc α+β+γ, where α= b -c .

Answer:

Let the position vectors of A, B and C with respect to some origin, O be a, b and c respectively.

Let D be the point on BC where bisectors of ∠A meets.
Let d be the position vector of D which divides CB internally in the ratio β and γ, where β=AC and γ=AB
Thus, β=c-a and γ=b-a
By section formula, the position vector of D is given by
OD=βb+γcβ+γLet α=b-c

Incentre is the concurrent point of angle bisectors and incentre divides the line AD in the ratio ∝: β + γ.

So, the position vector of incentre is given as,

αa+βb+γcβ+γ β+γα+β+γ=αa+βb+γcα+β+γ



Page No 23.36:

Question 1:

If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that OA+OB+OC=OD+OE+OF.

Answer:


Let D, E and F are the midpoints of BC, CA and AB respectively.
Therefore,
OB+OC2 = OD
OB + OC = 2 OD     .....1Similarly,OC+OA = 2 OE     .....2OA + OB = 2 OF     .....3

Adding (1), (2) and (3). We get,

      2 (OA + OB + OC) = 2 (OD + OE + OF). OA + OB + OC = OD + OE + OF .                                 
Hence Proved.

Page No 23.36:

Question 2:

Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

Answer:

Let a, b and c are the position vectors of the vertices A, B and C respectively.
Then we know that the position vector of the centroid O of the triangle is a+ b+ c3.
Therefore sum of the three vectors OA, OB and OC is
 OA + OB + OC = a - a+ b + c3 + b - a + b + c3 + c - a + b + c3                               =  a + b + c - 3 a + b + c3                               = 0
Hence, Sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.



Page No 23.37:

Question 3:

ABCD is a parallelogram and P is the point of intersection of its diagonals. If O is the origin of reference, show that OA+OB+OC+OD=4 OP.

Answer:



Given a parallelogram ABCD and P is the point of intersection of its diagonals. We know the diagonals of a parallelogram, bisect each other. Therefore,
OA+OC2= OP OA+OC=2 OP      .....1and OB+OD2=OPOB+OD=2 OP      .....2
Adding (1) and (2), We get,
OA+OB+OC+OD=4OP

Page No 23.37:

Question 4:

Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.

Answer:

Let ABCD is the quadrilateral and P, Q, R, S are mid points of the sides AB, BC, CD, DA respectively.

Join DB to form triangle ABD.

ASSD=APPBSP || DBandSP=12DB

In triangle BCD

CRRD=CQQBRQ || DBandRQ=12DB

In quadrilateral PQRS,
SP = RQ and SP || RQ
∴ PQRS is a parallelogram.

Diagonals of a parallelogram bisect each other.

∴ PR and QS bisect each other.

Page No 23.37:

Question 5:

ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD; BC and AD. Show that PA+PB+PC+PD=4 PQ, where P is any point.

Answer:



Let E, F, G and H are the midpoints of the sides AB, BC, CD and DA respectively of say quadrilateral ABCD.  By geometry of the figure formed by joining the midpoints E, F, G and H will be a parallelogram. Hence its diagonals will bisect each other, say at Q.
Now, F is the midpoint of BC.
PB + PC2 = PF PB + PC = 2PF     .....1
And, H is the midpoint of AD.
PA + PD2=PH PA + PD= 2PH     .....2
Adding (1) and (2). We get,
PA+ PB+ PC+ PD =2(PF + PH) = 2 (2PQ) =4 PQ. 

Page No 23.37:

Question 6:

Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.

Answer:



Let ABC be a triangle and α, β, γ be the position vectors of the vertices  A, B and C respectively. Let AD, BE and CF be the internal bisectors of A, B and C respectively.
We know that D divides BC in the ratio of AB : AC that is c : b.
Then,
P.V. of D is  cγ + bβc+b.
P.V. of E is cγ + aαc+a.
and   P.V. of F is aα+bβa+b.
 
The point dividing AD in the ratio b+c : a is aα + bβ + cγa+b+c.
The point dividing BE in the ratio of a+c : b is aα + bβ + cγa+b+c.
The point dividing CF in the ratio of a+b : c is aα + bβ + cγa+b+c.

Since the point aα+ bβ + cγa+b+c lies on all the three internal bisectors AD, BE and CF.
Hence the internal bisectors are concurrent .



Page No 23.4:

Question 1:

Represent the following graphically:
(i) a displacement of 40 km, 30° east of north
(ii) a displacement of 50 km south-east
(iii) a displacement of 70 km, 40° north of west.

Answer:

(i) The vector OP represents the required displacement vector.
(ii) The vector OQ represents the required vector.
(iii) The vector OR represents the required vector.

Page No 23.4:

Question 2:

Classify the following measures as scalars and vectors:
(i) 15 kg
(ii) 20 kg weight
(iii) 45°
(iv) 10 meters south-east
(v) 50 m/sec2

Answer:

The quantities which have only magnitude and which are not related to any
fixed direction in space are called scaler quantities or simply scalars.
The quantities which have both magnitude and direction are called vector quantities or simply vectors.

(i) Mass - Scalar
(ii) Weight(Force) - Vector
(iii) Angle - Scalar
(iv) Directed Disptance-  Vector
(v) Magnitude of acceleration - Scalar

Page No 23.4:

Question 3:

Classify the following as scalars and vector quantities:
(i) Time period
(ii) Distance
(iii) displacement
(iv) Force
(v) Work
(vi) Velocity
(vii) Acceleration

Answer:

The quantities which have only magnitude and which are not related to any
fixed direction in space are called scaler quantities or simply scalars.
The quantities which have both magnitude and direction are called vector quantities or simply vectors.

(i) Scalar
(ii) Scalar
(iii) Vector
(iv)Vector
(v) Scalar
(vi) Vector
(vii) Vector

Page No 23.4:

Question 4:

In Figure ABCD is a regular hexagon, which vectors are:
(i) Collinear
(ii) Equal
(iii) Coinitial
(iv) Collinear but not equal.
Figure

Answer:

(i) Vectors having the same or parallel supports are called collinear vector.
In the given figure the collinear vectors are
  a, d ; x, z, b ; c, y

(ii) Vectors having the same magnitude and direction are called equal vector.
In the given figure the equal vectors are
  b, x ; c, y ; a, d

(iii) Vectors having the same initial point are called co-initial vector.
In the given figure the co-initial vectors are
a, y, z 

(iv) The vectors which are collinear but not equal are   b, z ; x, z

Page No 23.4:

Question 5:

Answer the following as true or false:
(i) a  and a  are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Zero vector is unique.
(iv) Two vectors having same magnitude are collinear.
(v) Two collinear vectors having the same magnitude are equal.

Answer:

(i) True, As vectors having the same and parallel support are collinear.
(ii) False, Collinear vectors are parallel vector not equal vectors.
(iii) False.
(iv) False, Collinear vectors may not have a same magnitude.
(v) False, As two collinear vectors are equal only if they have same length and same sense.



Page No 23.42:

Question 1:

If the position vector of a point (−4, −3) be a, find  a .

Answer:

Given a point -4,-3 such that its position vector a is given by
 a = -4 i - 3 j
Then,
      a = -42 + -32 = 16 + 9 = 25 = 5

Page No 23.42:

Question 2:

If the position vector a of a point (12, n) is such that  a  = 13, find the value (s) of n.

Answer:

Given a position vector a of a point 12, n such that,
 a = 12 i + n j
Then,
a = 122 + n2
Also , a = 13   (given)
Thus, we get,

      122 + n2 = 13  122 + n2 = 169 n2 = 169 - 144 n2 = 25 n = ±5

Page No 23.42:

Question 3:

Find a vector of magnitude 4 units which is parallel to the vector 3i^+j^.

Answer:

Let a = 3 i^ + j^
Then, a = 32 + 1 = 3+1 = 4 = 2
A unit vector parallel to a = a^ = aa = 123 i^ + j^
Hence, Required vector = 4a^ = 4×123 i^ + j^ = 23 i^ + 2j^

Page No 23.42:

Question 4:

Express AB in terms of unit vectors i^ and j^, when the points are:
(i) A (4, −1), B (1, 3)
(ii) A (−6, 3), B (−2, −5)
Find AB in each case.

Answer:

(i) Given: A (4,-1) and B1, 3
Then the position vector AB is given by
AB = Position vector of B - Position vector of A
    = i + 3 j - 4 i - j= i + 3 j - 4 i + j= -3 i + 4 j
So, AB = -32 + 42 = 9+16 = 25 = 5


(ii) Given: A-6, 3 and B -2,-5
Then, the position vector AB is given by
AB  =Position vector of B - Position vector of A
       = -2 i-5 j - -6 i + 3 j= -2 i-5 j + 6 i - 3 j= 4 i - 8 j
So, AB = 42 + -82 = 16 + 64 = 80 = 45

Page No 23.42:

Question 5:

Find the coordinates of the tip of the position vector which is equivalent to AB, where the coordinates of A and B are (−1, 3) and (−2, 1) respectively.

Answer:

Let O be the origin. Let Px, y be the required point. Then, P is the tip of the position vector OP of the point P.
We have,
        OP = x i + y j.
and,  AB = Position vector of B - Position vector of A.
              = -2 i + j - - i + 3 j= -2 i + j + i-3 j= - i - 2 j

Given that OP = AB
So,  x i + y j = - i -2 j  x =-1 ,  y=-2
Hence, coordinated of the required point is -1.-2



Page No 23.43:

Question 6:

ABCD is a parallelogram. If the coordinates of A, B, C are (−2, −1), (3, 0) and (1, −2) respectively, find the coordinates of D.

Answer:

Let the coordinates of D is x, y.
Since, ABCD is a parallelogram.
∴ AB = DC
We have,

     AB = DC 3i - -2i - j = i - 2j - xi + yj5i + j= i 1-x + j -2-y 1-x = 5  and 1 =-2-y x=-4 and y =-3
Hence, the coordinates of D is -4,-3

Page No 23.43:

Question 7:

If the position vectors of the points A (3, 4), B (5, −6) and C (4, −1) are a,b,c respectively, compute a + 2b -3c .

Answer:

Let a, b, c are the position vectors of the points A3, 4B5,-6 and C4,-1.
Then,
   a = 3 i + 4 j
  b = 5 i - 6 j
  c= 4 i - j
Therefore,
       a + 2b - 3c =3 i + 4 j + 2 5 i - 6j - 3 4 i - j = 3 i + 4 j + 10 i - 12 j - 12 i + 3j = i  - 5 j

Page No 23.43:

Question 8:

If a be the position vector whose tip is (5, −3), find the coordinates of a point B such that AB=a , the coordinates of A being (4, −1).

Answer:

Let O be the origin and let P5,-3 be the tip of the position vector a. Then, a = OP = 5i^ - 3j^. Let the coordinate of B be x, y and A has coordinates 4,-1.
Therefore,
AB = Position vector of B - Position vector of A
    = xi^ + yj^ - 4i^ - j^= x-4i^ + y+1j¯
Now, 
    AB = ax-4i^ + y+1j^ = 5i^ - 3j^ x-4 = 5 and y+1 =- 3 x=9 and y=-4
Hence, the coordinates of B are 9,-4.

Page No 23.43:

Question 9:

Show that the points 2i^, − i^ − 4j^ and −i^ + 4j^ form an isosceles triangle.

Answer:

Given:- The points A, B, C with position vectors a, b, c respectively.
Also, a = 2i^
       b =-i^ - 4j^
      c =-i^ + 4j^
Then, 
      AB  = b - a AB = -i^- 4j^ - 2i^ AB = -3i^ - 4j^Now,  AB = -32 +- 42=9+16=25=5      BC = c - b BC = -i^ + 4j^ - -i^ - 4j^ BC =-i^ + 4j^ + i^ + 4j^ BC = 8j^
and 

     AC = c - aAC = -i^ + 4j^ - 2i^ AC = -3i^ + 4j^Now,  AC = -32 + 42=9+16=25=5

Since, the magnitude of AB and AC is equal.
Hence, the points 2i^, − i^ − 4j^ and −i^ + 4j^ form an isosceles triangle.

