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#### Question 1:

If P, Q and R are three collinear points such that and . Find the vector .

Given:  and $R$ are collinear such that  and . Then,

#### Question 2:

Give a condition that three vectors $\stackrel{\to }{a}$, $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$ form the three sides of a triangle. What are the other possibilities?

Let $ABC$ be a triangle such that  and . Then,

[âˆµ ]
[ Using triangle law]
[ By definition of null vector]

Other possibilities are

#### Question 3:

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are two non-collinear vectors having the same initial point. What are the vectors represented by $\stackrel{\to }{a}$ + $\stackrel{\to }{b}$ and $\stackrel{\to }{a}$$\stackrel{\to }{b}$.

Given:  are two non-collinear vectors having same initial points. Complete the parallelogram $ABCD$ such that  and
In $△ABC$,

In $△ABD$,

Therefore, $\stackrel{\to }{AC}$ and $\stackrel{\to }{DB}$ are the diagonals of a parallelogram whose adjacent sides are $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ respectively.

#### Question 4:

If $\stackrel{\to }{a}$ is a vector and m is a scalar such that m $\stackrel{\to }{a}$ = $\stackrel{\to }{0}$, then what are the alternatives for m and $\stackrel{\to }{a}$?

Given: $\stackrel{\to }{a}$ is a vector and $m$ is a scalar such that,
Then either   or,

#### Question 5:

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

(ii)

(iii)

(i) True.

Taking modulus on both sides of the equation, we get,

[ âˆµ  ]

(ii) False.
We cannot say
Consider an example,

(iii) False.
We cannot say
Consider an example,

#### Question 6:

ABCD is a quadrilateral. Find the sum the vectors and $\stackrel{\to }{DA}$.

Given: $ABCD$ is a quadrilateral.
We need to find the sum of
Consider,

[âˆµ  and ]

#### Question 7:

ABCDE is a pentagon, prove that
(i) $\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{CD}+\stackrel{\to }{DE}+\stackrel{\to }{EA}=\stackrel{\to }{0}$

(ii)

Given: $ABCDE$ is a pentagon.
(i) To Prove:
Proof: We have,

[ âˆµ   and  ]
[ âˆµ ]
= RHS
(ii) To Prove:
Proof: We have,

â€‹
[âˆµâ€‹   and  ]
=                       [âˆµ  ]

Hence proved.

#### Question 8:

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

Given: A regular octagon of eight sides with centre O.
To show:
Proof: We know centre of the regular octagon bisects all the diagonals passing through it.
and
and      ............(i)
Now,

Hence proved.

#### Question 9:

If P is a point and ABCD is a quadrilateral and $\stackrel{\to }{AP}+\stackrel{\to }{PB}+\stackrel{\to }{PD}=\stackrel{\to }{PC}$, show that ABCD is a parallelogram.

Given: $ABCD$ is a quadrilateral such that
To show: $ABCD$ is a parallelogram.
Proof: Consider,

[ âˆµ    and    ]
Again,

[ âˆµ     and   ]

Since, opposite sides of the quadrilateral are equal and parallel.
Hence, $ABCD$ is a parallelogram.

#### Question 10:

Five forces and act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is
6 $\stackrel{\to }{AO,}$ where O is the centre of hexagon.

Using (1) and (2),

$\stackrel{\to }{AB}+\stackrel{\to }{AE}+\stackrel{\to }{AC}+\stackrel{\to }{AF}+\stackrel{\to }{AD}\phantom{\rule{0ex}{0ex}}2\stackrel{\to }{AO}+2\stackrel{\to }{AO}+2\stackrel{\to }{AO}\phantom{\rule{0ex}{0ex}}=6\stackrel{\to }{AO}$

#### Question 1:

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

It is given that P and Q are two points with position vectors $\stackrel{\to }{\mathrm{OP}}=2\stackrel{\to }{a}+\stackrel{\to }{b}$ and $\stackrel{\to }{\mathrm{OQ}}=\stackrel{\to }{a}-2\stackrel{\to }{b}$, respectively.

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

Position vector of R = $\frac{1×\left(\stackrel{\to }{a}-2\stackrel{\to }{b}\right)+2×\left(2\stackrel{\to }{a}+\stackrel{\to }{b}\right)}{1+2}=\frac{5\stackrel{\to }{a}}{3}$

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

Position vector of R = $\frac{1×\left(\stackrel{\to }{a}-2\stackrel{\to }{b}\right)-2×\left(2\stackrel{\to }{a}+\stackrel{\to }{b}\right)}{1-2}=\frac{-3\stackrel{\to }{a}-4\stackrel{\to }{b}}{-1}=3\stackrel{\to }{a}+4\stackrel{\to }{b}$

#### Question 2:

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

Given:  and $\stackrel{\to }{d}$ are the position vectors of the four distinct points  and $D$.
Also, we have,

Again,

Consequently,  and . Thus two of its opposite sides are equal and parallel.
Hence, $ABCD$ is a parallelogram.

#### Question 3:

If 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.

Let the position vectors of C and D are $\stackrel{\to }{c}$ and $\stackrel{\to }{d}$ respectively. We have,

So C divides AB in the ratio of $3:2$ externally.

Position vector of point C is $3\stackrel{\to }{b}-2\stackrel{\to }{a}$
Moreover,

.
∴
Position vector of point D is $2\stackrel{\to }{a}-\stackrel{\to }{b}$

#### Question 4:

Show that the four points A, B, C, D with position vectors respectively such that are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

Let $AC$ and $BD$ intersects at a point $P$
We have,

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

#### Question 5:

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

Let the point of intersection of the line segments $PR$ and $QS$ is A. Then

$\frac{5\stackrel{\to }{p}+6\stackrel{\to }{r}}{5+6}=\frac{2\stackrel{\to }{q}+9\stackrel{\to }{s}}{2+9}$

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 .

#### Question 6:

The vertices A, B, C of triangle ABC have respectively position vectors , , with respect to a given origin O. Show that the point D where the bisector of ∠A meets BC has position vector .
Hence, deduce that the incentre I has position vector .

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

Let D be the point on BC where bisectors of ∠A meets.
Let $\stackrel{\to }{d}$ be the position vector of D which divides CB internally in the ratio β and γ, where
Thus,
By section formula, the position vector of D is given by

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,

#### 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 $\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}=\stackrel{\to }{OD}+\stackrel{\to }{OE}+\stackrel{\to }{OF}$.

Let D, E and F are the midpoints of BC, CA and AB respectively.
Therefore,

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

Hence Proved.

#### Question 2:

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

Let  and $\stackrel{\to }{c}$ are the position vectors of the vertices  and $C$ respectively.
Then we know that the position vector of the centroid $O$ of the triangle is .
Therefore sum of the three vectors  and $\stackrel{\to }{OC}$ is

Hence, Sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.

#### 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 .

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,

Adding (1) and (2), We get,
$\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}+\stackrel{\to }{OD}=4\stackrel{\to }{OP}$

#### Question 4:

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

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.

In triangle BCD

SP = RQ and SP || RQ
∴ PQRS is a parallelogram.

Diagonals of a parallelogram bisect each other.

∴ PR and QS bisect each other.

#### 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 , where P is any point.

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.

And, H is the midpoint of AD.

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

#### Question 6:

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

Let ABC be a triangle and  be the position vectors of the vertices  A, B and C respectively. Let AD, BE and CF be the internal bisectors of  and $\angle C$ respectively.
We know that D divides BC in the ratio of AB : AC that is c : b.
Then,
P.V. of D is  .
P.V. of E is .
and   P.V. of F is $\frac{a\stackrel{\to }{\alpha }+b\stackrel{\to }{\beta }}{a+b}$.

The point dividing AD in the ratio  is .
The point dividing BE in the ratio of  is .
The point dividing CF in the ratio of  is .

Since the point  lies on all the three internal bisectors AD, BE and CF.
Hence the internal bisectors are concurrent .

#### 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.

(i) The vector $\stackrel{\to }{OP}$ represents the required displacement vector.
(ii) The vector $\stackrel{\to }{OQ}$ represents the required vector.
(iii) The vector $\stackrel{\to }{OR}$ represents the required vector.

#### 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

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

#### 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

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

#### Question 4:

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

(i) Vectors having the same or parallel supports are called collinear vector.
In the given figure the collinear vectors are

(ii) Vectors having the same magnitude and direction are called equal vector.
In the given figure the equal vectors are

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

(iv) The vectors which are collinear but not equal are

#### Question 5:

Answer the following as true or false:
(i) and 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.

(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.

#### Question 1:

If the position vector of a point (−4, −3) be $\stackrel{\to }{a,}$ find .

