Rd Sharma Xi 2018 Solutions for Class 11 Humanities Math Chapter 28 Introduction To Three Dimensional Coordinate Geometry are provided here with simple step-by-step explanations. These solutions for Introduction To Three Dimensional Coordinate Geometry are extremely popular among Class 11 Humanities students for Math Introduction To Three Dimensional Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Humanities Math Chapter 28 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 28.10:

Question 9:

Show that the points (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of an isosceles right-angled triangle.

Answer:

Let A(0, 7, 10), B(-1, 6, 6), C(-4, 9, 6) be the coordinates of â–³ABC
Then, we have:
AB=0+12+7-62+10-62      =18      =32
BC=-1+42+6-92+6-62      =18      =32
AC=0+42+7-92+10-62      =36      =6Clearly, AB2+BC2=AC2  ABC=90°& AB=BC
Hence â–³ABC is a right-angled isosceles triangle.

Page No 28.10:

Question 10:

Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.

Answer:

Let A(3,3,3) , B(0,6,3) , C( 1,7,7) and D (4,4,7) are the vertices of quadrilateral ABCD

We have :

AB = 0-32+6-32+3-32
     =-32+32+02=9+9+0=18=32

BC = 1-02+7-62+7-32
     =12+12+42=1+1+16=18=32

CD = 4-12+4-72+7-72
     =32+-32+02=9+9+0=18=32

DA = 4-32+4-32+7-32
      =12+12+42=1+1+16=18=32

AB = BC = CD = DA

AC = 1-32+7-32+7-32
     =-22+42+42=4+16+16=36=6

BD = 4-02+4-62+7-32 
     =42+-22+42=16+4+16=36=6
AC = BD
Since, all sides and diagonals of quadrilateral ABCD are equal
Therefore, the points are the vertices of a square.    
                  
                     

Page No 28.10:

Question 11:

Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Answer:

Let A(1,3,0),B(-5,5,2), C(-9,-1,2) and D(-3,-3,0) be the coordinates of quadrilateralABCD

 AB=-5-12+5-32+2-02       =36+4+4       =  44BC=-9+52+-1-52+2-22       =16+36+0       =52    CD=-3+92+-3+12+0-22       =36+4+4       =44DA=1+32+3+32+0-02       =16+36+0       =52Here, we see that AB=CD& BC=DA
Since, opposite pair of sides are equal .

Therefore, ABCD is a parallelogram .

AC=-9-12+-1-32+2-02        =-102+-42+22        =100+16+4       =120 

BD=-3+52+-3-52+0-22        =22+-82+-22        =4+64+4        =72

AC BD
∴ ABCD is not a rectangle.

Page No 28.10:

Question 12:

Show that the points A(1, 3, 4), B(–1, 6, 10), C(–7, 4, 7) and D(–5, 1, 1) are the vertices of a rhombus.

Answer:

Let A(1,3,4) , B(-1,6,10) , C(-7,4,7) and D (-5,1,1) be the vertices of quadrilateral  ABCD
AB=-1-12+6-32+10-42      =4+9+36       =49      =7BC=-7+12+4-62+7-102      =36+4+9      =49      =7CD=-5+72+1-42+1-72      =4+9+36      =49      =7DA=1+52+3-12+4-12      =36+4+9      =49      =7 AB=BC=CD=DA

Hence, ABCD is a rhombus.

Page No 28.10:

Question 13:

Prove that the tetrahedron with vertices at the points O(0, 0, 0), A(0, 1, 1), B(1, 0, 1) and C(1, 1, 0) is a regular one.

Answer:

 
The faces of a regular tetrahedron are equilateral triangles.

Let us consider OAB
 OA=0-02+0-12+0-12      =2
 OB=1-02+0-02+1-02      =2
 AB=1-02+0-12+1-12      =2
Hence, this face is an equilateral triangle.
Similarly, OBC, OAC, ABC all are equilateral triangles.
Hence, the tetrahedron is a regular one.

Page No 28.10:

Question 14:

Show that the points (3, 2, 2), (–1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.

Answer:

Let the points be A (3, 2, 2), B (-1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere
whose centre be P (1, 3, 4).
Since, AP, BP, CP  and DP are radii.
Hence, AP = BP = CP = DP

AP=1-32+3-22+4-22      =-22+12+22      =4+1+4      =9      =3BP=1+12+3-42+4-22      =22+-12+22      =4+1+4      =9      =3CP=1-02+3-52+4-62      =12+-22+-22      =1+4+4      =9      =3DP=1-22+3-12+4-22      =-12+22+22      =1+4+4      =9      =3

Here, we see that AP = BP = CP = DP
Hence,  A (3, 2, 2), B (-1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere whose radius is 3 .

Page No 28.10:

Question 15:

Find the coordinates of the point which is equidistant  from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).

Answer:

Let P (x, y, z) be the required point which is equidistant from the points O(0,0,0), A(2,0,0)
B(0,3,0) and C(0,0,8)

Then,
OP = AP
OP2=AP2
 x2+y2+z2=x-22+y2+z2x2=x-22x2=x2-4x+44x=4 x=44x=1

Similarly, we have:
OP = BP
OP2=BP2
x2+y2+z2=x2+y-32+z2y2=y2-6y+96y=9 y=96 y=32

Similarly, we also have:

OP = CP
OP2=CP2
 x2+y2+z2=x2+y2+z-82z2=z2-16z+6416z=64 z=6416 z=4

Thus, the required point is P 1, 32, 4.

Page No 28.10:

Question 16:

If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.