Page No 23.43:

Question 10:

Find a unit vector parallel to the vector i^+3j^.

Answer:

Let a = i^ + 3 j^
Then, a= 12+ (3)2 = 1+3 = 4 = 2
Unit vector parallel to a = a^ = aa = 12i^ + 3 j^=12i^ + 32 j^

Page No 23.43:

Question 11:

The position vectors of points A, B and C are λi^+ 3j^, 12i^+μj^ and 11i^-3j^ respectively. If C divides the line segment joining A and B in the ratio 3:1, find the values of λ and μ

Answer:


The position vectors of points A, B and C are λi^+ 3j^, 12i^+μj^ and 11i^-3j^, respectively.

It is given that, C divides the line segment joining A and B in the ratio 3 : 1.

11i^-3j^=3×12i^+μj^+1×λi^+3j^3+111i^-3j^=36+λi^+3μ+3j^444i^-12j^=36+λi^+3μ+3j^
Equating the corresponding components, we get

36+λ=44 

λ=44-36=8 

and 

3μ+3=-12

3μ=-12-3=-15

μ=-5

Thus, the values of λ and μ are 8 and −5, respectively.



Page No 23.48:

Question 1:

Find the magnitude of the vector a=2i^+3j^-6k^.

Answer:

Given: a = 2i^ + 3j^ - 6k^
∴ Magnitude of the vector = a = 22+32+-62 = 4+9+36 = 49 = 7

Page No 23.48:

Question 2:

Find the unit vector in the direction of 3i^+4j^-12k^.

Answer:

Let a = 3i^ + 4j^ -12k^
Then, a = 32+ 42+ -122 = 9+16+144 =169 =13
So, a unit vector in the direction of a is given by
a^ = aa = 113 3i^ + 4j^ -12k^=313i^ + 413j^ -1213k^

Page No 23.48:

Question 3:

Find a unit vector in the direction of the resultant of the vectors i^-j^+3k^, 2i^+j^-2k^ and i^+2j^-2k^.

Answer:

Given: a = i^ - j^ + 3k^ ,  b = 2i^ + j^- 2k^  and  c=i^ + 2j^ - 2k^  are the position vectors.
Then, Resultant of the vectors = a + b + c
                                                 = i^ - j^ + 3k^ + 2i^ + j^ - 2k^ + i^ + 2j^ - 2k^= 4i^ + 2j^ - k^

So, a + b + c = 42+ 22 + 12 = 16 + 4 + 1 = 21
∴ Unit vector in the direction of the resultant vector = a + b + ca + b + c = 1214i^ + 2j^ - k^



Page No 23.49:

Question 4:

The adjacent sides of a parallelogram are represented by the vectors a=i^+j^-k^ and b=-2i^+j^+2k^. Find unit vectors parallel to the diagonals of the parallelogram.

Answer:



a=i^+j^k^b=2i^+j^+2k^

  AC=a+b=i^+2j^+k^

Let the unit vector along the diagonals AC and BD of the parallelogram be AC^ and BD^.AC^=i^+2j^+k^6

Similarly,BD=ba=-3i^+3k^

BD^=-3i^+3k^32=-i^+k2

Page No 23.49:

Question 5:

If a=3i^-j^-4k^, b=-2i^+4j^-3k^ and c=i^+2j^-k^, find  3a-2b+4c .

Answer:

Given: a = 3i^ - j^- 4k^,   b =-2i^ + 4j^ - 3k^ and c = i^ + 2j^ - k^.

Now, 3a - 2b+ 4c = 33i^ - j^ - 4k^ - 2-2i^ + 4j^ - 3k^ + 4i^ + 2j^ - k^                                     = 9i^ - 3j^ - 12k^ + 4i^ - 8j^ + 6k^ + 4i^ + 8j^ - 4k^                                     = 17i^ - 3j^ - 10k^
Hence, 3a - 2b + 4c = 172 + -32 + -102 = 289+9 + 100 = 398

Page No 23.49:

Question 6:

If PQ=3i^+2j^-k^ and the coordinates of P are (1, −1, 2), find the coordinates of Q.

Answer:

Given: PQ = 3i^ + 2j^- k^. Let the position vector of P1,-1, 2 is p such that p = i^-j^+2k^ and the position vector of Qx, y, z is q such that q = xi^ +yj^+ zk^.
Therefore,
  PQ = q - p 3i^ + 2j^ - k^  = xi^ + yj^ + zk^ - i^ - j^ + 2k^ 3i^ + 2j^ - k^ = x-1i^ + y+1j^ + z-2k^ x-1 = 3 , y+1 = 2, z-2 =-1 x= 4 , y=1, z=1
Hence, the coordinates of Q are 4, 1, 1

Page No 23.49:

Question 7:

Prove that the points i^-j^, 4i^+3j^+k^ and 2i^-4j^+5k^ are the vertices of a right-angled triangle.

Answer:

Given the points i^-j^, 4i^+3j^+k^ and 2i^-4j^+5k^  Are A, B and C respectively.
Then,
AB = 4i^+3j^+k^ - i^+j^ = 3i^+4j^+k^.BC = 2i^-4j^+5k^-4i^-3j^-k^ = -2i^-7j^+4k^.CA = i^-j^-2i^+4j^-5k^ =-i^+3j^-5k^.
AB+BC+CA = 3i^+4j^+k^-2i^-7j^+4k^-i^+3j^-5k^ = 0.
The given points forms a vertices of a triangle.
Now,
AB = 9+16+1 = 26.BC = 4+49+16 = 69.CA = 1+9+25 = 35.
AB2 + CA2 =26+35 = 61 ≠ BC2
The given triangle is not right-angled.

Page No 23.49:

Question 8:

If the vertices of a triangle are the points with position vectors a1i^+a2j^+a3k^, b1i^+b2j^+b3k^, c1i^+c2j^+c3k^, what are the vectors determined by its sides? Find the length of these vectors.

Answer:

Given the vertices of a triangle A, B and C with position vectors a1i^ + a2j^ + a3k^, b1i^+b2j^+b3k^ and c1i^+c2j^+c3k^ respectively. Then,
AB = (b1-a1)i^ + (b2-a2) j^ + (b3-a3)k^.BC = (c1-b1)i^ + (c2-b2) j^ + (b3-a3)k^.CA = (a1-c1)i^ + (a2-c2) j^ + (a3-c3) k^.
Therefore, the length of these vectors are:

AB = (b1-a1)2+ (b2-a2)2+ (b3-a3)2 .BC = (c1-b1)2+(c2-b2)2+(c3-b3)2.CA = (a1-c1)2+ (a2-c2)2 + (a3-c3)2.

Page No 23.49:

Question 9:

Find the vector from the origin O to the centroid of the triangle whose vertices are (1, −1, 2), (2, 1, 3) and (−1, 2, −1).

Answer:

Given the vertices of the triangle (1,-1, 2), (2,1, 3) and (-1, 2, -1). Then,
Position vectors are
a = i^-j^+2k^.b= 2i^+j^+3k^.c = -i^+2j^-k^.
The centroid of a triangle is given by a+b+c3
So, a+b+c3= i^-j^+2k^ + 2i^+j^+3k^- i^+2j^-k^3 = 2i^+2j^+4k^3=23i^+23j^+43k^.

Page No 23.49:

Question 10:

Find the position vector of a point R which divides the line segment joining points P i^+2j^+k^ and Q -i^+j^+k^ in the ratio 2:1.
(i) internally
(ii) externally

Answer:

(i) Given: R divides the line segment joining the points Pi^ + 2j^ +k^ , Q-i^ +j^+ k^ in the ratio 2 : 1 internally.
Therefore. position vector of R2-i^ + j^ + k^ + 1i^ + 2j^ + k^2+1
                                                  = 13-i^ + 4j^ + 3k^


(ii) Given:  R divides the line segment joining the points Pi^ + 2j^ +k^ , Q-i^ +j^+ k^ in the ratio 2 : 1 externally.
   Therefore. position vector of R2-i^ + j^ + k^ - 1i^ + 2j^ + k^2-1
                                                      = -3i^ + k^

Page No 23.49:

Question 11:

Find the position vector of the mid-point of the vector joining the points P 2i^-3j^+4k^ and Q 4i^+j^-2k^.

Answer:

Given: P2i^ -3j^ + 4k^ and Q4i^ + j^ - 2k
The position vector of the midpoint of the vector
joining these points =Position vector of P+Position vector of  Q2
                                = (2i^ - 3j^ + 4k^ )+ (4i^ + j^ - 2k^)2= 6i^ - 2j^ + 2k^2= 3i^ - j^ + k^

Page No 23.49:

Question 12:

Find the unit vector in the direction of vector PQ, where P and Q are the points (1, 2, 3) and (4, 5, 6).

Answer:

Let a and b are the position vectors of the points  P1, 2, 3 and Q4, 5, 6
Then,
 a = i^ + 2j^ + 3k^b = 4i^ + 5j^ + 6k^
So, 
PQ = b - a       = 4i^ + 5j^ + 6k^ - i^ - 2j^ - 3k^       = 3i^ + 3j^ + 3k^

Now, PQ= 32 + 32 + 32 = 9+9+9 = 33
Therefore, Unit vector parallel to PQ = PQPQ = 133 3i^ + 3j^ + 3k^ = 13i^+j^+k^

Page No 23.49:

Question 13:

Show that the points A 2i^-j^+k^, B i^-3j^-5k^, C 3i^-4j^-4k^ are the vertices of a right angled triangle.

Answer:

Given the points A2i^ -j^ + k^, Bi^- 3j^ - 5k^ and C3i^ - 4j^ - 4k^.
Then,
AB = Position vector of B- Position vector of A
     = i^ - 3j^ - 5k^ - 2i^ -j^ + k^= i^ - 3j^ - 5k^ - 2i^ + j^ - k^= -i^ - 2j^ -6k^

BC = Position vector of C - Position vector of B
     =3i^ - 4j^ - 4k^ - i^ - 3j^ - 5k^= 3i^ - 4j^ - 4k^ - i^ + 3j^ + 5k^= 2i^ - j^ + k^

CA = Position vector of A - Position vector of C
     = 2i^ - j^ + k^ - 3i^ - 4j^ - 4k^= 2i^ - j^ + k^ - 3i^ + 4j^ + 4k^= -i^ + 3j^ + 5k^
Clearly, AB + BC + CA = 0

Now,  AB=-12+-22+-62=1+4+36=41 BC=22+-12+12=4+1+1=6 CA=-12+32+52=1+9+25=35Clearly,  AB2= BC2+ CA2AB2=BC2+CA2

So, A, B , C forms a right angled triangle.

Page No 23.49:

Question 14:

Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q (4, 1, −2).

Answer:

Let p, q be the position vectors of the points P2, 3, 4, Q4, 1,-2
Then,
 p = 2i^+3j^+ 4k^  and   q= 4i^+j^-2k^
 Therefore, the position vector of the midpoint of the given points is p+q2
∴ p+q2 = (2i^+3j^+4k^)+(4i^+j^-2k^)2 = 3i^+2j^+k^

Page No 23.49:

Question 15:

Find the value of x for which x i^+j^+k^ is a unit vector.

Answer:

We have, xi^+j^+k^ is a unit vector.
x2+x2+x2=13x=1x=13x=±13

Page No 23.49:

Question 16:

If a =i^+j^+k^, b =2i^-j^+3k^ and c =i^-2j^+k^,find a unit vector parallel to 2 a -b +3 c.