Given a point $\left(-4,-3\right)$ such that its position vector $\stackrel{\to }{a}$ is given by

Then,

#### Question 2:

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

Given a position vector $\stackrel{\to }{a}$ of a point  such that,

Then,

Also ,    (given)
Thus, we get,

#### Question 3:

Find a vector of magnitude 4 units which is parallel to the vector $\sqrt{3}\stackrel{^}{i}+\stackrel{^}{j}$.

Let
Then,
A unit vector parallel to $\stackrel{\to }{a}$ =
Hence, Required vector =

#### Question 4:

Express $\stackrel{\to }{AB}$ in terms of unit vectors $\stackrel{^}{i}$ and $\stackrel{^}{j}$, when the points are:
(i) A (4, −1), B (1, 3)
(ii) A (−6, 3), B (−2, −5)
Find $\left|\stackrel{\to }{A}B\right|$ in each case.

(i) Given:  and
Then the position vector $\stackrel{\to }{AB}$ is given by
$\stackrel{\to }{AB}$ = Position vector of $B$ $-$ Position vector of $A$

So,

(ii) Given:  and
Then, the position vector $\stackrel{\to }{AB}$ is given by
Position vector of $B$ $-$ Position vector of $A$

So,

#### Question 5:

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

Let $O$ be the origin. Let  be the required point. Then, $\stackrel{\to }{P}$ is the tip of the position vector $\stackrel{\to }{OP}$ of the point $P$.
We have,

and,  Position vector of $B$ $-$ Position vector of $A$.

Given that
So,
Hence, coordinated of the required point is $\left(-1.-2\right)$

#### 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.

Let the coordinates of $D$ is .
Since, $ABCD$ is a parallelogram.
∴
We have,

Hence, the coordinates of $D$ is $\left(-4,-3\right)$

#### Question 7:

If the position vectors of the points A (3, 4), B (5, −6) and C (4, −1) are $\stackrel{\to }{a,}$$\stackrel{\to }{b,}$respectively, compute .

Let  are the position vectors of the points $B\left(5,-6\right)$ and $C\left(4,-1\right)$.
Then,

Therefore,

#### Question 8:

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

Let $O$ be the origin and let $P\left(5,-3\right)$ be the tip of the position vector $\stackrel{\to }{a}$. Then,  Let the coordinate of $B$ be  and $A$ has coordinates $\left(4,-1\right)$.
Therefore,
$\stackrel{\to }{AB}$ = Position vector of $B$ $-$ Position vector of $A$

Now,

Hence, the coordinates of $B$ are $\left(9,-4\right)$.

#### Question 9:

Show that the points 2$\stackrel{^}{i}$, − $\stackrel{^}{i}$ − 4$\stackrel{^}{j}$ and −$\stackrel{^}{i}$ + 4$\stackrel{^}{j}$ form an isosceles triangle.

Given:- The points  with position vectors  respectively.
Also,

Then,

and

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

#### Question 10:

Find a unit vector parallel to the vector $\stackrel{^}{i}+\sqrt{3}\stackrel{^}{j}$.

Let
Then,
Unit vector parallel to $\stackrel{\to }{a}$ =

#### Question 11:

The position vectors of points  and $C$ are 3$\stackrel{^}{j}$, 12$\stackrel{^}{i}+\mu$$\stackrel{^}{j}$ and 11$\stackrel{^}{i}-$3$\stackrel{^}{j}$ respectively. If C divides the line segment joining A and B in the ratio 3:1, find the values of $\lambda$ and $\mu$

The position vectors of points A, B and C are 3$\stackrel{^}{j}$, 12$\stackrel{^}{i}+\mu$$\stackrel{^}{j}$ and 11$\stackrel{^}{i}-$3$\stackrel{^}{j}$, respectively.

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

$11\stackrel{^}{i}-3\stackrel{^}{j}=\frac{3×\left(12\stackrel{^}{i}+\mu \stackrel{^}{j}\right)+1×\left(\lambda \stackrel{^}{i}+3\stackrel{^}{j}\right)}{3+1}\phantom{\rule{0ex}{0ex}}⇒11\stackrel{^}{i}-3\stackrel{^}{j}=\frac{\left(36+\lambda \right)\stackrel{^}{i}+\left(3\mu +3\right)\stackrel{^}{j}}{4}\phantom{\rule{0ex}{0ex}}⇒44\stackrel{^}{i}-12\stackrel{^}{j}=\left(36+\lambda \right)\stackrel{^}{i}+\left(3\mu +3\right)\stackrel{^}{j}$
Equating the corresponding components, we get

$36+\lambda =44$

$⇒\lambda =44-36=8$

and

$3\mu +3=-12$

$⇒3\mu =-12-3=-15$

$⇒\mu =-5$

Thus, the values of $\lambda$ and $\mu$ are 8 and −5, respectively.

#### Question 1:

Find the magnitude of the vector $\stackrel{\to }{a}=2\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}.$

Given:
∴ Magnitude of the vector =

#### Question 2:

Find the unit vector in the direction of $3\stackrel{^}{i}+4\stackrel{^}{j}-12\stackrel{^}{k}.$

Let
Then,
So, a unit vector in the direction of $\stackrel{\to }{a}$ is given by

#### Question 3:

Find a unit vector in the direction of the resultant of the vectors

Given:   and   are the position vectors.
Then, Resultant of the vectors =

So,
∴ Unit vector in the direction of the resultant vector =

#### Question 4:

The adjacent sides of a parallelogram are represented by the vectors Find unit vectors parallel to the diagonals of the parallelogram.

$\begin{array}{l}\stackrel{\to }{a}=\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\\ \stackrel{\to }{b}=-2\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}\end{array}$

$\begin{array}{ccc}\mathrm{Similarly}, \mathit{ }\stackrel{\mathit{\to }}{BD}& =& \stackrel{\to }{b}-\stackrel{\to }{a}\\ & =& -3\stackrel{^}{i}+3\stackrel{^}{k}\end{array}$

$⇒ \stackrel{^}{BD}=\frac{-3\stackrel{^}{i}+3\stackrel{^}{k}}{3\sqrt{2}}=\frac{-\stackrel{^}{i}+k}{\sqrt{2}}$

Given:  and

Hence,

#### Question 6:

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

Given:  Let the position vector of $P$ is $\stackrel{\to }{p}$ such that  and the position vector of  is $\stackrel{\to }{q}$ such that
Therefore,

Hence, the coordinates of $Q$ are

#### Question 7:

Prove that the points are the vertices of a right-angled triangle.

Given the points  and $2\stackrel{^}{i}-4\stackrel{^}{j}+5\stackrel{^}{k}$  Are A, B and C respectively.
Then,

The given points forms a vertices of a triangle.
Now,

≠ ${\left|\stackrel{\to }{BC}\right|}^{2}$
The given triangle is not right-angled.

#### Question 8:

If the vertices of a triangle are the points with position vectors what are the vectors determined by its sides? Find the length of these vectors.

Given the vertices of a triangle A, B and C with position vectors  and ${c}_{1}\stackrel{^}{i}+{c}_{2}\stackrel{^}{j}+{c}_{3}\stackrel{^}{k}$ respectively. Then,

Therefore, the length of these vectors are:

#### 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).

Given the vertices of the triangle  and . Then,
Position vectors are

The centroid of a triangle is given by $\frac{\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}}{3}$
So, .

#### Question 10:

Find the position vector of a point R which divides the line segment joining points in the ratio 2:1.
(i) internally
(ii) externally

(i) Given: R divides the line segment joining the points  in the ratio 2 : 1 internally.
Therefore. position vector of R
=

(ii) Given:  R divides the line segment joining the points  in the ratio 2 : 1 externally.
Therefore. position vector of R
=

#### Question 11:

Find the position vector of the mid-point of the vector joining the points

Given:  and
The position vector of the midpoint of the vector
joining these points =

#### Question 12:

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

Let $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are the position vectors of the points   and
Then,

So,

Now,
Therefore, Unit vector parallel to $\stackrel{\to }{PQ}$ =

#### Question 13:

Show that the points are the vertices of a right angled triangle.

Given the points  and
Then,
Position vector of $B$$-$ Position vector of A

Position vector of $C$ $-$ Position vector of $B$

Position vector of $A$ $-$ Position vector of $C$

Clearly,

So,  forms a right angled triangle.

#### 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).

Let  be the position vectors of the points
Then,
and
Therefore, the position vector of the midpoint of the given points is $\frac{\stackrel{\to }{p}+\stackrel{\to }{q}}{2}$
∴

#### Question 15:

Find the value of x for which is a unit vector.