Answer:

Let coordinates of point P be (x, y, z).
Given:
3PA = 2PB

3x+22+y-22+z-32=2x-132+y+32+z-1323x2+4x+4+y2-4y+4+z2-6z+9=2x2-26x+169+y2+6y+9+z2-26z+169Squaring both sides,9x2+y2+z2+4x-4y-6z+17=4x2+y2+z2-26x+6y-26z+3479x2+9y2+9z2+36x-36y-54z+153=4x2+4y2+4z2-104x+24y-104z+13885x2+5y2+5z2+140x-60y+50z-1235=05x2+y2+z2+140x-60y+50z-1235=05x2+y2+z2+140x-60y+50z-1235=0 is the locus of the point P.

Page No 28.10:

Question 17:

Find the locus of P if PA2 + PB2= 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).

Answer:

Let P (x, y, z) be the point if PA2+PB2=2k2

x-32+y-42+z-522+x+12+y-32+z+722=2k2x2-6x+9+y2-8y+16+z2-10z+25+x2+2x+1+y2-6y+9+z2+14z+49=2k22x2+2y2+2z2-4x-14y+4z+109-2k2=0

Hence, 2x2+2y2+2z2-4x-14y+4z+109-2k2=0 is the locus of point P.

Page No 28.10:

Question 18:

Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.

Answer:

Let A(a,b,c) , B(b,c,a) and C(c,a,b) be the vertices of ABC. Then,

AB = b-a2+c-b2+a-c2
     =b2-2ab+a2+c2-2bc+b2+a2-2ca+c2=2a2+2b2+2c2-2ab-2bc-2ca=2a2+b2+c2-ab-bc-ca

BC = c-b2+a-c2+b-a2
     =c2-2bc+b2+a2-2ca+c2+b2-2ab+a2=2a2+2b2+2c2-2ab-2bc-2ca=2a2+b2+c2-ab-bc-ca

CA = a-c2+b-a2+c-b2
     =a2-2ca+c2+b2-2ab+a2+c2-2bc+b2=2a2+2b2+2c2-2ab-2bc-2ca=2a2+b2+c2-ab-bc-ca
AB = BC = CA
Therefore,ABC is an equilateral triangle.

Page No 28.10:

Question 19:

Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?

Answer:

Let A(3,6,9), B(10,20,30) and C( 25,-41,5) are vertices of ABC

AB = 10-32+20-62+30-92
     =72+142+212=49+196+441=686=714

BC = 25-102+-41-202+5-302
     =152+-612+-252=225+3721+625=4571

CA= 3-252+6+412+9-52
     =-222+472+-42=484+2209+16=2709=3301

AB2+BC2=7142+45712                   =686+4571                   =5257CA2=2709 AB2+BC2CA2

A triangle ABC is right-angled at B if CA2=AB2+BC2.
But, CA2AB2+BC2
Hence, the points are not vertices of a right-angled triangle.

Page No 28.10:

Question 20:

Verify the following:
(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
(iv) (5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.

Answer:

(i) Let A(0, 7, -10) , B(1, 6, -6) , C(4, 9, -6) be the vertices of ABC.Then,
 AB = 1-02+6-72+-6+102
      =12+-12+42=1+1+16=18=32

BC = 4-12+9-62+-6+62
       =32+32+0=9+9=18=32

CA0-42+7-92+-10+62
    =16+4+16=36=6
Clearly, AB BC
Thus, the given points are the vertices of an isosceles triangle.


(ii) Let A(0,7,10) , B-1,6,6) and C-4,9,6) be the vertices of ABC. Then ,

AB = -1-02+6-72+6-102
          
     =-12+-12+-42=1+1+16=18=32
      
BC -4+12+9-62+6-62
     =-32+32+0=9+9=18=32

AC = -4-02+9-72+6-102
     =-42+22+-42=16+4+16=36=6
 AC2=AB2+BC2
Thus, the given points are the vertices of a right-angled triangle.

(iii) Let A(-1, 2, 1) , B(1, -2, 5) , C(4, -7, 8), D(2, -3, 4) be the vertices of quadrilateral ABCD

   AB=1+12+-2-22+5-12     =4+16+16     =36     =6BC=4-12+-7+22+8-52     =9+25+9     =43CD=2-42+-3+72+4-82     =4+16+16     =36     =6DA=-1-22+2+32+1-42     =9+25+9     =43 AB=CD and BC=DA
Since, each pair of opposite sides are equal.
Thus, quadrilateral ABCD is a parallelogram.

(iv) Let A(5, -1, 1) , B(7, -4, 7) , C(1, -6, 10), D(-1, -3, 4) be the vertices of quadrilateral ABCD

   AB=5-72+-1+42+1-72     =4+9+36     =49     =7BC=7-12+-4+62+7-102     =36+4+9     =49     =7CD=1+12+-6+32+10-42     =4+9+36     =49     =7DA=-1-52+-3+12+4-12     =36+4+9     =49     =7 AB=BC=CD=DA
Since, all the sides are equal.
Thus, quadrilateral ABCS is a rhombus.

Page No 28.10:

Question 21:

Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer:

Let P (x, y, z) be any point that is equidistant from the points A (1, 2, 3) and B (3, 2, −1).
Then, we have:
     PA = PB
   x-12+y-22+z-32=x-32+y-22+z+12    x2-2x-1+y2-4y+4+z2-6z+9=x2-6x+9+y2-4y+4+z2+2z+1-2x-4y-6z+14=-6x-4y+2z+14-2x+6x-4y+4y-6z-2z=14-144x-8z=0x-2z=0
Hence, the locus is x - 2z = 0

Page No 28.10:

Question 22:

Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(–4, 0, 0) is equal to 10.

Answer:

Let P (x, y, z) be any point , the sum of whose distance from the points A(4,0,0) and B(-4,0,0)
is equal to 10. Then,
PA + PB = 10

x-42+y-02+z-02+x+42+y-02+z-02=10x+42+y-02+z-02=10-x-42+y-02+z-02x2+8x+16+y2+z2=10-x2-8x+16+y2+z2
x2+8x+16+y2+z2=100+x2-8x+16+y2+z2-20x2-8x+16+y2+z216x-100=-20x2-8x+16+y2+z24x-25=-5x2-8x+16+y2+z216x2-200x+625=25x2-8x+16+y2+z216x2-200x+625=25x2-200x+400+25y2+25z29x2+25y2+25z2-225=0

 9x2+25y2+25z2-225=0 is the required locus . 