Answer:

We have, a= i^+j^+k^b= 2i^-j^+3k^  and c= i^-2j^+k^
∴ 2a-b+3c = 2i^+j^+k^ - 2i^-j^+3k^+ 3i^-2j^+k^ = 3i^-3j^+2k^.

A unit vector parallel to 2a-b+3c is given by 2a-b+3c2a-b+3c=3i^-3j^+2k^32+-32+22
                                                                                           = 3i^-3j^+2k^22 =322i^-322j^+222k^

Page No 23.49:

Question 17:

If a =i^+j^+k^, b =4i^-2j^+3k^ and c =i^-2j^+k^, find a vector of magnitude 6 units which is parallel to the vector 2a -b +3c .

Answer:

We have, a= i^+j^+k^, b=4 i^-2j^+3k^ and c = i^-2j^+k^.
Then, 
     2a-b+3c= 2i^+j^+k^ - 4i^-2j^+3k^ + 3i^-2j^+k^ = i^-2j^+2k^.                
∴ A unit vector parallel to 2a-b+3c is 2a-b+3c2a-b+3c=i^-2j^+2k^12+-22+22

                                                                                = i^-2j^+2k^9  = i^-2j^+2k^3
Hence, Required vector = 63i^-2j^+2k^ = 2i^-4j^+4k^.

Page No 23.49:

Question 18:

Find a vector of magnitude of 5 units parallel to the resultant of the vectors a =2i^+3j^-k^ and b =i^-2j^+k.^

Answer:

Given the position vectors a = 2i^+3j^-k^  and b = i^-2j^+k^
∴ Resultant Vector = a+b = 2i^+3j^-k^+i^-2j^+k^ = 3i^+j^
So, a unit vector parallel to the resultant vector is 3i^+j^3i^+j^=3i^+j^32+12 = 3i^+j^10
Hence, required vector = 5×3i^+j^10 = 523i^+j^

Page No 23.49:

Question 19:

The two vectors j^+k^ and 3i^-j^+4k^ represents the sides AB and AC respectively of a triangle ABC. Find the length of the median through A.             [CBSE 2015]

Answer:


Disclaimer: The question has been solved by taking the vector AB as j^+k^.


In ∆ABC, AB=j^+k^ and AC=3i^-j^+4k^.

Let the position vector of A be (0, 0, 0). Then, the position vectors of B and C are (0, 1, 1) and (3, −1, 4), respectively.



Suppose D be the mid-point of the line segment joining the points B(0, 1, 1) and C(3, −1, 4).

∴ Position vector of D =j^+k^+3i^-j^+4k^2=3i^+5k^2=32i^+52k^

Now,

Length of the median, AD = AD=32i^+52k^-0i^+0j^+0k^=32i^+52k^=322+02+522=344=172 units



Page No 23.60:

Question 1:

Show that the points A, B, C with position vectors a - 2b + 3c , 2a + 3b - 4c  and -7b + 10c  are collinear.

Answer:

We have, A, B ,C with position vectors  a-2b+3c,  2a+3b-4c, -7b+10c Then,
AB = Position Vector of B - Position Vector of A
     = 2a+3b-4c - a+2b-3c= a+5b-7c

BC = Position Vector of C - Position Vector of B
     =-7b+10c -2a-3b+4c=-2a-10b+14c=-2 a+5b-7c
∴ BC =-2AB
Hence, AB and BC are parallel vectors.
But B is a point common to them.
So, AB and BC are collinear.
Hence, points A, B and C  are collinear.

Page No 23.60:

Question 2:

If a, b, c are non-coplanar vectors, prove that the points having the following position vectors are collinear:
(i) a, b, 3a - 2b 

(ii) a + b + c , 4a + 3b , 10a + 7b - 2c 

Answer:

(i) Given: a, b, c are non coplanar vectors.
Let  the points be A, B, C respectively  with position vectors a, b, 3a-2b. Then,
AB= Position vector of B - Position vector of A
    = b-a

BC = Position vector of C - Position vector of B
     = 3a-2b - b= 3a - 3b=-3 b-a
∴ BC=-3AB
So, AB and BC are parallel vectors.
But B is a point common to them.
Hence, A, B and C are collinear.


 (ii) Given a, b, c are non coplanar vectors.
Let the points be A, B, C respectively with the position vectors a+b+c, 4a+3b, 10a+7b-2c. Then,
AB = Position vector of B - Position vector of A
      = 4a+3b - a-b-c= 3a+2b-c

BC= Position vector of C - Position vector of B
    =10a+7b-2c-4a-3b= 6a+4b-2c= 23a+2b-c

∴ BC= 2AB
So, AB and BC are parallel vectors.
But B is a point common to them.
So, AB and BC are collinear.
Hence, A, B and C are collinear.

Page No 23.60:

Question 3:

Prove that the points having position vectors i^+2j^+3k^, 3i^+4j^+7k^, -3i^-2i^-5k^ are collinear.

Answer:

Let A, B, C be the points with position vectors i^+2j^+3k^, 3i^+4j^+7k^, -3i^-2j^-5k^. Then,
AB= Position vector of B - Position vector of A
    =3i^+ 4j^+ 7k^- i^- 2j^- 3k^= 2i^ + 2j^ + 4k^

BC = Position vector of C - Position vector of B
     =-3i^- 2j^ - 5k^ - 3i^ - 4j^ - 7k^= -6i^ - 6j^ -12k^= -32i^ + 2j^ + 4k^
∴ BC =-3AB
 So, AB and BC are parallel vectors.
But B is a point common to them.
So, AB and BC are collinear.
Hence, A, B, C are collinear.

Page No 23.60:

Question 4:

If the points with position vectors 10i^+3j^, 12i^-5j^ and ai^+11j^ are collinear, find the value of a.

Answer:

Let A, B, C be the points with position vectors 10i^+3j^, 12i^-5j^, ai^+11j^. Then,
AB = Position vector of B - Position vector of A
     = 12i^ - 5j^ - 10i^ - 3j^= 2i^ - 8j^

BC = Position vector of C - Position vector of B
    =ai^ + 11j^ - 12i^+ 5j^= a-12i^ + 16j^

Since, A, B and C are collinear.
∴ AB =λBC.
 2i^ - 8j^ = λ a-12i^ + λ16j^ 2 = λ a-12, -8= λ16 2= λa-12, λ=-12 2=-12a-12- a+12 = 4 a=8



Page No 23.61:

Question 5:

If a, b are two non-collinear vectors, prove that the points with position vectors a + b, a - b and a + λb  are collinear for all real values of λ.

Answer:

Given: a,b are non collinear vectors.
Let the position vectors of points A, B and C be a+b, a-b, a+λb^ respectively.
Then,
AB = P.V. of B − P.V. of A.
      = a - b - a - b.= -2b.

BC = P.V. of C − P.V. of B.
     = a+λb - a+b.= bλ-1.

CA = P.V. of A − P.V. of C.
     = a + b - a - λb.= b 1-λ.
Now, the position vectors are collinear if and only if  AB and CA is some multiple of BC.
So,
           AB =β BC -2b=β  bλ-1-2 = β λ-1β =-2λ-1 

and BC=-CA.
Hence, for real values of λ, the given position vectors are parallel.

Page No 23.61:

Question 6:

If AO+OB=BO+OC, prove that A, B, C are collinear points.

Answer:

We have,
    AO + OB = BO + OC.AO - BO = OC - OB.OB - OA = OC - OB. AB = BC.
Hence A, B and C are collinear points.

Page No 23.61:

Question 7:

Show that the vectors 2i^-3j^+4k^ and -4i^+6j^-8k^ are collinear.

Answer:

Given the position vectors 2i^-3j^+4k^ and -4i^+6j^-8k^
Let a= 2i^-3j^+4k^  and  b = -4i^+6j^-8k^
Then,
    b =-4i^+6j^-8k^    =-22i^-3j^+4k^    =-2a
Hence, a, b are collinear.

Page No 23.61:

Question 8:

If the points A(m, −1), B(2, 1) and C(4, 5) are collinear, find the value of m.

Answer:


The given points are A(m, −1), B(2, 1) and C(4, 5).

Now,

AB=2i^+j^-mi^-j^=2-mi^+2j^

AC=4i^+5j^-mi^-j^=4-mi^+6j^

If A, B, C are collinear, then

AB=λAC2-mi^+2j^=λ4-mi^+6j^2-m=λ4-m and 2=6λ                   Equating coefficients of i^ and j^λ=13
and
2-m=134-m6-3m=4-m2m=2m=1

Thus, the value of m is 1.

Page No 23.61:

Question 9:

Show that the points (3, 4), (−5, 16) and (5, 1) are collinear.

Answer:


Let the given points be A(3, 4), B(−5, 16) and C(5, 1).

Now,

AB=-5i^+16j^-3i^+4j^=-8i^+12j^

AC=5i^+j^-3i^+4j^=2i^-3j^

Clearly, AB=-4AC

Therefore, AB and AC are parallel vectors. But, A is a common point of AB and AC.

Hence, the given points (3, 4), (−5, 16) and (5, 1) are collinear.

Page No 23.61:

Question 10:

If the vectors a=2i^-3j^ and b=-6i^+mj^ are collinear, find the value of m.

Answer:


It is given that the vectors a=2i^-3j^ and b=-6i^+mj^ are collinear.

b=λa for some scalar λ

-6i^+mj^=λ2i^-3j^-6i^+mj^=2λi^-3λj^-6=2λ and m=-3λm=-3×-62=9

Thus, the value of m is 9.

Page No 23.61:

Question 11:

Show that the points A (1, −2, −8), B (5, 0, −2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

Given points A1,-2,-8, B5, 0,-2, C11, 3, 7.
Therefore,  AB = 5i^+0j^-2k^ - i^+2j^+8k^ = 4i^+2j^+6k^
                 BC = 11i^+3j^+7k^- 5i^+2k^ = 6i^+3j^+9k^
and,           AC = 11i^+3j^+7k^ -i^+2j^+8k^ = 10i^+5j^+15k^
Clearly, AB+BC= AC
Hence A, B, C are collinear.
Suppose B divides in the ratio AC in the ratio λ:1. Then the position vector B is 
11λ+1λ+1i^ + 3λ-2λ+1j^ + 7λ-8λ+1k^
But the position vector of B is 5i^+0j^-2k^.

 11λ+1λ+1 = 5, 3λ-2λ+1 = 0 , 7λ-8λ+1 =-211λ+1 = 5λ+5, 3λ-2 =0, 7λ-8 =-2λ-2 6λ = 4, 3λ = 2, 9λ = 6 λ=23, λ=23, λ=23

Page No 23.61:

Question 12:

Using vectors show that the points A (−2, 3, 5), B (7, 0, −1) C (−3, −2, −5) and D (3, 4, 7) are such that AB and CD intersect at the point P (1, 2, 3).

Answer:

We have,
AP=position vector of P - position vector of AAP = (i^+2j^+3k^) - (-2i^+3j^+5k^)          =3i^-j^-2k^        PB=position vector of B - position vector of PPB = (7i^-0j^-k^) - (i^+2j^+3k^)           =  6i^-2j^-4k^   Since PB=2AP. So, vectors PB and AP are collinear. But P is a pointcommon to PB and AP.

Hence, P, A, B are collinear points.

Now, CP= (-3i^ - 2j^ - 5k^) - (i^ + 2j^ + 3k^)             =(-4i^ - 4j^ -8k^)PD = (i^ + 2j^ + 3k^) - (3i^ + 4j^ + 7k^)      =(-2i^ - 2j^ -4k^)Thus, CP = 2PD.So the vectors CP and PD are collinear.But P is a common point to CP and PD
Hence, C,P,D are collinear points. 
Thus A, B, C, D and P are points such that A,P,B and C,P,D are two sets of collinear points.
Hence, AB and CD intersect at point P.