We have, $x\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)$ is a unit vector.
$\therefore \sqrt{{x}^{2}+{x}^{2}+{x}^{2}}=1\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}\left|x\right|=1\phantom{\rule{0ex}{0ex}}⇒\left|x\right|=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

#### Question 16:

If find a unit vector parallel to

We have,   and
∴

A unit vector parallel to $2\stackrel{\to }{a}-\stackrel{\to }{b}+3\stackrel{\to }{c}$ is given by $\frac{2\stackrel{\to }{a}-\stackrel{\to }{b}+3\stackrel{\to }{c}}{\left|2\stackrel{\to }{a}-\stackrel{\to }{b}+3\stackrel{\to }{c}\right|}$$=\frac{\left(3\stackrel{^}{i}-3\stackrel{^}{j}+2\stackrel{^}{k}\right)}{\sqrt{{3}^{2}+{\left(-3\right)}^{2}+{2}^{2}}}$

#### Question 17:

If find a vector of magnitude 6 units which is parallel to the vector

We have,  and
Then,

∴ A unit vector parallel to $2\stackrel{\to }{a}-\stackrel{\to }{b}+3\stackrel{\to }{c}$ is $\frac{2\stackrel{\to }{a}-\stackrel{\to }{b}+3\stackrel{\to }{c}}{\left|2\stackrel{\to }{a}-\stackrel{\to }{b}+3\stackrel{\to }{c}\right|}$$=\frac{\left(\stackrel{^}{i}-2\stackrel{^}{j}+2\stackrel{^}{k}\right)}{\sqrt{{1}^{2}+{\left(-2\right)}^{2}+{2}^{2}}}$

Hence, Required vector =

#### Question 18:

Find a vector of magnitude of 5 units parallel to the resultant of the vectors

Given the position vectors   and
∴ Resultant Vector =
So, a unit vector parallel to the resultant vector is
Hence, required vector =

#### Question 19:

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

Disclaimer: The question has been solved by taking the vector $\stackrel{\to }{\mathrm{AB}}$ as $\stackrel{^}{j}+\stackrel{^}{k}$.

In âˆ†ABC, $\stackrel{\to }{\mathrm{AB}}=\stackrel{^}{j}+\stackrel{^}{k}$ and $\stackrel{\to }{\mathrm{AC}}=3\stackrel{^}{i}-\stackrel{^}{j}+4\stackrel{^}{k}$.

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 $=\frac{\left(\stackrel{^}{j}+\stackrel{^}{k}\right)+\left(3\stackrel{^}{i}-\stackrel{^}{j}+4\stackrel{^}{k}\right)}{2}=\frac{3\stackrel{^}{i}+5\stackrel{^}{k}}{2}=\frac{3}{2}\stackrel{^}{i}+\frac{5}{2}\stackrel{^}{k}$

Now,

Length of the median, AD = $\left|\stackrel{\to }{\mathrm{AD}}\right|=\left|\left(\frac{3}{2}\stackrel{^}{i}+\frac{5}{2}\stackrel{^}{k}\right)-\left(0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}\right)\right|=\left|\frac{3}{2}\stackrel{^}{i}+\frac{5}{2}\stackrel{^}{k}\right|=\sqrt{{\left(\frac{3}{2}\right)}^{2}+{0}^{2}+{\left(\frac{5}{2}\right)}^{2}}=\sqrt{\frac{34}{4}}=\sqrt{\frac{17}{2}}$ units

#### Question 1:

Show that the points A, B, C with position vectors and are collinear.

We have, A, B ,C with position vectors   Then,
Position Vector of B $-$ Position Vector of A

Position Vector of C $-$ Position Vector of B

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

#### Question 2:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are non-coplanar vectors, prove that the points having the following position vectors are collinear:
(i)

(ii)

(i) Given:  are non coplanar vectors.
Let  the points be  respectively  with position vectors  Then,
$\stackrel{\to }{AB}=$ Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

∴ $\stackrel{\to }{BC}=-3\stackrel{\to }{AB}$
are parallel vectors.
But $B$ is a point common to them.
Hence,  and $C$ are collinear.

(ii) Given  are non coplanar vectors.
Let the points be  respectively with the position vectors  Then,
Position vector of B $-$ Position vector of A

$\stackrel{\to }{BC}=$ Position vector of C $-$ Position vector of B

∴
are parallel vectors.
But $B$ is a point common to them.
So, $\stackrel{\to }{AB}$ and $\stackrel{\to }{BC}$ are collinear.
Hence,  and $C$ are collinear.

#### Question 3:

Prove that the points having position vectors are collinear.

Let  be the points with position vectors  Then,
$\stackrel{\to }{AB}=$ Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

∴
are parallel vectors.
But $B$ is a point common to them.
So, $\stackrel{\to }{AB}$ and $\stackrel{\to }{BC}$ are collinear.
Hence,  are collinear.

#### Question 4:

If the points with position vectors are collinear, find the value of a.

Let  be the points with position vectors . Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

Since,  and $C$ are collinear.
∴

#### Question 5:

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

Given: $\stackrel{\to }{a},\stackrel{\to }{b}$ are non collinear vectors.
Let the position vectors of points A, B and C be  respectively.
Then,
P.V. of B − P.V. of A.

P.V. of C − P.V. of B.

P.V. of A − P.V. of C.

Now, the position vectors are collinear if and only if  $\stackrel{\to }{AB}$ and $\stackrel{\to }{CA}$ is some multiple of $\stackrel{\to }{BC}$.
So,

and $\stackrel{\to }{BC}=-\stackrel{\to }{CA}.$
Hence, for real values of $\lambda$, the given position vectors are parallel.

#### Question 6:

If $\stackrel{\to }{AO}+\stackrel{\to }{OB}=\stackrel{\to }{BO}+\stackrel{\to }{OC},$ prove that A, B, C are collinear points.

We have,

Hence A, B and C are collinear points.

#### Question 7:

Show that the vectors are collinear.

Given the position vectors $2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}$ and $-4\stackrel{^}{i}+6\stackrel{^}{j}-8\stackrel{^}{k}$
Let   and
Then,

Hence,  are collinear.

#### Question 8:

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

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

Now,

$\stackrel{\to }{\mathrm{AB}}=\left(2\stackrel{^}{i}+\stackrel{^}{j}\right)-\left(m\stackrel{^}{i}-\stackrel{^}{j}\right)=\left(2-m\right)\stackrel{^}{i}+2\stackrel{^}{j}$

$\stackrel{\to }{\mathrm{AC}}=\left(4\stackrel{^}{i}+5\stackrel{^}{j}\right)-\left(m\stackrel{^}{i}-\stackrel{^}{j}\right)=\left(4-m\right)\stackrel{^}{i}+6\stackrel{^}{j}$

If A, B, C are collinear, then

and
$2-m=\frac{1}{3}\left(4-m\right)\phantom{\rule{0ex}{0ex}}⇒6-3m=4-m\phantom{\rule{0ex}{0ex}}⇒2m=2\phantom{\rule{0ex}{0ex}}⇒m=1$

Thus, the value of m is 1.

#### Question 9:

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

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

Now,

$\stackrel{\to }{\mathrm{AB}}=\left[\left(-5\stackrel{^}{i}+16\stackrel{^}{j}\right)-\left(3\stackrel{^}{i}+4\stackrel{^}{j}\right)\right]=-8\stackrel{^}{i}+12\stackrel{^}{j}$

$\stackrel{\to }{\mathrm{AC}}=\left[\left(5\stackrel{^}{i}+\stackrel{^}{j}\right)-\left(3\stackrel{^}{i}+4\stackrel{^}{j}\right)\right]=2\stackrel{^}{i}-3\stackrel{^}{j}$

Clearly, $\stackrel{\to }{\mathrm{AB}}=-4\stackrel{\to }{\mathrm{AC}}$

Therefore, $\stackrel{\to }{\mathrm{AB}}$ and $\stackrel{\to }{\mathrm{AC}}$ are parallel vectors. But, A is a common point of $\stackrel{\to }{\mathrm{AB}}$ and $\stackrel{\to }{\mathrm{AC}}$.

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

#### Question 10:

If the vectors $\stackrel{\to }{a}=2\stackrel{^}{i}-3\stackrel{^}{j}$ and $\stackrel{\to }{b}=-6\stackrel{^}{i}+m\stackrel{^}{j}$ are collinear, find the value of m.

It is given that the vectors $\stackrel{\to }{a}=2\stackrel{^}{i}-3\stackrel{^}{j}$ and $\stackrel{\to }{b}=-6\stackrel{^}{i}+m\stackrel{^}{j}$ are collinear.

$\therefore \stackrel{\to }{b}=\lambda \stackrel{\to }{a}$ for some scalar λ

Thus, the value of m is 9.