Page No 28.10:

Question 23:

Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

Answer:

To show that ABCD is a parallelogram, we need to show that its two opposite sides are equal.
AB=-1-12+-2-22+-1-32     =4+16+16     =36     =6BC=2+12+3+22+2+12     =9+25+9     =43CD=4-22+7-32+6-22     =4+16+16     =36     =6DA=1-42+2-72+3-62     =9+25+9     =43AB=CD and BC=DASince, opposite pairs of sides are equal .ABCD is a parallelogramAC=2-12+3-22+2-32     =1+1+1     =3BD=4+12+7+22+6+12     =25+81+49     =155Since, ACBD

Thus, ABCD is not a rectangle.

Page No 28.10:

Question 24:

Find the equation of the set of the points P such that its distances from the points A(3, 4, –5) and B(–2, 1, 4) are equal.

Answer:

Let P (x, y, z) be any point which is equidistant from A (3,4,5) and B (-2,1,4) .Then,
PA = PB
x-32+y-42+z+52=x+22+y-12+z-42x2-6x+9+y2-8y+16+z2+10z+25=x2+4x+4+y2-2y+1+z2-8z+16x2-6x+9+y2-8y+16+z2+10z+25=x2+4x+4+y2-2y+1+z2-8z+16-10x-6y+18z+29=010x+6y-18z-29=0

Hence, 10x + 6y -18z -29 = 0 is the required equation.



Page No 28.19:

Question 1:

The vertices of the triangle are A(5, 4, 6), B(1, –1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

Answer:



AB = 42+52+32=16+25+9=50=52
AC = 12+12+42=18=32
AD is the internal bisector of BAC.
BDDC=ABAC=53
Thus, D divides BC internally in the ratio 5:3.
 D5×4+3×15+3, 5×3+3-15+3, 5×2+3×35+3 D238, 128, 198 D238, 32, 198  AD=5-2382+4-1282+6-1982          =172+202+29282          =289+400+84182          =15308

Page No 28.19:

Question 2:

A point C with z-coordinate 8 lies on the line segment joining the points A(2, –3, 4) and B(8, 0, 10). Find its coordinates.

Answer:

Suppose C divides AB in the ratio λ:1.
Then, the coordinates of C are 8λ+2λ+1, -3λ+1, 10λ+4λ+1.
The z-coordinate of C is 8.
 10λ+4λ+1=810λ+4=8λ+82λ=4 λ=2
Hence, the coordinates of C are as follows:
  8λ+2λ+1, -3λ+1, 10λ+4λ+18×2+22+1, -32+1, 10×2+42+1183, -33, 243 6, -1, 8
Hence, the coordinates of C are (6, −1, 8).



Page No 28.20:

Question 3:

Show that the three points A(2, 3, 4), B(–1, 2 – 3) and C(–4, 1, –10) are collinear and find the ratio in which C divides AB.

Answer:

Suppose C divides AB in the ratio λ:1.
Then, the coordinates of C are -λ+2λ+1, 2λ+3λ+1, -3λ+4λ+1.
But, the coordinates of C are (-4, 1, -10).
 -λ+2λ+1=-4, 2λ+3λ+1=1, -3λ+4λ+1=-10-λ+2=-4λ-4,  2λ+3=λ+1, -3λ+4=-10λ-103λ=-6, λ=-2, 7λ=-14 λ=-2, λ=-2, λ=-2

From these three equations, we have:
λ=-2
So, C divides AB in the ratio 2:1 (externally).

Page No 28.20:

Question 4:

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Answer:

Let A(2, 4, 5) and B(3, 5, 4)
Let the line joining A and B be divided by the yz-plane at point P in the ratio λ:1.
Then, we have:
P3λ+2λ+1, 5λ+4λ+1, 4λ+5λ+1
Since P lies on the yz-plane, the x-coordinate of P will be zero.
3λ+2λ+1=0   3λ+2=0λ=-23
Hence, the yz-plane divides AB in the ratio 2:3 (externally).

Page No 28.20:

Question 5:

Find the ratio in which the line segment joining the points (2, –1, 3) and (–1, 2, 1) is divided by the plane x + y + z = 5.

Answer:

Suppose the plane x + y + z = 5 divides the line joining the points A (2, −1, 3) and B (-1, 2, 1) at a point C in the ratio λ:1.
Then, coordinates of C are as follows:
-λ+2λ+1, 2λ-1λ+1, λ+3λ+1
Now, the point C lies on the plane x + y + z = 5.
Therefore, the coordinates of C must satisfy the equation of the plane.
        -λ+2λ+1+2λ-1λ+1+λ+3λ+1=5-λ+2+2λ-1+λ+3=5λ+52λ+4=5λ+5-3λ=1λ=-13
So, the required ratio is 1:3 (externally).

Page No 28.20:

Question 6:

If the points A(3, 2, –4), B(9, 8, –10) and C(5, 4, –6) are collinear, find the ratio in which C divides AB.

Answer:

Suppose C divides AB in the ratio λ:1.
Then, coordinates of C are as follows:
9λ+3λ+1, 8λ+2λ+1, -10λ-4λ+1
But, the coordinates of C are (5, 4, -6).
 9λ+3λ+1=5, 8λ+2λ+1=4, -10λ-4λ+1=-6
These three equations give λ=12.
So, C divides AB in the ratio 1:2.