Page No 23.61:

Question 13:

Using vectors, find the value of λ such that the points (λ, −10, 3), (1, −1, 3) and (3, 5, 3) are collinear.            [NCERT EXEMPLAR]

Answer:


Let the given points be A(λ, −10, 3), B(1, −1, 3) and C(3, 5, 3).

AB=i^-j^+3k^-λi^-10j^+3k^=1-λi^+9j^

AC=3i^+5j^+3k^-λi^-10j^+3k^=3-λi^+15j^

If the points A, B, C are collinear, then

AB=kAC for some scalar k

1-λi^+9j^=k3-λi^+15j^1-λ=k3-λ and 9=15k                 Equating coefficients of i^ and j^1-λ=353-λ5-5λ=9-3λ2λ=-4λ=-2

Thus, the value of λ is −2.



Page No 23.65:

Question 1:

Show that the points whose position vectors are as given below are collinear:
(i) 2i^+j^-k^, 3i^-2j^+k^ and i^+4j^-3k^

(ii) 3i^-2j^+4k^, i^+j^+k^ and -i^+4j^-2k^

Answer:

(i) Let the points be A, B and C with position vectors 2i^ + j^ -k^,  3i^ -2j^ + k^  and i^ + 4j^ - 3k^. Then,
  AB = Position vector of B - Position vector of A
       = 3i^ -2j^ + k^ - 2i^ - j^ + k^= i^ - 3j^ + 2k^

BC = Position vector of C - Position vector of B
      = i^ + 4j^ - 3k^ - 3i^ + 2j^ - k^=-2i^+ 6j^ - 4k^=-2 i^ - 3j^ + 2k^

 AB =-2BC.
So, AB and BC are parallel vectors. But B is a point common to them.
Hence, A, B and C are collinear.

(ii) Let the points be A, B and C with position vectors 3i^-2j^ + 4k^,  i^ + j^ + k^  and -i^ + 4j^ -2k^ respectively. Then,
   AB = Position vector of B - Position vector of A
        = i^ + j^ + k^ - 3i^ + 2j^ - 4k^=-2i^ + 3j^ - 3k^

BC = Position vector of C - Position vector of B
      =-i^ + 4j^ - 2k^ - i^ -j^ - k^=-2i^ + 3j^ - 3k^

 AB = BC
So, AB and BC are parallel vectors.But B is a point common to them.
Hence, A, B and C are collinear.

Page No 23.65:

Question 2:

Using vector method, prove that the following points are collinear:
(i) A (6, −7, −1), B (2, −3, 1) and C (4, −5, 0)
(ii) A (2, −1, 3), B (4, 3, 1) and C (3, 1, 2)
(iii) A (1, 2, 7), B (2, 6, 3) and C (3, 10, −1)
(iv) A (−3, −2, −5), B (1, 2, 3) and C (3, 4, 7)

Answer:

(i) Given  the points A6,-7,-1, B2,-3, 1 and C4,-5, 0. Then,
  AB = Position vector of B - Position vector of A
        = 2i^ -3j^ + k^ - 6i^ + 7j^+ k^= -4i^ + 4j^ + 2k^=-22i^ -2j^ - k^

BC = Position vector of C - Position vector of B
      = 4i^ -5j^- 2i^ + 3j^ -k^= 2i^ -2j^-k^

 AB =-2BC
So, AB, BC  are parallel vectors. But B is a point common to them.
Hence, the given points A, B and C are collinear.

(ii) Given the points A2,-1, 3, B4, 3, 1 and C3, 1, 2. Then,
  AB = Position vector of B - Position vector of A
       = 4i^ + 3j^ + k^ - 2i^+ j^ - 3k^= 2i^ + 4j^ - 2k^= -2-i^ -2j^ + k^

BC = Position vector of C - Position vector of B
     = 3i^ + j^ + 2k^ - 4i^ - 3j^ - k^= -i^ - 2j^ + k^

       AB =-2BC
So, AB, BC  are parallel vectors. But B is a point common to them.
Hence, The given points A, B and C are collinear.

(iii) Given the points A1, 2, 7, B2, 6, 3 and C3, 10,-1. Then,
   AB =  Position vector of B - Position vector of A
        = 2i^ + 6j^+ 3k^ - i^ - 2j^ - 7k^= i^ + 4j^ - 4k^

BC = Position vector of C - Position vector of B.
      = 3i^ + 10j^ -k^ - 2i^ - 6j^ - 3k^= i^ + 4j^ - 4k^

AB = BC
So, AB, BC are parallel vectors. But B is a point common to them.
Hence, the given points A, B and C are collinear.

(iv) Given the points A-3,-2,-5, B1, 2, 3 and C3, 4, 7. Then,
   AB = Position vector of B - Position vector of A
        = i^ + 2j^ + 3k^ + 3i^ + 2j^ + 5k^= 4i^ + 4j^ + 8k^= 22i^ +2 j^ + 4k^

BC = Position vector of C - Position vector of B
     = 3i^ + 4j^ + 7k^ - i^ - 2j^ - 3k^=2i^ + 2j^ + 4k^

 AB = 2BC
So, AB, BC are parallel vectors. But B is a point common to them.
Hence, the given points  A, B and C are collinear.

Page No 23.65:

Question 3:

If a, b, c are non-zero, non-coplanar vectors, prove that the following vectors are coplanar:
(i) 5a + 6b + 7c, 7a - 8b + 9c  and 3a + 20b + 5c 

(ii) a - 2b + 3c ,- 3b + 5c  and -2a + 3b - 4c 

Answer:

(i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let 
 5a + 6b + 7c = x 7a -8b + 9c + y 3a + 20b + 5c.                         = a 7x + 3y + b-8x+20y + c 9x + 5y.
 7x+3y = 5,  -8x + 20y = 6  and 9x+5y = 7.
Solving first two of these equations, we get  x=12, y=12. Clearly, these values of x and y satisfies the third equation.
Hence, the given vectors are coplanar.

(ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let 
a-2b+3c = x (-3b+5c) + y (-2a+3b-4c).                   =a(-2y) + b (-3x+3y) + c (5x-4y).
-2y=1, -3x+3y=-2 and 5x-4y=3
Solving first two of these equations, we get  x=16,  y=-12.
These values of x and y does not satisfy the third equation.
Hence, the given vectors are not coplanar.

Page No 23.65:

Question 4:

Show that the four points having position vectors 6i^-7j^, 16i^-19j^-4k^, 3j^-6k^, 2i^-5j^+10k^ are coplanar.

Answer:

Let the given four points be P, Q, R and S respectively. Three points are coplanar if the vectors PQ, PR and PS are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of the other two. So, let
PQ = xPR + yPS.10i^ -12j^ - 4k^ = x -6i^ + 10j^ - 6k^ + y -4i^ + 2j^ + 10k^. 10i^ -12j^ - 4k^ = i^ -6x - 4y + j^ 10x + 2y + k^ -6x + 10y.
 -6x - 4y = 10,   10x + 2y =-12   and  -6x + 10y =-4.         [ Equating coefficients of i^, j^, k^ on both sides]
Solving the first of these three equations, we get  x=-1 and y=-1. These values also satisfy the third equation.
Hence, the given four points are coplanar.

Page No 23.65:

Question 5:

Prove that the following vectors are coplanar:
(i) 2i^-j^+k^, i^-3j^-5k^ and 3i^-4j^-4k^

(ii) i^+j^+k^, 2i^+3j^-k^ and -i^-2j^+2k^

Answer:

(i) Given the vectors P2i^ - j^ + k^, Qi^ - 3j^ - 5k^ and R3i^ - 4j^ - 4k^.
   We know the three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let,
2i^ - j^ + k^ = x i^ - 3j^ - 5k^ + y 3i^ - 4j^ - 4k^.                    = i^ x + 3y + j^ -3x-4y + k^ -5x-4y.
 x+ 3y = 2, -3x-4y =-1, -5x-4y=1                       [Equating the coefficients of i^, j^, k^ respectively]
Solving first two of these equation, we get x=-1 , y=1. Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.

(ii) Given the vectors Pi^ + j^ + k^,  Q2i^ + 3j^ - k^  and  R-i^ - 2j^ + 2k^.
    We know the three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let,
   i^ + j^ + k^ = x 2i^ + 3j^ - k^ + y -i^ - 2j^ + 2k^.                 = i^ 2x - y + j^ 3x - 2y + k^ -x + 2y.
 2x- y =1 , 3x- 2y = 1, -x + 2y = 1                             [ Equating the coefficients of i^, j^ , k^ respectvely]
Solving first two of these equation , we get x=1, y=1. Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.

Page No 23.65:

Question 6:

Prove that the following vectors are non-coplanar:
(i) 3i^+j^-k^, 2i^-j^+7k^ and 7i^-j^+23k^

(ii) i^+2j^+3k^, 2i^+j^+3k^ and i^+j^+k^

Answer:

(i) Let if possible the given vectors are coplanar. Then one of the given vector is expressible in terms of the other two.
We have,
3i^ + j^-k^ = x(2i^-j^+7k^) + y(7i^-j^+23k^).                = i^ (2x+7y) + j^(-x-y) + k^(7x+23y). 2x+7y = 3 ,      x+y=-1,    7x+23y =-1.By solving the first two equations, we get x=-2,  y=1.
Clearly these values of x and y does not satisfy the third equation.
Hence the given vectors are non-coplanar.


(ii) Let if possible the given vectors are coplanar. Then one of the given vector is expressible in terms of the other two.
We have,
i^+2j^+3k^ = x(2i^+j^+3k^) + y(i^+j^+k^).                = i^(2x+y) + j^(x+y)+k^(3x+y). 2x+y = 1,   x+y = 2,     3x+y=3.By solving the first two equation, we get x=-1,    y=3.
Clearly these values of x and y does not satisfy the third equation.
Hence the given vectors are non-coplanar.



Page No 23.66:

Question 7:

If a, b, c are non-coplanar vectors, prove that the following vectors are non-coplanar:
(i) 2a - b + 3c , a + b - 2c  and a + b - 3c 

(ii) a + 2b + 3c , 2a + b + 3c and a + b + c 

Answer:

(i) Let if possible the following vectors are coplanar. Then one of the vector is expressible in terms of the other two.
   We have,
      2a-b+3c = x(a+b-2c)  y(a+b-3c).                            = a(x+y) + b(x+y) + c(-2x-3y).  x+y =2 ,  x+y=-1 ,   -2x-3y=3. 
 which is not true, as x+y=2 ≠-1. Hence the given vectors are non-coplanar.

(ii) Let if possible the following vector are coplanar. Then one of the vector is expressible in terms of the other two.
We have,
  a+2b+3c = x(2a+b+3c) + y(a+b+c) .                      = a(2x+y) + b(x+y) + c(3x+y). 2x+y=1,  x+y=2, 3x+y=3.
On solving the first two equations we get x=-1, y=3. Clearly the values of x, y does not satisfy the third equation.
Hence the given vectors are non-coplanar.


 

Page No 23.66:

Question 8:

Show that the vectors a, b, c given by a=i^+2j^+3k^, b=2i^+j^+3k^ and c=i^+j^+k^ are non-coplanar. Express vector d=2i^-j^-3k^ as a linear combination of the vectors a, b  and c .