#### 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.

Given points .
Therefore,

and,
Clearly,
Hence  are collinear.
Suppose $B$ divides in the ratio AC in the ratio $\lambda :1$. Then the position vector $B$ is

But the position vector of $B$ is $5\stackrel{^}{i}+0\stackrel{^}{j}-2\stackrel{^}{k}.$

#### 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).

We have,

Hence, P, A, B are collinear points.

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.

#### 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]

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

$\stackrel{\to }{\mathrm{AB}}=\left(\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)-\left(\lambda \stackrel{^}{i}-10\stackrel{^}{j}+3\stackrel{^}{k}\right)=\left(1-\lambda \right)\stackrel{^}{i}+9\stackrel{^}{j}$

$\stackrel{\to }{\mathrm{AC}}=\left(3\stackrel{^}{i}+5\stackrel{^}{j}+3\stackrel{^}{k}\right)-\left(\lambda \stackrel{^}{i}-10\stackrel{^}{j}+3\stackrel{^}{k}\right)=\left(3-\lambda \right)\stackrel{^}{i}+15\stackrel{^}{j}$

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

$\stackrel{\to }{\mathrm{AB}}=k\stackrel{\to }{\mathrm{AC}}$ for some scalar k

Thus, the value of λ is −2.

#### Question 1:

Show that the points whose position vectors are as given below are collinear:
(i)

(ii)

(i) Let the points be  and $C$ with position vectors  and  Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

.
and $\stackrel{\to }{BC}$ are parallel vectors. But B is a point common to them.
Hence,  and C are collinear.

(ii) Let the points be  and $C$ with position vectors  and  respectively. Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

and $\stackrel{\to }{BC}$ are parallel vectors.But $B$ is a point common to them.
Hence,  and $C$ are collinear.

#### 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)

(i) Given  the points  and . Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

are parallel vectors. But $B$ is a point common to them.
Hence, the given points  and $C$ are collinear.

(ii) Given the points  and . Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

are parallel vectors. But $B$ is a point common to them.
Hence, The given points  and $C$ are collinear.

(iii) Given the points  and . Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B.

are parallel vectors. But $B$ is a point common to them.
Hence, the given points  and $C$ are collinear.

(iv) Given the points  and . Then,
Position vector of B $-$ Position vector of A

Position vector of C $-$ Position vector of B

are parallel vectors. But $B$ is a point common to them.
Hence, the given points   and $C$ are collinear.

#### Question 3:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are non-zero, non-coplanar vectors, prove that the following vectors are coplanar:
(i)

(ii)

(i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let

Solving first two of these equations, we get  . 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

Solving first two of these equations, we get  .
These values of x and y does not satisfy the third equation.
Hence, the given vectors are not coplanar.

#### Question 4:

Show that the four points having position vectors are coplanar.

Let the given four points be  and $S$ respectively. Three points are coplanar if the vectors  and $\stackrel{\to }{PS}$ are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of the other two. So, let

and           [ Equating coefficients of  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.

#### Question 5:

Prove that the following vectors are coplanar:
(i)

(ii)

(i) Given the vectors  and .
We know the three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let,

[Equating the coefficients of  respectively]
Solving first two of these equation, we get . Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.

(ii) Given the vectors   and  .
We know the three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let,

[ Equating the coefficients of  respectvely]
Solving first two of these equation , we get . Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.

#### Question 6:

Prove that the following vectors are non-coplanar:
(i)

(ii)

(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,

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,

Clearly these values of x and y does not satisfy the third equation.
Hence the given vectors are non-coplanar.

#### Question 7:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are non-coplanar vectors, prove that the following vectors are non-coplanar:
(i)

(ii)

(i) Let if possible the following vectors are coplanar. Then one of the vector is expressible in terms of the other two.
We have,

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,

On solving the first two equations we get . Clearly the values of  does not satisfy the third equation.
Hence the given vectors are non-coplanar.

#### Question 8:

Show that the vectors given by are non-coplanar. Express vector $\stackrel{\to }{d}=2\stackrel{^}{i}-\stackrel{^}{j}-3\stackrel{^}{k}$ as a linear combination of the vectors

Let the given vectors  and  are coplanar. Then one of the vector is expressible as a linear combination of the other two. Let,

On solving the first two equations we get . Clearly the values of  does not satisfy the third equation.
Hence the given vectors are non-coplanar.
Now,  which can be expressed as

= .

Hence $\stackrel{\to }{d}$ is expressible as the linear combination of $\stackrel{\to }{a},\stackrel{\to }{b}$ and $\stackrel{\to }{c}$.

#### Question 9:

Prove that a necessary and sufficient condition for three vectors $\stackrel{\to }{a}$, $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that

#### Question 10:

Show that the four points A, B, C and D with position vectors $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ and $\stackrel{\to }{d}$ respectively are coplanar if and only if

Necessary Condition: Firstly , let  are coplanar vectors.  Then, one of them is expressible as a linear combination of the other two. Let  for some scalars  Then,
for some scalars .
where .
Thus, if  are coplanar vectors, then there exists a scalars  not all zero simultaneously satisfying  where  are not all zero simultaneously.

Sufficient Condition: Let  are three scalars such that there exists scalars  not all zero simultaneously satisfying . We have to prove that  are coplanar vectors.
Now,

$⇒\stackrel{\to }{c}$ is a linear combination of $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$.
$⇒\stackrel{\to }{c}$ lies in a plane $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$.
Hence,  are coplanar vectors.

#### Question 1:

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

Yes,
Let a vector makes an angle  with   respectively. Let  be the direction cosines of the vector. Then,

So,

Since, the vector has direction cosines such that
Hence, a vector can have direction angles

#### Question 2:

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

Let  be the direction cosines of a straight line. Then
≠ 1
Since direction cosines of a line which makes equal angle with the axes  must satisfy .
Hence  cannot be the direction cosines of a straight line.

#### Question 3:

A vector makes an angle of $\frac{\mathrm{\pi }}{4}$ with each of x-axis and y-axis. Find the angle made by it with the z-axis.

Let the vector $\stackrel{\to }{OP}$ makes an angle  with  respectively. Suppose $\stackrel{\to }{OP}$ is inclined at angle $\gamma$ to $OZ$.
Let  be the direction cosines of $\stackrel{\to }{OP}$. Then,
,       and
Now, we have,

Hence, the angle made by it with the  $z-$axis is $\frac{\mathrm{\pi }}{2}$.

#### Question 4:

A vector $\stackrel{\to }{r}$ is inclined at equal acute angles to x-axis, y-axis and z-axis. If $\left|\stackrel{\to }{r}\right|$ = 6 units, find $\stackrel{\to }{r}$.

Suppose, vector $\stackrel{\to }{r}$ makes an angle $\alpha$ with each of the axis  and $OZ$. Then, its direction cosines are   and  i.e.

Therefore,

#### Question 5:

A vector $\stackrel{\to }{r}$ is inclined to x-axis at 45° and y-axis at 60°. If $\left|\stackrel{\to }{r}\right|$ = 8 units, find $\stackrel{\to }{r}$.

Here,  $\stackrel{\to }{r}$ makes an angle $45°$ with $OX$ and $60°$ with $OY$. So,

Therefore,

#### Question 6:

Find the direction cosines of the following vectors:
(i) $2\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}$

(ii) $6\stackrel{^}{i}-2\stackrel{^}{j}-3\stackrel{^}{k}$

(iii) $3\stackrel{^}{i}-4\stackrel{^}{k}$

(i) We have,
The direction cosines are  or,

(ii)  We have,
The direction cosines are   or,

(iii) We have,
The direction cosines are    or,

#### Question 7:

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

(ii) $\stackrel{^}{j}-\stackrel{^}{k}$

(iii) $4\stackrel{^}{i}+8\stackrel{^}{j}+\stackrel{^}{k}$

(i) Let $\stackrel{\to }{r}$ be the given vector, and let it make an angle  with  respectively. Then, its direction cosines are . So, direction ratios of $\stackrel{\to }{r}$ =  are proportional to . Therefore,
Direction cosine of $\stackrel{\to }{r}$ are
or,  .
∴

(ii) Let $\stackrel{\to }{r}$ be the given vector, and let it make an angles  with  respectively. Then, its direction cosines are . So, direction ratios of $\stackrel{\to }{r}$  are proportional to . Therefore, direction cosines of $\stackrel{\to }{r}$ are
or,

∴

(iii) Let $\stackrel{\to }{r}$ be the given vector, and let it make an angle   with   respectively. Then, its direction cosines are . So, direction ratio of $\stackrel{\to }{r}$  are proportional to
Therefore, direction ratio of $\stackrel{\to }{r}$ are
or,  .