Page No 28.20:

Question 7:

The mid-points of the sides of a triangle ABC are given by (–2, 3, 5), (4, –1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Answer:

Let D x1, y1, z1, E x2 ,y2, z2 and F x3, y3, z3 be the vertices of the given triangle.
And, let A -2, 3, 5, B 4, -1, 7 and C 6, 5, 3 be the mid-points of the sides EF, FA and DE, respectively.
Now, A is the mid-point of EF.
x2+x32=-2, y2+y32=3, z2+z32=5x2+x3=-4, y2+y3=6, z2+z3=10                   i
B is the mid-point of FD.
x1+x32=4, y1+y32=-1, z1+z32=7x1+x3=8, y1+y3=-2, z1+z3=14                ii

C is the mid-point of DE.
x1+x22=6, y1+y22=5, z1+z22=3x1+x2=12, y1+y2=10, z1+z2=6                    iii
Adding the first three equations in (i), (ii) and (iii):
2x1+x2+x3=-4+8+12x1+x2+x3=8
Solving the first three equations in (i), (ii) and (iii) with x1+x2+x3=8:
x1=12, x2=0, x3=-4
Adding the next three equations in (i), (ii) and (iii):
2y1+y2+y3=6-2+10y1+y2+y3=7
Solving the next three equations in (i), (ii) and (iii) with y1+y2+y3=7:
y1=1, y2=9, y3=-3
Adding the last three equations in (i), (ii) and (iii):
2z1+z2+z3=10+14+6z1+z2+z3=15
Solving the last three equations in (i), (ii) and (iii) with z1+z2+z3=15:
z1=5, z2=1, z3=9
Thus, the vertices of the triangle are (12, 1, 5), (0, 9, 1), (−4, −3, 9).

Page No 28.20:

Question 8:

A(1, 2, 3), B(0, 4, 1), C(–1, –1, –3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

Answer:



AB = -12+22+-22=1+4+4=9=3
AC = -22+-32+-62=4+9+36=49=7

AD is the internal bisector of BAC.
BDDC=ABAC=37
Thus, D divides BC internally in the ratio 3:7.
D3-1+3×03+7, 3-1+743+7, 3-3+7×13+7 D-310, 2510, -210 D-310, 52, -15 

Page No 28.20:

Question 9:

Find the ratio in which the sphere x2 + y2+ z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).

Answer:

Let the sphere  meet the line joining the given points at the point (x1, y1, z1).

Then, we have:

Suppose the point (x1, y1, z1) divides the line joining the points  (12, – 4, 8) and (27, – 9, 18) in the ratio λ:1.

Substitute these values in equation (1):

Thus, the sphere divides the line joining the given points in the ratio 2:3 and 2:– 3.

Hence, the given sphere divides the line joining the points (12, – 4, 8) and (27, – 9, 18) internally in the ratio 2:3 and externally in the ratio −2:3.

Page No 28.20:

Question 10:

Show that the plane ax + by + cz + d = 0 divides the line joining the points (x1, y1, z1) and (x2, y2, z2) in the ratio -ax1+by1+cz1+dax2+by2+cz2+d.

Answer:

Let:
A = (x1, y1, z1
B = (x2, y2, z2)

Now, let the line joining A and B be divided by the plane ax + by + cz + d = 0 at point P in the ratio λ:1.

P = λx2+x1λ+1, λy2+y1λ+1, λz2+z1λ+1
Since P lies on the given plane,

ax + by + cz + d = 0

Thus,

aλx2+x1λ+1+bλy2+y1λ+1+cλz2+z1λ+1+d=0aλx2+x1+bλy2+y1+cλz2+z1+dλ+1=0λax2+by2+cz2+d+ax1+by1+cz1+d=0λax2+by2+cz2+d=-ax1+by1+cz1+dλ=-ax1+by1+cz1+dax2+by2+cz2+dλ=-ax1+by1+cz1+dax2+by2+cz2+dThus, the given plane divides the line joining x1, y1, z1 and x2, y2, z2 in the ratio -ax1+by1+cz1+dax2+by2+cz2+d. 

Page No 28.20:

Question 11:

Find the centroid of a triangle, mid-points of whose sides are (1, 2, –3), (3, 0, 1) and (–1, 1, –4).

Answer:

Let Ax1, y1, z1, Bx2, y2, z2 and Cx3, y3, z3 be the vertices of the given triangle,and let D1, 2,-3 , E3, 0, 1 and F-1, 1,-4 be the mid points of the sides BC, CA and AB respectively.D is the mid point of BC x2+x32=1, y2+y32=2 and z2+z32=-3x2+x3=2, y2+y3=4 and z2+z3=-6          ...1E is the mid point of CA x1+x32=3, y1+y32=0 and z1+z32=1x1+x3=6, y1+y3=0 and z1+z3=2          ...2F is the mid point of AB x1+x22=-1, y1+y22=1 and z1+z22=-4x1+x2=-2, y1+y2=2 and z1+z2=-8          ...2Adding first three equations we get,2x1+x2+x3=6, 2y1+y2+y3=6 and 2z1+z2+z3=-12x1+x2+x3=3, y1+y2+y3=3 and z1+z2+z3=-6The coordinate of the centroid is given by x1+x2+x33, y1+y2+y33,z1+z2+z331, 1, -2

Page No 28.20:

Question 12:

The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, –6) respectively, find the coordinates of the point C.

Answer:



Let G be the centroid of ABC.
Given: G1, 1, 1
Let Cx, y, z
      Then, 1=3-1+x3      3=3-1+x      3=2+x x=1and 1=-5+7+y3    3=2+y    y=1and 1=7-6+z3   3=1+z   z=2

C1, 1, 2

Page No 28.20:

Question 13:

Find the coordinates of the points which tisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6).