Answer:

Let the given vectors a = i^ + 2j^ + 3k^,    b = 2i^ + j^ + 3k^ and c = i^ + j^ + k^˙˙ are coplanar. Then one of the vector is expressible as a linear combination of the other two. Let,
i^ + 2j^ + 3k^ = x 2i^ +j^ + 3k^ + y i^ + j^ + k^.                      = i^ 2x + y + j^ x + y + k^ 3x + y.
 2x + y =1,  x+ y = 2,     3x + y = 3.
On solving the first two equations we get x=-1, y=3. Clearly the values of x, y does not satisfy the third equation.
Hence the given vectors are non-coplanar.
Now, d = 2i^-j^-3k^ which can be expressed as
2i^-j^-3k^ = x(i^+2j^+3k^) + y(2i^+j^+3k^) + z(i^+j^+k^).
               = i^(x+2y+z) +j^(2x+y+z) + k^(3x+3y+z) .
   
x+2y+z=2, 2x+y+z=-1, 3x+3y+z=-3. x=-83, y=13, z=4
Hence d is expressible as the linear combination of a,b and c.

Page No 23.66:

Question 9:

Prove that a necessary and sufficient condition for three vectors a, b and c to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that la + mb + nc =0 .

Answer:


Necessary Condition: First let a,b,c be three coplanar vectors. Then one of them is expressable as a linear combination of the other two. Let c=xa+yb for some scalars x,y. Then, c= xa+ yb for some scalars x,y la+mb+nc= 0, where l=x, m=y, n=-1Thus, if a,b,c are coplanar vectors, then there exists scalars l, m, n such that  la+mb+nc= 0 where l, m, n are all non zero simultaneously.

Sufficient Condition: Let a, b, c be three vectors such that there exists scalars l, m,n not all zero simulataneously satisfying la+ mb+ nc=0. We have tp prove that a, b, c are coplanar vectors.Now, la+ mb+ nc=0nc = -la- mbc =-1na+ -mnbc is a linear combination of a and  b .Hence a, b, c are coplanar vectors.

Page No 23.66:

Question 10:

Show that the four points A, B, C and D with position vectors a, b, c and d respectively are coplanar if and only if 3a - 2b + c - 2d = 0 .

Answer:

Necessary Condition: Firstly , let a, b, c are coplanar vectors.  Then, one of them is expressible as a linear combination of the other two. Let c= xa + yb for some scalars x, y. Then,
c = xa + yb for some scalars x, y.
 la + mb + nc = 0,  where l=x, m=y, n=-1.
Thus, if a, b, c are coplanar vectors, then there exists a scalars l, m, n not all zero simultaneously satisfying la + mb + nc = 0 where l, m, n are not all zero simultaneously.

Sufficient Condition: Let a, b, c are three scalars such that there exists scalars l, m, n not all zero simultaneously satisfying la + mb + nc= 0. We have to prove that a, b, c are coplanar vectors. 
Now, 
  la + mb + nc = 0. nc =-la - mb. c = -lna + -mnb.
c is a linear combination of a and b.
c lies in a plane a and b.
Hence, a, b, c are coplanar vectors.
 



Page No 23.73:

Question 1:

Can a vector have direction angles 45°, 60°, 120°?

Answer:

Yes,
Let a vector makes an angle α = 45° , β= 60° , γ = 120° with  OX, OY, OZ respectively. Let l, m, n be the direction cosines of the vector. Then,
l = cos 45° = 12 ,  m = cos 60° = 12 ,   n = cos 120° = -12
So,
               l2 + m2 + n2 = 12 + 14 + 14 = 1
Since, the vector has direction cosines such that l2 + m2 + n2 = 1
Hence, a vector can have direction angles 45°, 60°, 120°

Page No 23.73:

Question 2:

Prove that 1, 1, 1 cannot be direction cosines of a straight line.

Answer:

Let 1, 1, 1 be the direction cosines of a straight line. Then 
12+12+12 = 3 ≠ 1
Since direction cosines of a line which makes equal angle with the axes  must satisfy l2+m2+n2 = 1.
Hence 1, 1, 1 cannot be the direction cosines of a straight line.

Page No 23.73:

Question 3:

A vector makes an angle of π4 with each of x-axis and y-axis. Find the angle made by it with the z-axis.

Answer:

Let the vector OP makes an angle α=45° and β = 45° with OX, OY respectively. Suppose OP is inclined at angle γ to OZ.
Let l, m, n be the direction cosines of OP. Then,
l = cos π4 = 12     ,     m = cos π4 = 12  and n = cos γ
Now, we have,

       l2 + m2 + n2 = 1 12 + 12 + n2 = 1 n2 = 0 n = 0 cos γ = cos π2 γ = π2
Hence, the angle made by it with the  z-axis is π2.

Page No 23.73:

Question 4:

A vector r is inclined at equal acute angles to x-axis, y-axis and z-axis. If r = 6 units, find r.

Answer:

Suppose, vector r makes an angle α with each of the axis OX, OY and OZ. Then, its direction cosines are l = cos α ,  m = cos α  and n = cos α i.e. l = m = n.
 
  Now,  l2 + m2 + n2 = 1 l2 + l2 + l2 = 1 3l2 = 1 l2 = 13 l =± 13Since, r makes acute angle with the axis.Hence, we take only positive value.
Therefore,   

 r = r  l i^ + m j^ + n k^ r = 6 13 i^ + 13 j^ + 13 k^   = 23  i^ + j^ + k^

 

Page No 23.73:

Question 5:

A vector r is inclined to x-axis at 45° and y-axis at 60°. If r = 8 units, find r.

Answer:

Here,  r makes an angle 45° with OX and 60° with OY. So,
l = cos 45° = 12  and m = cos 60° = 12
Now,  l2+m2+n2=112+14+n2 = 1 n2 = 14 n=±12
Therefore, 
   r = r l i^ + m j^ + n k^    = 8 12i^ + 12 j^ ± 12k^    = 4 2 i^ + j^ ± k^

Page No 23.73:

Question 6:

Find the direction cosines of the following vectors:
(i) 2i^+2j^-k^

(ii) 6i^-2j^-3k^

(iii) 3i^-4k^

Answer:

(i) We have,  2i^ + 2j^ - k^
The direction cosines are 222+22+-12,   222+22+-12 , -122+22+-12 or,  23 , 23 , -13

(ii)  We have, 6i^ - 2j^ - 3k^
  The direction cosines are 662+-22+-32 ,  -262+-22+-32 ,  -362+-22+-32  or,  67, -27, -37

(iii) We have,  3i^ - 4k^
  The direction cosines are 332+0+-42 , 032+0+-42 , -432+0+-42   or,    35, 0, -45

Page No 23.73:

Question 7:

Find the angles at which the following vectors are inclined to each of the coordinate axes:
(i) i^-j^+k^

(ii) j^-k^

(iii) 4i^+8j^+k^

Answer:

(i) Let r be the given vector, and let it make an angle α, β, γ with OX, OY, OZ respectively. Then, its direction cosines are cos α, cos β, cos γ. So, direction ratios of r = i^ - j^ + k^ are proportional to 1,-1, 1. Therefore,
Direction cosine of r are 112+ -12 + 12 ,   -112+ -12 + 12 , 112+ -12 + 12
or,  13, -13, 13.
∴ cos α = 13, cos β =-13, cos γ = 13
α = cos-113 ,  β = cos-1-13 , γ = cos-113

(ii) Let r be the given vector, and let it make an angles α, β, γ with OX, OY, OZ respectively. Then, its direction cosines are cos α,  cos β,  cos γ. So, direction ratios of r = j^ - k^ are proportional to 0, 1,-1. Therefore, direction cosines of r are
00+12+-12 ,  10+12+-12 , -10+12+-12  or,  0, 12 , -12

∴ cos α=0, cos β = 12  , cos γ = -12
 α = cos-10 , β = cos-112 , γ = cos-1-12 α = π2 ,  β = π4 , γ = 3π4

(iii) Let r be the given vector, and let it make an angle α, β, γ  with  OX, OY, OZ respectively. Then, its direction cosines are cos α , cos β ,   cos γ. So, direction ratio of r = 4i^ + 8j^ + k^ are proportional to 4, 8, 1
Therefore, direction ratio of r are
 442+82+12,   842 + 82 + 12 ,  142+82+12   or,  49, 89, 19.

∴  α = cos-149 ,  β = cos-189 ,   γ = cos-119.

Page No 23.73:

Question 8:

Show that the vector i^+j^+k^ is equally inclined with the axes OX, OY and OZ.

Answer:

Let r = i^ + j^ + k^ and it make an angle α, β, γ with OX, OY, OZ respectively.Then its direction cosines are cos α, cos β and cos γ. So, Direction ratios of r = i^ + j^ + k^ are proportional 1, 1, 1. Therefore, direction cosines of r are 
112 + 12 + 12 , 112 + 12 + 12 , 112 + 12 + 12   or,  13, 13, 13.
Thus, 
 cos α = 13  ,  cos β = 13   and   cos γ = 13.
α = β = γ.
Hence, all are equally inclined with the coordinate axis.

Page No 23.73:

Question 9:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are 13,13,13.

Answer:

Suppose the vector makes equal angle α with the coordinate axis.
Then, its direction cosines are l = cos α ,  m = cos α  ,   n = cos α. Therefore, l = m = n
    l2 + m2 + n2 = 1l2 + l2 + l2 = 1 3l2 = 1 l2 = 13l=13
Hence, direction cosines are 13, 13 , 13.

Page No 23.73:

Question 10:

If a unit vector a makes an angle π3 with i^,π4 with j^ and an acute angle θ with k^, then find θ and hence, the components of a.

Answer:

The Direction cosines of vector a are l = cos π3=12,    m = cos π4=12 , n = cos θ

Therefore,  
     l2 + m2 + n2 = 114+ 12+ n2 = 1 n2 = 1 - 34 n2 =  14 n = 12          a makes acute angle with k^
 cos θ = 12  θ = π3
Since, a is the unit vector.
 a = li^ + mj^ + nk^ a = 12i^ + 12j^ + 12k^
Hence, components of a are 12i^ + 12j^ + 12k^



Page No 23.74:

Question 11:

Find a vector r of magnitude 32 units which makes an angle of π4 and π2 with y and z-axes respectively.              [NCERT EXEMPLAR]

Answer:


Suppose vector r makes an angle α with the x-axis.

Let l, m, n be the direction cosines of r. Then,

l=cosα, m=cosπ4=12, n=cosπ2=0

Now,

l2+m2+n2=1cos2α+12+0=1cos2α=1-12=12cosα=±12

We know that

r=rli^+mj^+nk^r=32±12i^+12j+0k^                        r=32r=±3i^+3j^

Page No 23.74:

Question 12:

A vector r is inclined at equal angles to the three axes. If the magnitude of r is 23, find r.                 [NCERT EXEMPLAR]

Answer:


Let l, m, n be the direction cosines of r.

Now, r is inclined at equal angles to the three axes.

l=m=n                         α=β=γcosα=cosβ=cosγSo, l2+m2+n2=13l2=1l=±13l=m=n=±13
We know that

r=rli^+mj^+nk^r=23±13i^±13j^±13k^r=2±i^±j^±k^



Page No 23.75:

Question 1:

Define "zero vector".

Answer:

A vector whose initial and terminal point are coincident is called a zero vector or null vector. The null vector is denoted by 0. The magnitude of null vectors is zero.

Page No 23.75:

Question 2:

Define unit vector.

Answer:

A vector whose modulus is unity is called a unit vector.The unit vector in the direction of a vector a is denoted by a^.
Thus, a^=1

Page No 23.75:

Question 3:

Define position vector of a point.

Answer:

A point O is fixed as origin in space (or plane) and P is any point, then OP is called a position vector of P with respect to O.

Page No 23.75:

Question 4:

Write PQ+RP+QR in the simplified form.

Answer:

We have, PQ + RP + QR= PQ + QR + RP
                                     = PR + RP                    [∴ PQ + QR = PR]
                                     = 0

Page No 23.75:

Question 5:

If a and b are two non-collinear vectors such that xa + yb =0 , then write the values of x and y.

Answer:

We have, xa + yb = 0  
 x= 0 and y=0   [a and b are non-collinear vectors]

Page No 23.75:

Question 6:

If a and b represent two adjacent sides of a parallelogram, then write vectors representing its diagonals.