∴

#### Question 8:

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

Let  and it make an angle  with  respectively.Then its direction cosines are  and . So, Direction ratios of  are proportional . Therefore, direction cosines of $\stackrel{\to }{r}$ are
or,  .
Thus,
and   .
.
Hence, all are equally inclined with the coordinate axis.

#### Question 9:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$

Suppose the vector makes equal angle $\alpha$ with the coordinate axis.
Then, its direction cosines are . Therefore,

Hence, direction cosines are .

#### Question 10:

If a unit vector $\stackrel{\to }{a}$ makes an angle $\frac{\mathrm{\pi }}{3}$ with $\stackrel{^}{i},\frac{\mathrm{\pi }}{4}$ with $\stackrel{^}{j}$ and an acute angle θ with $\stackrel{^}{k}$, then find θ and hence, the components of $\stackrel{\to }{a}$.

The Direction cosines of vector $\stackrel{\to }{a}$ are

Therefore,

Since, $\stackrel{\to }{a}$ is the unit vector.

Hence, components of $\stackrel{\to }{a}$ are

#### Question 11:

Find a vector $\stackrel{\to }{r}$ of magnitude $3\sqrt{2}$ units which makes an angle of $\frac{\mathrm{\pi }}{4}$ and $\frac{\mathrm{\pi }}{2}$ with y and z-axes respectively.              [NCERT EXEMPLAR]

Suppose vector $\stackrel{\to }{r}$ makes an angle α with the x-axis.

Let l, m, n be the direction cosines of $\stackrel{\to }{r}$. Then,

Now,

${l}^{2}+{m}^{2}+{n}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\alpha +\frac{1}{2}+0=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\alpha =1-\frac{1}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\alpha =±\frac{1}{\sqrt{2}}$

We know that

#### Question 12:

A vector $\stackrel{\to }{r}$ is inclined at equal angles to the three axes. If the magnitude of $\stackrel{\to }{r}$ is $2\sqrt{3}$, find $\stackrel{\to }{r}$.                 [NCERT EXEMPLAR]

Let l, m, n be the direction cosines of $\stackrel{\to }{r}$.

Now, $\stackrel{\to }{r}$ is inclined at equal angles to the three axes.

We know that

$\stackrel{\to }{r}=\left|\stackrel{\to }{r}\right|\left(l\stackrel{^}{i}+m\stackrel{^}{j}+n\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}=2\sqrt{3}\left(±\frac{1}{\sqrt{3}}\stackrel{^}{i}±\frac{1}{\sqrt{3}}\stackrel{^}{j}±\frac{1}{\sqrt{3}}\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}=2\left(±\stackrel{^}{i}±\stackrel{^}{j}±\stackrel{^}{k}\right)$

#### Question 1:

Define "zero vector".

A vector whose initial and terminal point are coincident is called a zero vector or null vector. The null vector is denoted by $\stackrel{\to }{0}$. The magnitude of null vectors is zero.

#### Question 2:

Define unit vector.

A vector whose modulus is unity is called a unit vector.The unit vector in the direction of a vector $\stackrel{\to }{a}$ is denoted by $\stackrel{^}{a}.$
Thus, $\left|\stackrel{^}{a}\right|=1$

#### Question 3:

Define position vector of a point.

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

#### Question 4:

Write $\stackrel{\to }{PQ}+\stackrel{\to }{RP}+\stackrel{\to }{QR}$ in the simplified form.

We have,
[∴ ]

#### Question 5:

If and are two non-collinear vectors such that then write the values of x and y.

We have,
[$\because$$\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are non-collinear vectors]

#### Question 6:

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ represent two adjacent sides of a parallelogram, then write vectors representing its diagonals.

Let $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ represents two adjacent sides of a parallelogram $ABCD$.
∴

and
In $△ABC$,

In $△ABD$,

#### Question 7:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ represent the sides of a triangle taken in order, then write the value of

Let $ABC$ be a triangle such that   and Then,

[âˆµ ]

#### Question 8:

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

Given:  and $\stackrel{\to }{c}$ are the position vectors of  and $C$ respectively. Then,

Consider,

#### Question 9:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are position vectors of the points A, B and C respectively, write the value of $\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{AC}.$

Given:  are the position vectors of  respectively. Then,

Therefore,

#### Question 10:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are the position vectors of the vertices of a triangle, then write the position vector of its centroid.

Let $ABC$ be a triangle and  and $F$ are the midpoints of the sides  and $AB$ respectively. Also, Let   are the position vectors of  respectively. Then the position vectors of  are  respectively.
The position vector of a point divides $AD$ in the ratio of 2 ;  is
Similarly, Position vectors of the points divides  in the ratio of $2:1$ are equal to .
Thus, the point dividing $AD$ in the ratio 2 : 1 also divides  in the same ratio.
Hence, the medians of a triangle are concurrent and the position vector of the centroid is .

#### Question 11:

If G denotes the centroid of âˆ†ABC, then write the value of $\stackrel{\to }{GA}+\stackrel{\to }{GB}+\stackrel{\to }{GC}.$

Let  be the position vectors of the vertices  respectively. Then, the position vector of the centroid $G$ is
Thus,

#### Question 12:

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ 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.

Given: $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are the position vectors of points $A$ and $B$ respectively and $C$ is a point on $AB$ such that
Let $\stackrel{\to }{c}$ is the position vector of C
Now,

Consider,

Hence, the position vector of $C$ is

#### Question 13:

If D is the mid-point of side BC of a triangle ABC such that write the value of λ.

Given: $D$ is the midpoint of the side $BC$ of a triangle $ABC$ such that
Let are the position vectors of AB, BC and CA.
Now, the position vector of $D$ is . Then,

Now, we have,

#### 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 $\stackrel{\to }{AD}+\stackrel{\to }{BE}+\stackrel{\to }{CF}.$

Given:  are the midpoints of the sides  respectively. Then, the position vectors of the midpoints  are given by

#### Question 15:

If $\stackrel{\to }{a}$ is a non-zero vector of modulus a and m is a non-zero scalar such that m $\stackrel{\to }{a}$ is a unit vector, write the value of m.

Given $\stackrel{\to }{a}$ a non zero vector with modulus $a$. Also, $m\stackrel{\to }{a}$ is the unit vector. Therefore,

#### Question 16:

If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then write the value of

Let, ABC be a given equilateral triangle and its vertices are A($\stackrel{⇀}{a}$), B($\stackrel{⇀}{b}$) and C($\stackrel{⇀}{c}$).
Also, O($\stackrel{⇀}{0}$) be the orthocentre of triangle ABC.
We know that centroid and orthocentre of equilateral triangle coincide at one point.

#### Question 17:

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

Suppose $\stackrel{\to }{r}$ makes an angle $\alpha$ with each of the axis $ΟΧ$,$ΟΥ$ and $ΟΖ$.
Then, its direction cosines are .
Now,

Since, the angle is acute Hence, we take only positive value
Therefore, unit vector is

#### 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.

Suppose, a vector $\stackrel{\to }{OP}$ makes an angle  with  respectively. Then, direction cosines of the vector are given by

Consider,

[âˆµ ]
= 2

#### 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.

Suppose a vector $\stackrel{\to }{r}$ makes an angle 45$°$ with $ΟΧ$, 60$°$ with $ΟΥ$ and having magnitude 12 units.

Therefore,

#### Question 20:

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

Given: Projection on the coordinate axes are  units. Therefore,
Length of vector $=\sqrt{{12}^{2}+{3}^{2}+{4}^{2}}$
= $\sqrt{169}$
= 13

#### Question 21:

Write the position vector of a point dividing the line segment joining points A and B with position vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ externally in the ratio 1 : 4, where

The position vectors of $A$ and $B$ are
$\stackrel{\to }{a}=2\stackrel{^}{i}+3j+4\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}=-\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$
Let $C$ divides $AB$ in the ratio such that AB : CB = 1 : 4
Position vector of $C$ = $\frac{1\left(-\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)-4\left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)}{1-4}$
= $\frac{-\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}-8\stackrel{^}{i}-12\stackrel{^}{j}-16\stackrel{^}{k}}{-3}$
= $\frac{-9\stackrel{^}{i}-11\stackrel{^}{j}-15\stackrel{^}{k}}{-3}$
= $3\stackrel{^}{i}+\frac{11\stackrel{^}{j}}{3}+5\stackrel{^}{k}$

#### Question 22:

Write the direction cosines of the vector $\stackrel{\to }{r}=6\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}.$

Given:
Then, direction cosines of $\stackrel{⏜}{r}$ are  or,

#### Question 23:

If write unit vectors parallel to

Given:
Now, $\stackrel{\to }{a}+\stackrel{\to }{b}-2\stackrel{\to }{c}=\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{j}+\stackrel{^}{k}-2\stackrel{^}{k}-2\stackrel{^}{i}$
$=-\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}$
Unit vector parallel to
$=\frac{-\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}}{\sqrt{6}}$

#### Question 24:

If write a unit vector along the vector

Given:
Therefore,

Hence, Unit vector along

#### Question 25:

Write the position vector of a point dividing the line segment joining points having position vectors externally in the ratio 2:3.