Answer:

Let P(4, 2, -6) and Q(10, -16, 6)
Let A and B be the point of trisection.
Then, we have:
PA = AB = BQ
∴ PA:AQ = 1:2
Thus, A divides PQ internally in the ratio 1:2.
∴ A1×10+2×41+2, 1×-16+2×21+2, 1×6+2×-61+2
                    10+83, -16+43, 6-123183, -123, -636, -4, -2
A(6, -4, -2)
AB = BQ       


                                                                                    
Therefore, B is the mid-point of AQ.
∴ B6+102, -4-162, -2+62
                  162, -202, 428, -10, 2
B(8, -10, 2)

Page No 28.20:

Question 14:

Using section formula, show that he points A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) are collinear.

Answer:

Let, C divides AB in the ratio λ:1.
Then, coordinates of C are -λ+2λ+1,2λ-3λ+1,λ+4λ+1.
But, the coordinates of C are 0, 13, 2.
-λ+2λ+1=0 , 2λ-3λ+1=13, λ+4λ+1 = 2
From each of these equations, we get λ=2.
Since each of these equations give the same value of λ.
Therefore, the given points are collinear.

Page No 28.20:

Question 15:

Given that  P(3, 2, –4), Q(5, 4, –6) and R(9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Answer:

Let Q divide PR in the ratio λ:1
Thus, the coordinates of Q are as follows:
9λ+3λ+1, 8λ+2λ+1, -10λ-4λ+1
But, the coordinates of Q are (5, 4, −6) .
 9λ+3λ+1=5 , 8λ+2λ+1=4 , -10λ-4λ+1=-6
These three equation gives λ=12.
So, Q divides PR in the ratio 12:1 or 1:2

Page No 28.20:

Question 16:

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, –8) is divided by the yz-plane.

Answer:

Let A = (4, 8, 10) , B = (6, 10, -8)
Let the line joining A and B be divided by the YZ-plane at point P in the ratio λ:1.
P=6λ+4λ+1, 10λ+8λ+1, -8λ+10λ+1
Since P lies on the YZ-plane, the x-coordinate of P will be zero.
6λ+4λ+1=06λ+4=06λ=-4 λ=-23
Hence, the YZ-plane divides AB in the ratio 2:3 (externally)



Page No 28.22:

Question 1:

Write the distance of the point P (2, 3,5) from the xy-plane.

Answer:

The distance of the point P (2, 3, 5) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 5.
Hence, the distance of the point P (2, 3,5) from the xy-plane is 5.

Page No 28.22:

Question 2:

Write the distance of the point P(3, 4, 5) from z-axis.

Answer:

The distance of the point P(3, 4, 5) from z-axis is given by
32+42=25=5

Page No 28.22:

Question 3:

If the distance between the points P(a, 2, 1) and Q (1, −1, 1) is 5 units, find the value of a.

Answer:

PQ = 5
1-a2+-1-22+1-12=51-a2+-32=251-2a+a2+9-25=0a2-2a-15=0a-5a+3=0a=5, -3

Page No 28.22:

Question 4:

The coordinates of the mid-points of sides AB, BC and CA of  â–³ABC are D(1, 2, −3), E(3, 0,1) and F(−1, 1, −4) respectively. Write the coordinates of its centroid.

Answer:

Let the coordintes of the triangles be A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3).
Now,
Mid point of AB is D(1, 2, −3)
x1+x22=1, y1+y22=2 and z1+z22=-3x1+x2=2, y1+y2=4 and z1+z2=-6       .....1

Mid point of BC is E(3, 0,1)
x2+x32=3, y2+y32=0 and z2+z32=1x2+x3=6, y2+y3=0 and z2+z3=2       .....2

Mid point of AC is F(−1, 1, −4)

x1+x32=-1, y1+y32=1 and z1+z32=-4x1+x3=-2, y1+y3=2 and z1+z3=-8      .....3

Adding (1), (2) and (3),we get
2x1+x2+x3=6, 2 y1+y2+y3=6 and 2z1+z2+z3=-12x1+x2+x3=3, y1+y2+y3=3 and z1+z2+z3=-6x1+x2+x33=1, y1+y2+y33=1 and z1+z2+z33=-2
Hence, the centroid of the traingle ABC is (1, 1, −2).

Page No 28.22:

Question 5:

Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.

Answer:

We know that the x  and z coordinates on y - axis are 0
The coordinates of the foot of the perpendicular from a point (1, 2, 3) on y - axis are (0, 2, 0)

Page No 28.22:

Question 6:

Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.

Answer:

The distance of the point P(3, 5, 12) from x-axis is given by
52+122=169=13

Page No 28.22:

Question 7:

Write the coordinates of third vertex of a triangle having centroid at the origin and two vertices at (3, −5, 7) and (3, 0, 1).

Answer:

Let the coordinates of third vertex  be (x1, y1, z1)
Now,
x1+3+33=0, y1-5+03=0 and z1+7+13=0x1=-6, y1=5 and z1=-8
Hence, the coordinates of third vertex of a triangle  is (−6, 5, −8).

Page No 28.22:

Question 8:

What is the locus of a point for which y = 0, z = 0?

Answer:

We know that on x - axis both y = 0, z = 0.
Hence, the locus of a point for which y = 0, z = 0 is x - axis.

Page No 28.22:

Question 9:

Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.


 

Answer:

Let the yz - plane divide the line sgement joining the points (2, 4,5) and (3, −5, 4) in m : 1.
Now, we know that on yz-plane the coordinate of x is 0.
m×3+1×2m+1=03m+2=0m=-23
Hence, yz - plane divide the line sgement joining the points (2, 4,5) and (3, −5, 4) in 2 : 3 externally.

Page No 28.22:

Question 10:

Find the point on y-axis which is at a distance of 10 units from the point (1, 2, 3).

Answer:

We know that the x and z coordinates of the point on the y-axis are 0.
So, let the required point be (0, y, 0)
Now,
1-02+2-y2+3-02=101+4-4y+y2+9=10y2-4y+4=0y-22=0y=2, 2
Hence, the required point is (0, 2, 0)

Page No 28.22:

Question 11:

Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).