Answer:

Let a and b represents two adjacent sides of a parallelogram ABCD.
∴ AB = DC and AD= BC

DC = AB = a   and   AD = BC = b
In ABC,
      AB + BC =AC a + b = AC
In ABD,
     AD + DB = AB b + DB = a DB = a - b

Page No 23.75:

Question 7:

If a, b, c represent the sides of a triangle taken in order, then write the value of a + b + c .

Answer:

Let ABC be a triangle such that BC = a,  CA = b  and AB = c.Then,
a + b + c = BC + CA + AB
                = BA + AB=0                      [∵ BC + CA = BA]
                                                    

Page No 23.75:

Question 8:

If a, b, c are position vectors of the vertices A, B and C respectively, of a triangle ABC, write the value of AB+BC+CA.

Answer:

Given: a, b and c are the position vectors of A, B and C respectively. Then,
AB = b - aBC = c - bCA = a -c
Consider,
 AB + BC + CA = b - a + c - b + a - c                              = 0

Page No 23.75:

Question 9:

If a, b, c are position vectors of the points A, B and C respectively, write the value of AB+BC+AC.

Answer:

Given: a, b, c are the position vectors of A, B, C respectively. Then,
AB = b - aBC = c - bAC = c - a
Therefore,
 AB + BC + AC = b - a + c - b + c - a                              = 2 c - a

Page No 23.75:

Question 10:

If a, b, c are the position vectors of the vertices of a triangle, then write the position vector of its centroid.

Answer:

Let ABC be a triangle and D, E and F are the midpoints of the sides BC, CA and AB respectively. Also, Let  a, b, c are the position vectors of A, B, C respectively. Then the position vectors of D, E, F are b + c2,  c + a2,  a + b2 respectively.
The position vector of a point divides AD in the ratio of 2 ;  is 1 . a + 2b + c22 = a + b + c3.
Similarly, Position vectors of the points divides BE, CF in the ratio of 2:1 are equal to a + b + c3.
Thus, the point dividing AD in the ratio 2 : 1 also divides BE, CF in the same ratio.
Hence, the medians of a triangle are concurrent and the position vector of the centroid is a + b + c3.

Page No 23.75:

Question 11:

If G denotes the centroid of ∆ABC, then write the value of GA+GB+GC.

Answer:

Let a, b, c be the position vectors of the vertices A, B, C respectively. Then, the position vector of the centroid G is a + b + c3
Thus, 
GA + GB + GC = a - a + b + c3 + b - a + b + c3 + c - a + b + c3                               = a + b + c  - 3 a + b + c3                               = a + b + c- a + b + c                               = 0

Page No 23.75:

Question 12:

If a and b denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB, then write the position vector of C.

Answer:

Given: a and b are the position vectors of points A and B respectively and C is a point on AB such that 3 AC= 2AB.
Let c is the position vector of C
Now,
AB = b - a
AC = c - a
Consider,
     3 AC= 2AB 3 c - a = 2 b - a 3c - 3a = 2b - 2a 3c = 2b + ac = 13 2b+ ac = 13  a+2b

Hence, the position vector of C is 13 a+2b 

Page No 23.75:

Question 13:

If D is the mid-point of side BC of a triangle ABC such that AB+AC=λ AD, write the value of λ.

Answer:

Given: D is the midpoint of the side BC of a triangle ABC such that AB + BC = λAD.
Let a, b, c  are the position vectors of AB, BC and CA.
Now, the position vector of D is b + c2. Then,
AB = b - aAC = c - aAD = b + c2 - a
Now, we have,
      AB + AC = λ AD b - a + c - a = λ b + c2 -ab + c - 2a = λ b + c - 2a2 λ = 2

Page No 23.75:

Question 14:

If D, E, F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC, write the value of AD+BE+CF.

Answer:

Given: D, E, F are the midpoints of the sides BC, CA, AB respectively. Then, the position vectors of the midpoints D, E,  F are given by b + c2, c + a2, a + b2
Now, AD + BE + CF= b + c2- a  + c + a2- b + a + b2- c                             =2a +b + c2 -a +b + c                             =a +b + c-a +b + c                             = 0

Page No 23.75:

Question 15:

If a is a non-zero vector of modulus a and m is a non-zero scalar such that m a is a unit vector, write the value of m.

Answer:

Given a a non zero vector with modulus a. Also, ma is the unit vector. Therefore,
  m a = 1m a = 1 m a = 1 m= 1a m =± 1a

Page No 23.75:

Question 16:

If a, b, c are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then write the value of a + b + c .

Answer:

Let, ABC be a given equilateral triangle and its vertices are A(a), B(b) and C(c).
Also, O(0) be the orthocentre of triangle ABC.
We know that centroid and orthocentre of equilateral triangle coincide at one point.
Orthocentre of ABC=0Centroid  ABC=0a+b+c3=0 a+b+c=0

Page No 23.75:

Question 17:

Write a unit vector making equal acute angles with the coordinates axes.

Answer:

Suppose r makes an angle α with each of the axis ΟΧ,ΟΥ and ΟΖ.
Then, its direction cosines are l=cos α, m=cosα, n=cosα.
Now,
     Ι2+m2+n2=1l2 + l2 + l2=1     l=m=n3l2 =1 l2 = 13l = ±13
Since, the angle is acute Hence, we take only positive value
Therefore, unit vector is 13i^ + 13j^ + 13k^.

Page No 23.75:

Question 18:

If a vector makes angles α, β, γ with OX, OY and OZ respectively, then write the value of sin2 α + sin2 β + sin2 γ.

Answer:

Suppose, a vector OP makes an angle α, β, γ with OX, OY, OZ respectively. Then, direction cosines of the vector are given by
l = cos α ,  m = cos β and n = cos γ
Consider,
 sin2α + sin2 β + sin2 γ = 1 - cos2α + 1 - cos2β + 1- cos2γ                                    = 3 - cos2α + cos2β + cos2γ                                    = 3 - l2+m2+n2
                             = 3 - 1                               [∵ l2 + m2 + n2 = 1]
                             = 2

Page No 23.75:

Question 19:

Write a vector of magnitude 12 units which makes 45° angle with X-axis, 60° angle with Y-axis and an obtuse angle with Z-axis.

Answer:

Suppose a vector r makes an angle 45° with ΟΧ, 60° with ΟΥ and having magnitude 12 units.
l = cos 45° = 12  and m = cos 60° = 12
Now,  l2+m2+n2=112+14+n2 = 1 n2 = 14 n=-12     The angle with the z-axis is obtuse
Therefore, 
   r = r l i^ + m j^ + n k^    =12 12i^ + 12 j^ - 12k^    = 6 2 i^ + j^ - k^



Page No 23.76:

Question 20:

Write the length (magnitude) of a vector whose projections on the coordinate axes are 12, 3 and 4 units.

Answer:

Given: Projection on the coordinate axes are 12, 3, 4 units. Therefore,
Length of vector =122+32+42
                            = 169
                            = 13

Page No 23.76:

Question 21:

Write the position vector of a point dividing the line segment joining points A and B with position vectors a and b externally in the ratio 1 : 4, where a=2i^+3j^+4k^ and b=-i^+j^+k^.

Answer:

The position vectors of A and B are
a=2i^+3j+4k^b=-i^+j^+k^
Let C divides AB in the ratio such that AB : CB = 1 : 4
 Position vector of C = 1-i^+j^+k^-42i^+3j^+4k^1-4
                                = -i^+j^+k^-8i^-12j^-16k^-3
                                = -9i^-11j^-15k^-3
                               = 3i^+11j^3+5k^
                  
                          

Page No 23.76:

Question 22:

Write the direction cosines of the vector r=6i^-2j^+3k^.

Answer:

Given: r = 6i^ - 2j^ + 3k^
Then, direction cosines of r are 662 + -22 + 32 , -262 + -22 + 32 , 362 + -22 + 32 or, 67, -27, 37

Page No 23.76:

Question 23:

If a=i^+j^, b=j^+k^ and c=k^+i^, write unit vectors parallel to a + b - 2c .

Answer:

Given: a=i^+j^ , b=j^+k^ , c=k^+i^
Now, a+b-2c=i^+j^+j^+k^-2k^-2i^
              =-i^+2j^-k^
Unit vector parallel to a+b-2c=-i^+2j^-k^-12+22+-12  
                                                 =-i^+2j^-k^6

Page No 23.76:

Question 24:

If a =i^+2j^, b =j^+2k^, write a unit vector along the vector 3a - 2b .

Answer:

 Given: a=i^+2j^ , b=j^+2k^ 
Therefore,
 3a-2b=3i^+6j^-2j^-4k^              =3i^+4j^-4k^ 

Hence, Unit vector along 3a-2b=3i^+4j^-4k^32+42+-42=3i^+4j^-4k^9+16+16=141 3i^+4j^-4k^

Page No 23.76:

Question 25:

Write the position vector of a point dividing the line segment joining points having position vectors i^+j^-2k^ and 2i^-j^+3k^ externally in the ratio 2:3.

Answer:

Let A and B be the points with position vectors a=i^+j^-2k^  , b=2i^-j^+3k^ respectively.
Let C divide AB externally in the ratio 2 : 3 such that AC : CB = 2 : 3
∴ Position vector of C = 22i^-j^+3k^-3i^+j^-2k^2-3
                                     = 4i^-2j^+6k^-3i^-3j^+6k^-1
                                     = i^-5j^+12k^-1
                                     = -i^+5j^-12k^
   

Page No 23.76:

Question 26:

If a =i^+j^, b =j^+k^, c =k^+i^, find the unit vector in the direction of a + b + c .

Answer:

Let a=i^+j^ , b=j^+k^ , c=k^+i^
Then, a+b+c =i^+j^+j^+k^+k^+i^ =2i^+j^+k^
∴ a+b+c=22+22+22=12=23 
Therefore, unit vector in the direction of a+b+c=2i^+j^+k^23=13i^+j^+k^

Page No 23.76:

Question 27:

If a =3i^-j^-4k^, b =-2i^+4j^-3k^ and c =i^+2j^-k^, find  3a -2b +4c .

Answer:

Given  a=3i^-j^-4k^ , b=-2i^+4j^-3k^ , c=i^+2j^-k^
Now,  3a-2b+4c=33i^-j^-4k^-2-2i^+4j^-3k^+4i^+2j^-k^
                          = 9i^-3j^-12k^+4i^-8j^+6k^+4i^+8j^-4k^
                          = 17i^-3j^-10k^
3a-2b+4c=172+-32+-102=289+9+100=398 

Page No 23.76:

Question 28:

A unit vector r makes angles π3 and π2 with j^ and k^ respectively and an acute angle θ with i^. Find θ.

Answer:

A unit vector makes an angle π3 and π2 with  j^ and k^
Let Ι, m, n  be its direction cosines
 ∴ l=cosθ , m=cosπ3=12  , n=cosπ2=0
Now
∴  l2+m2+n2=1
 l2+14+0=1
l2=1-14=34
  l=±32
∴ r makes an angle 30° , 150° with i
Since, angle θ is acute.
∴ θ=30°

Page No 23.76:

Question 29:

Write a unit vector in the direction of a =3i^+2j^+6k^.

Answer:

We have,
 a=3i^-2j^+6k^ a=32+-22+62 =9+4+36 =49=7 

∴ Unit vector in the direction of a =  a^=aa=17 3i^-2j^+6k^=37i^-27j^+67k^

Page No 23.76:

Question 30:

If a =i^+2j^-3k^ and b =2i^+4j^+9k^, find a unit vector parallel to a +b .