Let $A$ and $B$ be the points with position vectors respectively.
Let $C$ divide $AB$ externally in the ratio 2 : 3 such that
∴ Position vector of $C$ = $\frac{2\left(2\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)-3\left(\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}\right)}{2-3}$
= $\frac{4\stackrel{^}{i}-2\stackrel{^}{j}+6\stackrel{^}{k}-3\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}}{-1}$
= $\frac{\stackrel{^}{i}-5\stackrel{^}{j}+12\stackrel{^}{k}}{-1}$
= $-\stackrel{^}{i}+5\stackrel{^}{j}-12\stackrel{^}{k}$

#### Question 26:

If find the unit vector in the direction of .

Let
Then,
∴
Therefore, unit vector in the direction of $\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\frac{2\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)$

#### Question 27:

Given
Now,  $3\stackrel{\to }{a}-2\stackrel{\to }{b}+4\stackrel{\to }{c}=3\left(3\stackrel{^}{i}-\stackrel{^}{j}-4\stackrel{^}{k}\right)-2\left(-2\stackrel{^}{i}+4\stackrel{^}{j}-3\stackrel{^}{k}\right)+4\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)$
= $9\stackrel{^}{i}-3\stackrel{^}{j}-12\stackrel{^}{k}+4\stackrel{^}{i}-8\stackrel{^}{j}+6\stackrel{^}{k}+4\stackrel{^}{i}+8\stackrel{^}{j}-4\stackrel{^}{k}$
= $17\stackrel{^}{i}-3\stackrel{^}{j}-10\stackrel{^}{k}$
$\therefore$

#### Question 28:

A unit vector $\stackrel{\to }{r}$ makes angles $\frac{\mathrm{\pi }}{3}$ and $\frac{\mathrm{\pi }}{2}$ with respectively and an acute angle θ with $\stackrel{^}{i}$. Find θ.

A unit vector makes an angle $\frac{\mathrm{\pi }}{3}$ and $\frac{\mathrm{\pi }}{2}$ with  $\stackrel{^}{j}$ and $\stackrel{^}{k}$
Let  be its direction cosines
∴
Now
∴  ${l}^{2}+{m}^{2}+{n}^{2}=1$
$⇒$ ${l}^{2}+\frac{1}{4}+0=1$
$⇒$${l}^{2}=1-\frac{1}{4}=\frac{3}{4}$
$⇒$   $l=±\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}$
∴ $\stackrel{\to }{r}$ makes an angle  with $\stackrel{⏜}{i}$
Since, angle $\theta$ is acute.
∴ $\theta =30°$

#### Question 29:

Write a unit vector in the direction of

We have,

∴ Unit vector in the direction of $\stackrel{\to }{a}$ =

#### Question 30:

If find a unit vector parallel to .

Given:

Unit vector parallel to

#### Question 31:

Write a unit vector in the direction of .

Given:

∴ Unit vector = $\frac{\stackrel{\to }{b}}{\left|\stackrel{\to }{b}\right|}=\frac{1}{3}\left(2\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}\right)=\frac{2}{3}\stackrel{^}{i}+\frac{1}{3}\stackrel{⏜}{j}+\frac{2}{3}\stackrel{^}{k}$

#### 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).

Given: A (3, 4, −2) and B(1, 2, 4)
Let C is the mid point of AB
∴ Position vector of C =$\frac{3\stackrel{^}{i}+4\stackrel{^}{j}-2\stackrel{^}{k}+\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{^}{k}}{2}\phantom{\rule{0ex}{0ex}}$

#### Question 33:

Find a vector in the direction of which has magnitude of 6 units.

Given:

∴ Required Vector

#### Question 34:

What is the cosine of the angle which the vector $\sqrt{2}\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$ makes with y-axis?

Given  .
Therefore , direction cosines are  or $\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{2}$

So, cosine angle with respect to y-axis is $\frac{1}{2}$

#### Question 35:

Write two different vectors having same magnitude.

Let  and $\stackrel{\to }{b}=-2\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}$
It can be observed that

Hence, $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are two vectors having same magnitude.

#### Question 36:

Write two different vectors having same direction.

Let  and
Then, direction cosines of $\stackrel{\to }{p}$ are

Direction cosines of $\stackrel{\to }{q}$ are

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

#### Question 37:

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

Given:
$\stackrel{\to }{a}=5\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}\right|=\sqrt{{5}^{2}+{\left(-1\right)}^{2}+{2}^{{}^{2}}}$

∴ Position Vector in the direction of â€‹vector $=8×\frac{\stackrel{\to }{a}}{\left|\stackrel{\to }{a}\right|}=\frac{8}{\sqrt{30}}\left(5\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)$

#### Question 38:

Write the direction cosines of the vector $\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$.

Given: $\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$
Then, direction cosines are
or, .

#### Question 39:

Find a unit vector in the direction of .

Given:

Unit vector =

#### Question 40:

For what value of 'a' the vectors are collinear?

Given: Two vectors , let $\stackrel{\to }{p}=$   and $\stackrel{\to }{q}=$ $a\stackrel{^}{i}+6\stackrel{^}{j}-8\stackrel{^}{k}$
Since the given vectors are collinear, we have,

#### Question 41:

Write the direction cosines of the vectors $-2\stackrel{^}{i}+\stackrel{^}{j}-5\stackrel{^}{k}$.

Given:
Then, its direction cosines are:
or,

#### Question 42:

Find the sum of the following vectors:

Given

So, Sum of the three vectors = $\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\stackrel{^}{i}-2\stackrel{^}{j}+2\stackrel{^}{i}-3\stackrel{^}{j}+2\stackrel{^}{i}+3\stackrel{^}{k}$
= $5\stackrel{^}{i}-5\stackrel{^}{j}+3\stackrel{^}{k}$

#### Question 43:

Find a unit vector in the direction of the vector .

Given: $\stackrel{\to }{a}=3\stackrel{^}{i}-2\stackrel{^}{j}+6\stackrel{^}{k}$
Then,
$\stackrel{\to }{\left|a\right|}=\sqrt{{3}^{2}+{\left(-2\right)}^{2}+{6}^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$

∴ Unit vector = $\frac{\stackrel{\to }{a}}{\left|\stackrel{\to }{a}\right|}=\frac{3\stackrel{^}{i}-2\stackrel{^}{j}+6\stackrel{^}{k}}{7}$

#### Question 44:

If are two equal vectors, then write the value of x + y + z.

Given:  and
Since the two vectors are equal. We have,

∴

#### Question 45:

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

We have, $\stackrel{\to }{a}=2\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}$ and $\stackrel{\to }{b}=2\stackrel{^}{i}+\stackrel{^}{j}-7\stackrel{^}{k}$.

$\therefore \stackrel{\to }{a}+\stackrel{\to }{b}=\left(2\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}\right)+\left(2\stackrel{^}{i}+\stackrel{^}{j}-7\stackrel{^}{k}\right)=4\stackrel{^}{i}+3\stackrel{^}{j}-12\stackrel{^}{k}$

$⇒\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=\sqrt{{4}^{2}+{3}^{2}+{\left(-12\right)}^{2}}=\sqrt{16+9+144}=\sqrt{169}=13$

∴ Required unit vector = $\frac{\stackrel{\to }{a}+\stackrel{\to }{b}}{\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|}=\frac{4\stackrel{^}{i}+3\stackrel{^}{j}-12\stackrel{^}{k}}{13}=\frac{4}{13}\stackrel{^}{i}+\frac{3}{13}\stackrel{^}{j}-\frac{12}{13}\stackrel{^}{k}$

#### Question 46:

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

Let $\stackrel{\to }{a}=3\stackrel{^}{i}+2\stackrel{^}{j}+9\stackrel{^}{k}$ and $\stackrel{\to }{b}=\stackrel{^}{i}-2p\stackrel{^}{j}+3\stackrel{^}{k}$ be the two given vectors.

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are parallel, then

$\stackrel{\to }{b}=\lambda \stackrel{\to }{a}$ for some scalar λ

Thus, the value of p is $-\frac{1}{3}$.