Answer:

We know that the y and z coordinates of the point on the x-axis are 0.
So, let the required point be C (x, y, z)
Now, CA = CB

3-x2+2-02+2-02=5-x2+5-02+4-029-6x+x2+4+4=25-10x+x2+25+1617-6x+x2=66-10x+x24x=49x=494
Hence, the required point is 494, 0, 0

Page No 28.22:

Question 12:

Find the coordinates of a point equidistant from the origin and points A (a, 0, 0), B (0, b, 0) andC(0, 0, c).

Answer:

Let the point be P(x, y, z).
Now, PO = PA
0-x2+0-y2+0-z2=a-x2+0-y2+0-zx2+y2+z2=a2-2ax+x2+y2+z20=a2-2axx=a2
Also, PO = PB
0-x2+0-y2+0-z2=0-x2+b-y2+0-zx2+y2+z2=x2+b2-2by+y2+z20=b2-2byy=b2
Again, PO = PC
0-x2+0-y2+0-z2=0-x2+0-y2+c-zx2+y2+z2=x2+y2+c2-2cz+z20=c2-2czz=c2

Hence, the coordinates of the point is a2, b2, c2

Page No 28.22:

Question 13:

Write the coordinates of the point P which is five-sixth of the way from A(−2, 0, 6) to B(10, −6, −12).

Answer:

Let the coordinates of the point be P(x, y, z).
Now
PA=56PBPAPB=56x=5×10-6×25+6, y=5×-6+6×05+6, z=5×-12+6×65+6x=3811, y=-2411, z=-3411
Hence, the coordinates of the point is 3811, -2411, -3411

Page No 28.22:

Question 14:

If a parallelopiped is formed by the planes drawn through the points (2,3,5) and (5, 9, 7) parallel to the coordinate planes, then write the lengths of edges of the parallelopiped and length of the diagonal.

Answer:

Lengths of edges of the parallelopiped are given by
5 − 2, 9 − 3, 7 − 5
= 3, 6, 2
Length of the diagonal is given by
32+62+22=49=7 units

Page No 28.22:

Question 15:

Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3,2) and C(0, 0,1).

Answer:

The coordiante of x point on yz-plane is 0
Let the point be P(0, y, z).
Now, PA = PB
2-02+0-y2+3-z2=0-02+3-y2+2-z4+y2+9-6z+z2=9-6y+y2+4-4z+z2-6z=-6y-4z3y-z=0               .....1 
Also, PA = PC
 2-02+0-y2+3-z2=0-02+0-y2+1-z24+y2+9-6z+z2=y2+1-2z+z213-6z=1-2z-4z=-12z=3      .....2 
Solving (1) and (2),we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).

Page No 28.22:

Question 16:

If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(−2, b −5) and C (4, 7, c), find the values of a, b, c.

Answer:


We have A(a, 1, 3), B(−2, b −5) and C (4, 7, c)
Now,
a-2+43=0, 1+b+73=0 and 3-5+c3=0a=-2, b=-8 and c=2

Page No 28.22:

Question 1:

The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
(a) 2 : 3
(b) 3 : 2
(c) –2 : 3
(d) 4 : –3

Answer:

(c) -2 : 3

Let A(2, 4, 5) and B(3, 5, -9)
Let the line joining A and B be divided by the yz-plane at point P in the ratio λ:1.
Then, we have,

P3λ+2λ+1, 5λ+4λ+1, -9λ+5λ+1

Since P lies on the yz-plane, the x-coordinate of P will be zero.

3λ+2λ+1=03λ+2=0λ=-23
Hence, the yz-plane divides AB in the ratio -2 : 3

Page No 28.22:

Question 2:

The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
(a) a : b
(b) b : c
(c) c : a
(d) c : b

Answer:

(d) c : b

Let A(a, b, c) and B(-a, -c, -b)
Let the line joining A and B be divided by the xy-plane at point P in the ratio λ:1.
Then, we have,

P-aλ+aλ+1, -cλ+bλ+1, -bλ+cλ+1

Since P lies on the xy-plane, the z-coordinate of P will be zero.

-bλ+cλ+1=0-bλ+c=0λ=cb
Hence, the xz-plane divides AB in the ratio c : b

Page No 28.22:

Question 3:

If P(0, 1, 2), Q(4, –2, 1) and O(0, 0, 0) are three points, then POQ =

(a) π6

(b) π4

(c) π3

(d) π2

Answer:

(d) π2

PQ2=4-02+-2-12+1-22=16+9+1=26OP2=0-02+1-02+2-02=0+1+4=5QO2=0-42+0+22+0-12=16+1+4=21Since, PQ2=OP2+QO2Hence, POQ=π2

Page No 28.22:

Question 4:

If the extremities of the diagonal of a square are (1, –2, 3 and (2, –3, 5), then the length of the side is

(a) 6

(b) 3

(c) 5

(d) 7

Answer:

(b) 3

Length of the diagonal = 2-12+-3+22+5-32=1+1+4=6

∴ Length of the side = Length of diagonal2=62=3



Page No 28.23:

Question 5:

The points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) are the vertices of
(a) a rectangle
(b) a square
(c) a parallelogram
(d) none of these

Answer:

(d) None of these

Suppose:
A(5, –4, 2)
B(4, –3, 1)
C(7, 6, 4)
D(8, –7, 5)

AB=4-52+-3+42+1-22      =-12+12+-12      =1+1+1=3BC=7-42+6+32+4-12      =32+92+32      =9+81+9=99=311CD=8-72+-7-62+5-42      =12+-132+12      =1+169+1=171DA=8-52+-7+42+5-22      =32+-32+32      =9+9+9=27=33

We see that none of the sides are equal. 