Answer:

Given:
a=i^+2j^-3k^ , b=2i^+4j^+9k^ Now, a+b=3i^+6j^+6k^ a+b=32+62+62=9+36+36 =81=9 

Unit vector parallel to a+b=a+ba+b=3i^+6j^+6k^9 =19×3i^+2j^+2k^=13i^+2j^+2k^

Page No 23.76:

Question 31:

Write a unit vector in the direction of b =2i^+j^+2k^.

Answer:

Given:
b=2i^+j^+2k^b=22+12+22=4+1+4=9=3 
∴ Unit vector = bb=132i^+j^+2k^=23i^+13j+23k^

Page No 23.76:

Question 32:

Find the position vector of the mid-point of the line segment AB, where A is the point (3, 4, −2) and B is the point (1, 2, 4).

Answer:

Given: A (3, 4, −2) and B(1, 2, 4)
Let C is the mid point of AB
∴ Position vector of C =3i^+4j^-2k^+i^+2j^+4k^2
                               =4i^+6j^+2k^2 =2i^+3j^+k^ 
                          

Page No 23.76:

Question 33:

Find a vector in the direction of a =2i^-j^+2k^, which has magnitude of 6 units.

Answer:

Given:
a=2i^-j^+2k^a=22+-12+22=4+1+4=9=3 
∴ Required Vector =6×aa=6×2i^-j^+2k^3=4i^-2j^+4k^ 

Page No 23.76:

Question 34:

What is the cosine of the angle which the vector 2i^+j^+k^ makes with y-axis?

Answer:

Given 2 i^+j^+k^ .
Therefore , direction cosines are 222+12+12 ,122+12+12 , 122+12+12 or 12,12,12

So, cosine angle with respect to y-axis is 12

Page No 23.76:

Question 35:

Write two different vectors having same magnitude.

Answer:

Let a= 2i^-j^+3k^ and b=-2i^+j^-3k^
It can be observed that 
a = 22+-12+32= 14 b = -22+12+-32 = 14 

Hence, a and b are two vectors having same magnitude.

Page No 23.76:

Question 36:

Write two different vectors having same direction.

Answer:

Let p =i^+2j^+3k^  and q =2i^+4j^+6k^
Then, direction cosines of p are
l = 112+22+32 = 214 ,   m = 212+22+32= 214 and n = 312+22+32 = 314
Direction cosines of q are
l =222+42+62 =2214=114 ,  m = 422+42+62 =4214= 214 and n = 622+42+62 =6214= 314

The direction cosines of two vectors are same. Hence the two diffrent vectors p, q have same directions.

Page No 23.76:

Question 37:

Write a vector in the direction of vector 5i^-j^+2k^ which has magnitude of 8 unit.

Answer:

Given:
a=5i^-j^+2k^a=52+-12+22
   =25+1+4=30 
∴ Position Vector in the direction of â€‹vector =8×aa=8305i^-j^+2k^

Page No 23.76:

Question 38:

Write the direction cosines of the vector i^+2j^+3k^.

Answer:

Given: i^+2j^+3k^
Then, direction cosines are
112+22+32,  212+22+32 , 312+22+32 or, 114, 214, 314.

Page No 23.76:

Question 39:

Find a unit vector in the direction of a =2i^-3j^+6k^.

Answer:

Given:
a=2i^-3j^+6k^ a=22+-32+62=4+9+36=49=7 
Unit vector = aa=2i^-3j^+6k^7 =27i^-37j^+67k^

Page No 23.76:

Question 40:

For what value of 'a' the vectors 2i^-3j^+4k^ and ai^+6j^-8k^ are collinear?

Answer:

Given: Two vectors , let p= 2i^ -3j^+4k^  and q= ai^+6j^-8k^
Since the given vectors are collinear, we have,
p= λq
2i^-3j^+4k^ = λ ai^+6j^-8k^2i^-3j^+4k^= aλi^+6λj^-8λk^

 λa =2,  6λ=-3  and -8λ=4 λ =-12 and a=-4

Page No 23.76:

Question 41:

Write the direction cosines of the vectors -2i^+j^-5k^.

Answer:

Given: -2i^ + j^ - 5k^
Then, its direction cosines are:
-2-22+12+-52 ,    1-22+12+-52 ,   -5-22+12+-52  or,     -230 , 130 , -530

Page No 23.76:

Question 42:

Find the sum of the following vectors: a =i^-2j^, b =2i^-3j^, c =2i^+3k^.

Answer:

Given a=i^-2j^ , b=2i^-3j^ , c=2i^+3k^

So, Sum of the three vectors = a+b+c=i^-2j^+2i^-3j^+2i^+3k^
                          = 5i^-5j^+3k^

Page No 23.76:

Question 43:

Find a unit vector in the direction of the vector a =3i^-2j^+6k^.

Answer:

Given: a=3i^-2j^+6k^
Then,
a=32+-22+62=9+4+36=49=7

∴ Unit vector = aa=3i^-2j^+6k^7 = 37i^ -27j^ + 67k^

Page No 23.76:

Question 44:

If a =xi^+2j^-zk^ and b =3i^-yj^+k^ are two equal vectors, then write the value of x + y + z.

Answer:

Given: a = xi^+2j^-zk^ and b= 3i^-yj^+k^
Since the two vectors are equal. We have,
    xi^+2j^-zk^ = 3i^-yj^+k^ x=3, y=-2, z=-1
∴ x+y+z = 3-2-1=0



Page No 23.77:

Question 45:

Write a unit vector in the direction of the sum of the vectors a=2i^+2j^-5k^ and b=2i^+j^-7k^.                       [CBSE 2014]

Answer:


We have, a=2i^+2j^-5k^ and b=2i^+j^-7k^.    

a+b=2i^+2j^-5k^+2i^+j^-7k^=4i^+3j^-12k^

a+b=42+32+-122=16+9+144=169=13

∴ Required unit vector = a+ba+b=4i^+3j^-12k^13=413i^+313j^-1213k^

Page No 23.77:

Question 46:

Find the value of 'p' for which the vectors 3i^+2j^+9k^ and i^-2pj^+3k^ are parallel.              [CBSE 2014]

Answer:


Let a=3i^+2j^+9k^ and b=i^-2pj^+3k^ be the two given vectors.

If a and b are parallel, then

b=λa for some scalar λ

i^-2pj^+3k^=λ3i^+2j^+9k^i^-2pj^+3k^=3λi^+2λj^+9λk^1=3λ and -2p=2λ                      Equating coefficients of i^, j^,k^p=-λ=-13

Thus, the value of p is -13.

Page No 23.77:

Question 47:

Find a vector a of magnitude 52, making an angle of π4 with x-axis, π2 with y-axis and an acute angle θ with z-axis.         [CBSE 2014]

Answer:


It is given that vector a makes an angle of π4 with x-axis, π2 with y-axis and an acute angle θ with z-axis.
l=cosπ4=12, m=cosπ2=0, n=cosθ
Now,

l2+m2+n2=112+0+cos2θ=1cos2θ=1-12=12cosθ=12                     θ is acute

We know that

a=ali^+mj^+nk^a=5212i^+0j^+12k^a=5i^+0j^+k^

Page No 23.77:

Question 48:

Write a unit vector in the direction of PQ, where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively.              [CBSE 2014]

Answer:


P(1, 3, 0) and Q(4, 5, 6) are the given points.

PQ=4i^+5j^+6k^-i^+3j^+0k^=3i^+2j^+6k^PQ=32+22+62=9+4+36=49=7

∴ Unit vector in the direction of PQ = PQPQ=3i^+2j^+6k^7=173i^+2j^+6k^

Page No 23.77:

Question 49:

Find a vector in the direction of vector 2i^-3j^+6k^ which has magnitude 21 units.                [CBSE 2014]

Answer:


Let a=2i^-3j^+6k^.

a=22+-32+62=4+9+36=49=7

Unit vector in the direction of a=aa=2i^-3j^+6k^7

∴ Vector in the direction of vector a which has magnitude 21 units
=21×2i^-3j^+6k^7=32i^-3j^+6k^=6i^-9j^+18k^ 

Page No 23.77:

Question 50:

If a=4 and -3λ2, then write the range of λa.

Answer:


It is given that

-3λ2-3×aλa2×a-3×4λa2×4                      ka=ka, k is scalar-12λa8        

Thus, the range of λa is [−12, 8].

Page No 23.77:

Question 51:

In a triangle OAC, if B is the mid-point of side AC and OA=a, OB=b, then what is OC?            [CBSE 2015]

Answer:


In ∆OAC, OA=a and OB=b.



It is given that B is the mid-point of AC.

∴ Position vector of B = Position vector of A+Position vector of C2

OB=OA+OC2b=a+OC2a+OC=2bOC=2b-a

Page No 23.77:

Question 52:

Write the position vector of the point which divides the join of points with position vectors 3a-2b and 2a+3b in the ratio 2 : 1.

Answer:

Suppose R be the point which divides the line joining the points with position vectors 3a-2b and 2a+3b in the ratio 2 : 1
And, OA = 3a-2b and OB =2a+3b
Here, m : n = 2 : 1
Therefore, position vector OR is as follows:
OR =mOB + nOA m+n=2(2a+3b)+1(3a-2b)2+1=7a+4b3=73a+43b



Page No 23.78:

Question 1:

If in a ∆ABC, A = (0, 0), B = (3, 33), C = (−33, 3), then the vector of magnitude 22 units directed along AO, where O is the circumcentre of ∆ABC is
(a) 1-3i^+1+3j^

(b) 1+3i^+1-3j^

(c) 1+3i^+3-1j^

(d) none of these

Answer:

(a) 1-3i^+1+3j^



|AO|=22|AO|=|BO|=|CO|=22=RLet the position vector ofObe xi^+yj^.|AO|=x2+y2x2+y2=8                                 .....(1)Also,|BO|=|CO|x32+y332=x+332+y32x26x+9+y263y+27=x2+63x+27+y26y+9y663=x63+6y=x1+313                               .....(2)

Substituting y from (2) in (1) we get,

132x2+1+32 x2=8132x2×8=8132x=13y=1+3The position vector of O is 13 i^+1+3 j.^AO=13 i^+1+3 j^

Page No 23.78:

Question 2:

If a , b are the vectors forming consecutive sides of a regular hexagon ABCDEF, then the vector representing side CD is
(a) a + b 

(b) a - b 

(c) b - a 

(d) -a + b 

Answer:

(c) b - a 
Let ABCDEF be a regular hexagon such that AB = a and BC =b.We know, AD is parallel to BC such that AD=2BC.
∴ AD =2 BC = 2b
In ABC, we have
     AB + BC = ACa+b = AC

In ACD, we have
     AC + CD = AD CD = AD - AC CD = 2b - a+b CD = b - a

Page No 23.78:

Question 3:

Forces 3 O A, 5 O B act along OA and OB. If their resultant passes through C on AB, then

(a) C is a mid-point of AB
(b) C divides AB in the ratio 2 : 1
(c) 3 AC = 5 CB
(d) 2 AC = 3 CB

Answer:

(c) 3 AC = 5 CB

Draw ON, the perpendicular to the line AB



Let i be the unit vector along ON
The resultant force R=3OA+5OB                .....1

The angles between i and the forces R, 3OA, 5OB are ∠CON, ∠AON, ∠BON respectively.