#### Question 47:

Find a vector $\stackrel{\to }{a}$ of magnitude $5\sqrt{2}$, making an angle of $\frac{\mathrm{\pi }}{4}$ with x-axis, $\frac{\mathrm{\pi }}{2}$ with y-axis and an acute angle θ with z-axis.         [CBSE 2014]

It is given that vector $\stackrel{\to }{a}$ makes an angle of $\frac{\mathrm{\pi }}{4}$ with x-axis, $\frac{\mathrm{\pi }}{2}$ with y-axis and an acute angle θ with z-axis.

Now,

We know that

$\stackrel{\to }{a}=\left|\stackrel{\to }{a}\right|\left(l\stackrel{^}{i}+m\stackrel{^}{j}+n\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{a}=5\sqrt{2}\left(\frac{1}{\sqrt{2}}\stackrel{^}{i}+0\stackrel{^}{j}+\frac{1}{\sqrt{2}}\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{a}=5\left(\stackrel{^}{i}+0\stackrel{^}{j}+\stackrel{^}{k}\right)$

#### Question 48:

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

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

$\therefore \stackrel{\to }{\mathrm{PQ}}=\left(4\stackrel{^}{i}+5\stackrel{^}{j}+6\stackrel{^}{k}\right)-\left(\stackrel{^}{i}+3\stackrel{^}{j}+0\stackrel{^}{k}\right)=3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{\mathrm{PQ}}\right|=\sqrt{{3}^{2}+{2}^{2}+{6}^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$

∴ Unit vector in the direction of $\stackrel{\to }{\mathrm{PQ}}$ = $\frac{\stackrel{\to }{\mathrm{PQ}}}{\left|\stackrel{\to }{\mathrm{PQ}}\right|}=\frac{3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}}{7}=\frac{1}{7}\left(3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}\right)$

#### Question 49:

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

Let $\stackrel{\to }{a}=2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}$.

$\therefore \left|\stackrel{\to }{a}\right|=\sqrt{{2}^{2}+{\left(-3\right)}^{2}+{6}^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$

Unit vector in the direction of $\stackrel{\to }{a}=\frac{\stackrel{\to }{a}}{\left|\stackrel{\to }{a}\right|}=\frac{2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}}{7}$

∴ Vector in the direction of vector $\stackrel{\to }{a}$ which has magnitude 21 units
$=21×\left(\frac{2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}}{7}\right)\phantom{\rule{0ex}{0ex}}=3\left(2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=6\stackrel{^}{i}-9\stackrel{^}{j}+18\stackrel{^}{k}$

#### Question 50:

If $\left|\stackrel{\to }{a}\right|=4$ and $-3\le \lambda \le 2$, then write the range of $\left|\lambda \stackrel{\to }{a}\right|$.

It is given that

Thus, the range of $\left|\lambda \stackrel{\to }{a}\right|$ is [−12, 8].

#### Question 51:

In a triangle OAC, if B is the mid-point of side AC and , then what is $\stackrel{\to }{\mathrm{OC}}$?            [CBSE 2015]

In âˆ†OAC, $\stackrel{\to }{\mathrm{OA}}=\stackrel{\to }{a}$ and $\stackrel{\to }{\mathrm{OB}}=\stackrel{\to }{b}$.

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

∴ Position vector of B =

$⇒\stackrel{\to }{\mathrm{OB}}=\frac{\stackrel{\to }{\mathrm{OA}}+\stackrel{\to }{\mathrm{OC}}}{2}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\mathit{\to }}{\mathit{b}}=\frac{\stackrel{\mathit{\to }}{\mathit{a}}+\stackrel{\to }{\mathrm{OC}}}{2}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\mathit{\to }}{\mathit{a}}+\stackrel{\to }{\mathrm{OC}}=2\stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{\mathrm{OC}}=2\stackrel{\to }{b}-\stackrel{\mathit{\to }}{a}$

#### Question 52:

Write the position vector of the point which divides the join of points with position vectors in the ratio 2 : 1.

Suppose R be the point which divides the line joining the points with position vectors  in the ratio 2 : 1
And,
Here, m : n = 2 : 1
Therefore, position vector $\stackrel{\to }{\mathrm{OR}}$ is as follows:

#### Question 1:

If in a âˆ†ABC, A = (0, 0), B = (3, 3$\sqrt{3}$), C = (−3$\sqrt{3}$, 3), then the vector of magnitude 2$\sqrt{2}$ units directed along AO, where O is the circumcentre of âˆ†ABC is
(a) $\left(1-\sqrt{3}\right)\stackrel{^}{i}+\left(1+\sqrt{3}\right)\stackrel{^}{j}$

(b) $\left(1+\sqrt{3}\right)\stackrel{^}{i}+\left(1-\sqrt{3}\right)\stackrel{^}{j}$

(c) $\left(1+\sqrt{3}\right)\stackrel{^}{i}+\left(\sqrt{3}-1\right)\stackrel{^}{j}$

(d) none of these

(a) $\left(1-\sqrt{3}\right)\stackrel{^}{i}+\left(1+\sqrt{3}\right)\stackrel{^}{j}$

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

#### Question 2:

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

(b)

(c)

(d)

(c)
Let $ABCDEF$ be a regular hexagon such that  and We know, $AD$ is parallel to $BC$ such that $AD=2BC$.
∴
In $△ABC$, we have

In $△ACD$, we have

#### Question 3:

Forces 3 O $\stackrel{\to }{A}$, 5 O $\stackrel{\to }{B}$ 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

(c) 3 AC = 5 CB

Draw ON, the perpendicular to the line AB

Let $\stackrel{\to }{i}$ be the unit vector along ON
The resultant force

The angles between $\stackrel{\to }{i}$ and the forces are ∠CON, ∠AON, ∠BON respectively.

$\stackrel{\to }{R}·\stackrel{\to }{i}=3\stackrel{\to }{OA}·\stackrel{\to }{i}+5\stackrel{\to }{OB}·\stackrel{\to }{i}$

⇒ R⋅1⋅ cos ∠CON = 3 $\stackrel{\to }{OA}$⋅1⋅cos∠AON + 5$\stackrel{\to }{OB}$⋅1⋅cos∠BON

$R·\frac{ON}{OC}=3OA×\frac{ON}{OA}+5OB\frac{ON}{OB}\phantom{\rule{0ex}{0ex}}\frac{R}{OC}=\left(3+5\right)$
R = 8$\stackrel{\to }{OC}$

We know that,

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

$\begin{array}{ccccccc}\begin{array}{ccc}3\stackrel{\to }{OA}& +& 5\stackrel{\to }{OB}\end{array}& =& 8\stackrel{\to }{OC}& +& 3\stackrel{\to }{\mathrm{CA}}& +& 5\stackrel{\to }{\mathrm{CB}}\\ \stackrel{\to }{\mathrm{R}}& =& 8\stackrel{\to }{OC}& +& 3\stackrel{\to }{\mathrm{CA}}& +& 5\stackrel{\to }{\mathrm{CB}}\\ 8\stackrel{\to }{OC}& =& 8\stackrel{\to }{OC}& +& 3\stackrel{\to }{\mathrm{CA}}& +& 5\stackrel{\to }{\mathrm{CB}}\end{array}\phantom{\rule{0ex}{0ex}}\left|3\stackrel{\to }{\mathrm{AC}}\right|=\left|5\stackrel{\to }{\mathrm{CB}}\right|\phantom{\rule{0ex}{0ex}}⇒3\mathrm{AC}=5\mathrm{CB}$

#### Question 4:

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

(b)

(c)

(d) none of these

(d) None of these

$\stackrel{\to }{a}+\stackrel{\to }{b}$ is collinear with $\stackrel{\to }{c}$

where x is scalar and x ≠ 0.

$\stackrel{\to }{b}+\stackrel{\to }{c}$ is collinear with $\stackrel{\to }{a}$

y is scalar and y ≠ 0

Substracting (2) from (1) we get,

$\begin{array}{l}\stackrel{\to }{a} -\stackrel{\to }{c}= x\stackrel{\to }{c}-y\stackrel{\to }{a}\\ \stackrel{\to }{a}\left(1+y\right)=\left(1+x\right)\stackrel{\to }{c}\end{array}$

As given are not collinear,

∴ 1 + y = 0 and 1 + x = 0

y = −1 and x = −1

Putting value of x in equation (1)

$\begin{array}{l}\stackrel{\to }{a}+\stackrel{\to }{b}=-\stackrel{\to }{c}\\ \stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=0\end{array}$

#### Question 5:

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

(b) −40
Given: Three points  and  are collinear. Then,
We have,

Therefore,

#### Question 6:

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then $O\stackrel{\to }{A}+O\stackrel{\to }{B}+O\stackrel{\to }{C}+O\stackrel{\to }{D}=$
(a)

(b)

(c)

(d)

(b)

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

Also, G is the mid point BD

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

$2\stackrel{\to }{OG}+2\stackrel{\to }{OG}=\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}+\stackrel{\to }{OD}\phantom{\rule{0ex}{0ex}}4\stackrel{\to }{OG}=\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}+\stackrel{\to }{OD}\phantom{\rule{0ex}{0ex}}\therefore \stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}+\stackrel{\to }{OD}=4\stackrel{\to }{OG}$

#### Question 7:

The vector cos α cos β$\stackrel{^}{i}$ + cos α sin β$\stackrel{^}{j}$ + sin α$\stackrel{^}{k}$ is a
(a) null vector
(b) unit vector
(c) constant vector
(d) none of these

(b) unit vector
Given: The vector
Then,

Hence, the given vector is a unit vector.