Page No 28.23:

Question 6:

In a three dimensional space the equation x2 – 5x + 6 = 0 represents
(a) points
(b) planes
(c) curves
(d) pair of straight lines

Answer:

(c) curves

Since, there is only one variable in the given equation. Also, it is quadratic equation.
Hence, It represents curves in yz plane.

Page No 28.23:

Question 7:

Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to
(a) 2
(b) 3
(c) 6
(d) 7

Answer:

(b) 3

Suppose d is the diameter of the sphere. Then

d2=-1-32+2-42+3+12d2=-42+-22+42d2=16+4+16d2=36 d=6

Hence, radius of the sphere is 3 units.

Page No 28.23:

Question 8:

XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
(a) 3 : 7
(b) 2 : 7
(c) –3 : 7
(d) –2 : 7

Answer:

(c) −3:7

Let A(2, 3, 1) and B(6, 7, 1)
Let the line joining A and B be divided by the xz-plane at point P in the ratio λ:1.
Then, we have,

P6λ+2λ+1, 7λ+3λ+1, λ+1λ+1

Since P lies on the xz-plane, the y-coordinate of P will be zero.

7λ+3λ+1=07λ+3=0λ=-37
Hence, the xz-plane divides AB in the ratio -3 : 7

Page No 28.23:

Question 9:

What is the locus of a point for which y = 0, z = 0?
(a) x - axis
(b) y - axis
(c) z - axis
(d) yz - plane

Answer:

We know that on x - axis both y = 0, z = 0.
Hence, the correct answer is option (a)

Page No 28.23:

Question 10:

The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz - plane are
(a) (3, 4, 0)
(b) (0, 4, 5)
(c) (3, 0, 5)
(d) (3, 0, 0)

Answer:

We know that the x - coordinate on yz - plane is 0.
The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz - plane are (0, 4, 5).
Hence, the correct answer is option (b).

Page No 28.23:

Question 11:

The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are

(a) (6, 0, 0)
(b) (0, 7, 0)
(c) (0, 0, 8)
(d) (0, 7, 8)

Answer:

We know that the y  and z coordinates on x - axis are 0
The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are (6, 0, 0)
Hence, the correct answer is option (a).

Page No 28.23:

Question 12:

The perpendicular distance of the point P (6, 7, 8) from xy - plane is

(a) 8
(b) 7
(c) 6
(d) 10

Answer:

The distance of the point P (6, 7, 8) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 8.
Hence, the correct answer is option (a).

Page No 28.23:

Question 13:

The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is
(a) 10
(b) 34
(c) 113
(d) 512

Answer:

The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is given by

32+52=34
Hence, the correct answer is option (b)

Page No 28.23:

Question 14:

The perpendicular distance of the point P(3, 3,4) from the x-axis is

(a) 32
(b) 5
(c) 3
(d) 4

Answer:

The perpendicular distance of the point P(3, 3,4) from the x-axis is given by

32+42=25=5
Hence, the correct answer is option (b)

Page No 28.23:

Question 15:

The length of the perpendicular drawn from the point P(a, b, c) from z-axis is

(a) a2+b2

(b) b2+c2

(c) a2+c2

(d) a2+b2+c2

Answer:

The length of the perpendicular drawn from the point P(x, y, z) from z-axis is given by y2+x2
Thus, the length of the perpendicular drawn from the point P(a, b, c) from z-axis is a2+b2
Hence, the correct answer is option (a)



Page No 28.6:

Question 1:

Name the octants in which the following points lie:
(i) (5, 2, 3)
(ii) (–5, 4, 3)
(iii) (4, –3, 5)
(iv) 7, 4, –3)
(v) (–5, –4, 7)
(vi) (–5, –3, –2)
(vii) (2, –5, –7)
(viii) (–7, 2 – 5)

Answer:

(i)  The x-coordinate, the y-coordinate and the z-coordinate of the point (5, 2, 3) are all positive.
      Therefore, this point lies in XOYZ octant.

(ii) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, 4, 3) are negative, positive and positive, respectively.
      Therefore, this point lies in X'OYZ octant.

(iii) The x-coordinate, the y-coordinate and the z-coordinate of the point (4, −3, 5) are positive, negative and positive, respectively.
      Therefore, this point lies in XOY'Z octant.

(iv) The x-coordinate, the y-coordinate and the z-coordinate of the point (7, 4, −3) are positive, positive and negative, respectively.
      Therefore, this point lies in XOYZ' octant.

(v) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, −4, 7) are negative, negative and positive, respectively.
      Therefore, this point lies in X'OY'Z octant .

(vi) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, −3, −2) are all negative.
      Therefore, this point lies in X'OY'Z' octant

(vii) The x-coordinate, the y-coordinate and the z-coordinate of the point (2, −5, −7) are positive, negative and negative, respectively.
      Therefore, this point lies in XOY'Z' octant.

(viii) The x-coordinate, the y-coordinate and the z-coordinate of the point(−7, 2, −5) are negative, positive and negative, respectively.
      Therefore, this point lies in X'OYZ' octant.

Page No 28.6:

Question 2:

Find the image  of:
(i) (–2, 3, 4) in the yz-plane.
(ii) (–5, 4, –3) in the xz-plane.
(iii) (5, 2, –7) in the xy-plane.
(iv) (–5, 0, 3) in the xz-plane.
(v) (–4, 0, 0) in the xy-plane.

Answer:

(i) 2, 3, 4
(ii) -5,-4,-3
(iii) 5, 2, 7
(iv) -5, 0, 3
(v) -4, 0, 0

Page No 28.6:

Question 3:

A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.