  R·i=3OA·i+5OB·i

⇒ R⋅1⋅ cos ∠CON = 3 OA⋅1⋅cos∠AON + 5OB⋅1⋅cos∠BON

R·ONOC=3OA×ONOA+5OBONOBROC=3+5
R = 8OC

We know that,

OA=OC+CA3OA=3OC+3CA             .....iOB=OC+CB5OB=5OC+5CB            .....ii

on adding (i) and (ii) we get,

3OA+5OB=8OC+3CA+5CBR=8OC+3CA+5CB8OC=8OC+3CA+5CB3AC=5CB3AC=5CB

Page No 23.78:

Question 4:

If a , b , c are three non-zero vectors, no two of which are collinear and the vector a + b  is collinear with c , b + c  is collinear with a , then a + b + c =
(a) a 

(b) b 

(c) c 

(d) none of these

Answer:

(d) None of these

a+b is collinear with c

a+b=xc           .....(1)

where x is scalar and x ≠ 0.

b+c is collinear with a

b+c=ya           .....(2)

y is scalar and y ≠ 0

Substracting (2) from (1) we get,

ac=xcyaa(1+y)=(1+x)c

As given a, c are not collinear,

∴ 1 + y = 0 and 1 + x = 0

y = −1 and x = −1

Putting value of x in equation (1)

a+b=ca+b+c=0
 

Page No 23.78:

Question 5:

If points A (60i^ + 3j^), B (40i^ − 8j^) and C (ai^ − 52j^) are collinear, then a is equal to
(a) 40
(b) −40
(c) 20
(d) −20

Answer:

(b) −40
Given: Three points A60i^ + 3j^, B40i^ - 8j^  and Cai^ - 52j^ are collinear. Then, AB = λ BC.
We have,
AB = 40i^ - 8j^ - 60i^ + 3j^ = -20i^ - 11j^
BC = ai^ - 52j^ - 40i^ - 8j^ = a-40i^ - 44j^
Therefore,
    AB =λ BC-20i^ - 11j^ = λ a-40i^ - λ44j^ λ a-40 =-20  , -44λ =-11   λ = 14 a-40 = -80a=-40     

Page No 23.78:

Question 6:

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then OA+OB+OC+OD=
(a) 2 OG

(b) 4 OG

(c) 5 OG

(d) 3 OG

Answer:

(b) 4 OG

Let us consider the point O as origin.
G is the mid point of AC.




OG=OA+OC22OG=OA+OC                       .....1

Also, G is the mid point BD

OG=OB+OD22OG=OB+OD                   .....2

On adding (1) and (2) we get,

2OG+2OG=OA+OB+OC+OD4OG=OA+OB+OC+ODOA+OB+OC+OD=4OG
 

Page No 23.78:

Question 7:

The vector cos α cos βi^ + cos α sin βj^ + sin αk^ is a
(a) null vector
(b) unit vector
(c) constant vector
(d) none of these

Answer:

(b) unit vector
Given: The vector cosα cosβ i^ + cosα sin β j^ + sin α k^.
Then,
 cosα cos β i^+ cosα sinβ j^ + sinα k^ = cos2α cos2β + cos2α sin2β + sin2α = cos2α + sin2α = 1
Hence, the given vector is a unit vector.

Page No 23.78:

Question 8:

In a regular hexagon ABCDEF, AB = a, BC = b and CD=c . Then, AE =
(a) a + b + c 

(b) 2a +b +c 

(c) b + c 

(d) a + 2b + 2c 

Answer:

Option(c)
Given a regular hexagon ABCDEF such that AB=a , BC= b and CD= c. Then,
In ABC, we have
AC= a+b.
In ACD, we have
AC+CD = AD.AD = AC+c.AD = a+b+c.
Again, in ADE, we have
AE= AD+DE.AE = a+b+c-a. AE = b+c.
Hence option (c).

Page No 23.78:

Question 9:

The vector equation of the plane passing through a , b , c , is r = αa +βb + γc , provided that
(a) α + β + γ = 0
(b) α + β + γ =1
(c) α + β = γ
(d) α2 + β2 + γ2 = 1

Answer:

(b) α + β + γ =1
Given: A plane passing through a ,b, c.

⇒ Lines a-b and c-a lie on the plane.

The parmetric equation of the plane can be written as:

r=a+λ1(ab)+λ2(ca)r=a(1+λ1λ2)λ1b+λ2cGiventhat r=αa+βb+γcα+β+γ=1+λ1λ2λ1+λ2α+β+γ=1
 

Page No 23.78:

Question 10:

If O and O' are circumcentre and orthocentre of ∆ ABC, then OA+OB+OC equals
(a) 2 OO'
(b) OO'
(c) O'O
(d) 2O'O

Answer:

Option (b).
Given: O be the circumcentre and O' be the orthocentre of ABC. Let G be the centroid of the triangle.
We know that O, G and H are collinear and by geometry O'G= 2 OG. This yields,
O'O = O'G+GO = 2GO + GO = 3 GO. 
In other words OO' =3OG.
Since, OG = a+b+c3.

∴ OO' = 3 × a+b+c3 = a+b+c = OA+OB+OC.
 

Page No 23.78:

Question 11:

If a, b, c and d are the position vectors of points A, B, C, D such that no three of them are collinear and a+c=b+d, then ABCD is a
(a) rhombus
(b) rectangle
(c) square
(d) parallelogram

Answer:

Given:
 a+c = b+dc-d = b-aAB=DCAnd a+c = b+d c -b= d-aAD=BCAlso, since a+c = b+d12(a+c) =12(b+d)so, position vector of mid point of BD = position vector of mid point of AC.hence diagonals bisect each other.the given ABCD is a parallelogram.
Option (d).



Page No 23.79:

Question 12:

Let G be the centroid of ∆ ABC. If AB=a, AC=b, then the bisector AG, in terms of a and b  is
(a) 23a+b

(b) 16a+b

(c) 13a+b

(d) 12a+b

Answer:

(c) 13a+b

Taking A as origin. Then, position vector of A, B and C are 0, a and b respectively. Then, 
Centroid G has position vector 0+a+b3 = a+b3
Therefore, AG = a+b3 - 0 = a+b3

Page No 23.79:

Question 13:

If ABCDEF is a regular hexagon, then AD+EB+FC equals
(a) 2AB
(b) 0
(c) 3AB
(d) 4AB

Answer:

(d) 4AB


AD=2BCEB=2FAFC=2AB

AD+EB=2BC+FA=2AO+FA            BC=AO

In triangle AOF,

FA+AO+FO=0FA+AO=FOAD+EB=2FO

And AB=FO

AD+EB=2ABAD+EB+FC=2AB+2AB=4AB

Page No 23.79:

Question 14:

The position vectors of the points A, B, C are 2i^+j^-k^, 3i^-2j^+k^ and i^+4j^-3k^ respectively. These points
(a) form an isosceles triangle
(b) form a right triangle
(c) are collinear
(d) form a scalene triangle

Answer:

(a) form an isosceles triangle
Given: Position vectors of A, B, C are 2i^+j^-k^,   3i^-2j^+k^ and i^+4j^-3k^. Then,
AB = 3i^-2j^+k^ - 2i^+j^-k^ = i^-3j^+2k^BC = i^+4j^-3k^ - 3i^-2j^+k^ = -2i^+6j^-4k^CA = 2i^+j^-k^ - i^+4j^-3k^ =i^-3j^+2k^

 Now, AB=12+-32+22=1+9+4=14 CA=12+-32+22=1+9+4=14BC=-22+62+-42=4+36+16=56AB=CA
Hence, the triangle is isosceles as two of its sides are equal.

Page No 23.79:

Question 15:

If three points A, B and C have position vectors i^+xj^+3k^, 3i^+4j^+7k^ and yi^-2j^-5k^ respectively are collinear, then (x, y) =
(a) (2, −3)
(b) (−2, 3)
(c) (−2, −3)
(d) (2, 3)

Answer:

(a) (2, −3)
Given position vectors of A, B and C are i^+xj^+3k^,  3i^+4j^+7k^ and yi^-2j^-5k^. Then,
AB= 3i^+4j^+7k^-i^-xj^-3k^ = 2i^+4-xj^+4k^BC = yi^-2j^-5k^-3i^-4j^-7k^ = y-3i^-6j^-12k^
Since, the given vectors are collinear.
 AB=λBC2i^ +4-xj^+ 4k^ = λ y-3i^ -6λ j^-12λ k^2=λ y-3 , 4-x = -6λ, 4=-12λ 2=λ y-3 , 4-x = -6λ,  λ=-132=-13y-3 , 4-x = -6×-13-6=y-3 , 4-x = 2 y=-3 ,  x= 2

Page No 23.79:

Question 16:

ABCD is a parallelogram with AC and BD as diagonals. Then, AC-BD=
(a) 4 AB
(b) 3 AB
(c) 2 AB
(d) AB

Answer:

(c) 2 AB
Given: ABCD, a parallelogram with diagonals AC and BD. Then,
AC=AB+BCAD = AB+BD  BD = AD-AB

∴ AC - BD = AB+BC-AD+AB = 2AB                        [∵ AD=BC]

Page No 23.79:

Question 17:

If OACB is a parallelogram with OC=a and AB=b, then OA=
(a) a+b

(b) a-b

(c) 12b-a

(d) 12a-b

Answer:

(d) 12a-b
Given a parallelogram OACB such that OC = a,    AB=b. Then,
 OB+BC = OC OB = OC-BC
 OB = OC - OA                           [∵ BC=OA]
OB= a-OA      ...1                 
   
Therefore,
   OA + AB = OB OA + b = a - OA      Using 12OA = a-b OA = 12 a-b

Page No 23.79:

Question 18:

If a and b are two collinear vectors, then which of the following are incorrect?
(a) b =λa  for some scalar λ
(b) a =±b 
(c) the respective components of a and b  are proportional
(d) both the vectors a and b have the same direction but different magnitudes

Answer:

(d) both the vectors a and b have the same direction but different magnitudes
If a and b are collinear vectors, then they are paprallel. Therefore, we have
b = λa  , for some scalar λ.
If λ=±1  a = ±b.
If b = b1i^ + b2 j^+ b3 k^ and  a = a1i^ + a2 j^ + a3 k^. Then,
     b = λa.b1i^+ b2 j^ +b3 k^ = λ a1i^+a2 j^+a3k^.b1 i^+b2 j^+b3k^ = λa1i^+λa2j^+λa3k^. b1=λa1 ,  b2=λa2 ,  b3=λa3. b1a1=b2a2=b3a3 = λ.

Thus, the respective components of a and b can have different directions. Hence, the statement given in (d) is incorrect.

Page No 23.79:

Question 19:

In Figure, which of the following is not true?

(a) AB+BC+CA=0

(b) AB+BC-AC=0

(c) AB+BC-CA=0

(d) AB-CB+CA=0
Figure

Answer:

(c) AB+BC-CA=0


We have, LHS =  AB+BC-CA = AC-CA                    [∵ AB+BC = AC]
                                              =-CA-CA=-2CA
So, LHSRHS
Hence, It is not true.



Page No 24.43:

Question 12:

Find the components along the coordinate axes of the position vector of each of the following points:
(i) P(3, 2)
(ii) Q(–5, 1)
(iii) R(–11, –9)
(iv) S(4, –3)

Answer:

(i) Let O be the origin.
    The position vector of point P(3,2), OP=3i^+2j^
    Component of OP along x-axis = a vector of magnitude 3 having its direction along the positive direction of x-axis.
    Component of OP along y-axis = a vector of magnitude 2 having its direction along the positive direction of y-axis.

(ii) The position vector of point Q(-5,1), OQ=-5i^+j^
    Component of OQ along x-axis = a vector of magnitude 5 having its direction along the negative direction of x-axis.
    Component of OQ along y-axis = a vector of magnitude 1 having its direction along the positive direction of y-axis.

(iii) The position vector of point R(-11,-9), OR=-11i^-9j^
     Component of OR along x-axis = a vector of magnitude 11 having its direction along the negative direction of x-axis.
     Component of OR along y-axis = a vector of magnitude 9 having its direction along the negative direction of y-axis.

(iv) The position vector of point S(4,-3), OS=4i^-3j^
     Component of OS along x-axis = a vector of magnitude 4 having its direction along the positive direction of x-axis.
     Component of OS along y-axis = a vector of magnitude 3 having its direction along the negative direction of y-axis.



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