#### Question 8:

In a regular hexagon ABCDEF, A$\stackrel{\to }{B}$ = a, B$\stackrel{\to }{C}$ = . Then, $\stackrel{\to }{AE}$ =
(a)

(b)

(c)

(d)

Option(c)
Given a regular hexagon $ABCDEF$ such that  and . Then,
In $△ABC$, we have

In $△ACD$, we have

Again, in $△ADE$, we have

Hence option (c).

#### Question 9:

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

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

⇒ Lines $\stackrel{\to }{a}-\stackrel{\to }{b}$ and $\stackrel{\to }{c}-\stackrel{\to }{a}$ lie on the plane.

The parmetric equation of the plane can be written as:

#### Question 10:

If O and O' are circumcentre and orthocentre of âˆ† ABC, then $\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}$ equals
(a) 2 $\stackrel{\to }{OO}$'
(b) $O\stackrel{\to }{O\text{'}}$
(c) $\stackrel{\to }{O\text{'}O}$
(d) $2\stackrel{\to }{O\text{'}O}$

Option (b).
Given: $O$ be the circumcentre and $O\text{'}$ be the orthocentre of $△ABC$. Let $G$ be the centroid of the triangle.
We know that  and $H$ are collinear and by geometry  This yields,

In other words
Since, .

∴

#### Question 11:

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

Given:

Option (d).

#### Question 12:

Let G be the centroid of âˆ† ABC. If then the bisector $\stackrel{\to }{AG},$ in terms of is
(a) $\frac{2}{3}\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)$

(b) $\frac{1}{6}\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)$

(c) $\frac{1}{3}\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)$

(d) $\frac{1}{2}\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)$

(c) $\frac{1}{3}\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)$

Taking $A$ as origin. Then, position vector of  and $C$ are  and $\stackrel{\to }{b}$ respectively. Then,
Centroid $G$ has position vector
Therefore,

#### Question 13:

If ABCDEF is a regular hexagon, then $\stackrel{\to }{AD}+\stackrel{\to }{EB}+\stackrel{\to }{FC}$ equals
(a) $2\stackrel{\to }{AB}$
(b) $\stackrel{\to }{0}$
(c) $3\stackrel{\to }{AB}$
(d) $4\stackrel{\to }{AB}$

(d) $4\stackrel{\to }{AB}$

$\begin{array}{l}\stackrel{\to }{\mathrm{AD}}=2\stackrel{\to }{\mathrm{BC}}\\ \stackrel{\to }{\mathrm{EB}}=2\stackrel{\to }{\mathrm{FA}}\\ \stackrel{\to }{\mathrm{FC}}=2\stackrel{\to }{\mathrm{AB}}\end{array}$

In triangle AOF,

$\begin{array}{l}\stackrel{\to }{FA}+\stackrel{\to }{AO}+\stackrel{\to }{FO}=0\\ \therefore \stackrel{\to }{FA}+\stackrel{\to }{AO}=-\stackrel{\to }{FO}\\ \therefore \stackrel{\to }{AD}+\stackrel{\to }{EB}=-2\stackrel{\to }{FO}\end{array}$

And $\stackrel{\to }{AB}=-\stackrel{\to }{FO}$

$\begin{array}{l}\therefore \stackrel{\to }{AD}+\stackrel{\to }{EB}=2\stackrel{\to }{AB}\\ \therefore \stackrel{\to }{AD}+\stackrel{\to }{EB}+\stackrel{\to }{FC}=2\stackrel{\to }{AB}+2\stackrel{\to }{AB}=4\stackrel{\to }{AB}\end{array}$

#### Question 14:

The position vectors of the points A, B, C are respectively. These points
(a) form an isosceles triangle
(b) form a right triangle
(c) are collinear
(d) form a scalene triangle

(a) form an isosceles triangle
Given: Position vectors of  are  and $\stackrel{^}{i}+4\stackrel{^}{j}-3\stackrel{^}{k}$. Then,

Hence, the triangle is isosceles as two of its sides are equal.

#### Question 15:

If three points A, B and C have position vectors respectively are collinear, then (x, y) =
(a) (2, −3)
(b) (−2, 3)
(c) (−2, −3)
(d) (2, 3)

(a) (2, −3)
Given position vectors of  and $C$ are  and $y\stackrel{^}{i}-2\stackrel{^}{j}-5\stackrel{^}{k}.$ Then,

Since, the given vectors are collinear.

#### Question 16:

ABCD is a parallelogram with AC and BD as diagonals. Then, $\stackrel{\to }{AC}-\stackrel{\to }{BD}=$
(a)
(b)
(c)
(d) $\stackrel{\to }{AB}$

(c)
Given: $ABCD$, a parallelogram with diagonals $AC$ and $BD$. Then,

∴                         [âˆµ $\stackrel{\to }{AD}=\stackrel{\to }{BC}$]

#### Question 17:

If OACB is a parallelogram with then $\stackrel{\to }{OA}=$
(a) $\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)$

(b) $\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)$

(c) $\frac{1}{2}\left(\stackrel{\to }{b}-\stackrel{\to }{a}\right)$

(d) $\frac{1}{2}\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)$

(d) $\frac{1}{2}\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)$
Given a parallelogram $OACB$ such that . Then,

[âˆµ $\stackrel{\to }{BC}=\stackrel{\to }{OA}$]

Therefore,

#### Question 18:

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

(d) both the vectors have the same direction but different magnitudes
If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are collinear vectors, then they are paprallel. Therefore, we have
, for some scalar $\lambda$.
If $\lambda =±1$
If  and  . Then,

Thus, the respective components of $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ can have different directions. Hence, the statement given in (d) is incorrect.

#### Question 19:

In Figure, which of the following is not true?

(a) $\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{CA}=\stackrel{\to }{0}$

(b) $\stackrel{\to }{AB}+\stackrel{\to }{BC}-\stackrel{\to }{AC}=\stackrel{\to }{0}$

(c) $\stackrel{\to }{AB}+\stackrel{\to }{BC}-\stackrel{\to }{CA}=\stackrel{\to }{0}$

(d) $\stackrel{\to }{AB}-\stackrel{\to }{CB}+\stackrel{\to }{CA}=\stackrel{\to }{0}$
Figure

(c) $\stackrel{\to }{AB}+\stackrel{\to }{BC}-\stackrel{\to }{CA}=\stackrel{\to }{0}$

We have, LHS =                     [âˆµ ]
$=-\stackrel{\to }{CA}-\stackrel{\to }{CA}\phantom{\rule{0ex}{0ex}}=-2\stackrel{\to }{CA}$
So, LHS$\ne$RHS
Hence, It is not true.

#### 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)

(i) Let O be the origin.
The position vector of point P(3,2), $\stackrel{\to }{OP}=3\stackrel{^}{i}+2\stackrel{^}{j}$
Component of $\stackrel{\to }{OP}$ along x-axis = a vector of magnitude 3 having its direction along the positive direction of x-axis.
Component of $\stackrel{\to }{OP}$ 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), $\stackrel{\to }{OQ}=-5\stackrel{^}{i}+\stackrel{^}{j}$
Component of $\stackrel{\to }{OQ}$ along x-axis = a vector of magnitude 5 having its direction along the negative direction of x-axis.
Component of $\stackrel{\to }{OQ}$ 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), $\stackrel{\to }{OR}=-11\stackrel{^}{i}-9\stackrel{^}{j}$
Component of $\stackrel{\to }{OR}$ along x-axis = a vector of magnitude 11 having its direction along the negative direction of x-axis.
Component of $\stackrel{\to }{OR}$ 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), $\stackrel{\to }{OS}=4\stackrel{^}{i}-3\stackrel{^}{j}$
Component of $\stackrel{\to }{OS}$ along x-axis = a vector of magnitude 4 having its direction along the positive direction of x-axis.
Component of $\stackrel{\to }{OS}$ along y-axis = a vector of magnitude 3 having its direction along the negative direction of y-axis.

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