Answer:

Let P(1, 0, −1)
The length of each side of the cube is 5.
The three edges from vertex of the cube are drawn from P towards the negative x and y axes and the positive z-axis.
Therefore, the coordinates of the vertex of the cube will be as follows:
x-coordinate = 1, 1-5 = -4, i.e. 1, -4
y-coordinate = 0, 0-5 = -5, i.e. 0, -5
z-coordinate = -1, -1 + 5 = 4, i.e. -1, 4
Hence, the remaining seven vertices of the cube are as follows:
1, 0, 41, -5, -11, -5, 4-4, 0, -1-4, 0, 4-4, -5, -1-4, -5, 4

Page No 28.6:

Question 4:

 Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.

Answer:

 
Let P(3, 0, −1),  Q(−2, 5, 4)
PE = Distance between the parallel planes ABCP and FQDE
    = 4+1=5
                                       (These planes are perpendicular to the z-axis)
PA = Distance between the parallel planes ABQF and PCDE
     = -2-3=5
                                        (These planes are perpendicular to the x-axis)
Similarly, PC = 5-0=5
Thus, the length of the edges of the parallelepiped are 5, 5 and 5



Page No 28.7:

Question 5:

 Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.

Answer:


Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.

PA = Distance between planes PBEC and QDAF
           = 5 - 3
           = 2
PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and -2, respectively, from the zx-plane.
PB = 0 - (-2)
          = 2
PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.
PC = 2 - 5
           = -3

Page No 28.7:

Question 6:

Find the distances of the point P(–4, 3, 5) from the coordinate axes.

Answer:



Let PQ be the perpendicular to the xy-plane and QA be perpendicular from Q to the y-axis.
PA will be perpendicular to the x-axis. 
Also, QA = 3 and PQ = 5
Now, distance of P from the x-axis:
PB = BQ2+QP2
     =32+52=9+25=34
Similarly,
From the right-angled PAQ:
distance of P from the y-axis:
PA = AQ2+QP2
    =-42+52=16+25=41
Similarly, the length of the perpendicular from P to the z-axis =-42+32
                                                                                          =16+9=25=5

Page No 28.7:

Question 7:

The coordinates of a point are (3, –2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

Answer:

The seven coordinates are as follows:
-3, 2, 53, 2, -5-3, -2, 53, -2, -5-3, 2, -5-3, -2, -53, 2, 5



Page No 28.9:

Question 1:

Find the distance between the following pairs of points:
(i) P(1, –1, 0) and Q(2, 1, 2)
(ii) A(3, 2, –1) and B(–1, –1, –1).

Answer:

(i) PQ = x2-x12+y2-y12+z2-z12
         = 2-12+1+12+2-02
         = 12+22+22
         = 1+4+4
         = 9
         = 3

(ii) AB = x2-x12+y2-y12+z2-z12
           =-1-32+-1-22+-1+12=-42+-32+02=16+9+0=25=5

Page No 28.9:

Question 2:

Find the distance between the points P and Q having coordinates (–2, 3, 1) and (2, 1, 2).

Answer:

PQ = x2-x12+y2-y12+z2-z12         
=2+22+1-32+2-12=42+-22+12=16+4+1=21 units       

Page No 28.9:

Question 3:

Using distance formula prove that the following points are collinear:
(i) A(4, –3, –1), B(5, –7, 6) and C(3, 1, –8)
(ii) P(0, 7, –7), Q(1, 4, –5) and R(–1, 10, –9)
(iii) A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6)

Answer:

(i)  AB = 5-42+-7+32+6+12
           =12+-42+72=1+16+49=66

      BC = 3-52+1+72+-8-62
           = -22+82+-142
           =4+64+196=264=266

      AC = 3-42+1+32+-8+12
           =-12+42+-72=1+16+49=66

 Here , AB+AC=66+66                           =266                           =BC
Hence, the points are collinear.

(ii) PQ =  1-02+4-72+-5+72
           =12+-32+22=1+9+4=14

      QR = -1-12+10-42+-9+52
           =-22+62+-42=4+36+16=56=214

      PR=-1-02+10-72+-9+72
            =-12+32+-22=1+9+4=14

Here, PQ+PR=14+14                           =214                           =QR
Hence, the points are collinear.

(iii) AB = -1-32+0+52+8-12
            =-42+52+72=16+25+49=90=310

       BC = 7+12+-10-02+-6-82
            =82+-102+-142=64+100+196=360=610

      AC = 7-32+-10+52+-6-12
           =42+-52+-72=16+25+49=90=310

Here, AB+AC=310+310                          =610                          =BC
Hence, the points are collinear.

Page No 28.9:

Question 4:

Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

Answer:

(i) We know that the z-coordinate of every point on the xy-plane is zero.
 So, let P (x, y, 0) be a point on the xy-plane such that PA = PB = PC
Now, PA = PB
  PA2=PB2x-12+y+12+0-02 =x-22+y-12+0-22x2-2x+1+y2+2y+1=x2-4x+4+y2-2y+1+4-2x+4x+2y+2y+2-9=02x+4y-7=02x+4y=7                 ...(1) PB=PC PB2=PC2x-22+y-12+0-22=x-32+y-22+0+12x2-4x+4+y2-2y+1+4=x2-6x+9+y2-4y+4+1-4x+6x-2y+4y+9-14=02x+2y-5=0x+y=52x=52-y                           ...2Putting the value of x in equation 1: 252-y+4y=7 5-2y+4y=7 5+2y=7 2y=2  y=22y=1Putting the value of y in equation 2:x=52-1x=5-22x=32Hence, the required point is 32, 1, 0. 

(ii)  We know that the x-coordinate of every point on the yz-plane is zero.
So, let P (0, y, z) be a point on the yz-plane such that PA = PB = PC
Now, PA = PB
0-12+y+12+z-02=0-22+y-12+z-22
1+y2+2y+1+z2=4+y2-2y+1+z2-4z+42y+2=-2y-4z+92y+2y-4z=9-24y-4z=7y-z=74           ... 1PB=PCPB2=PC20-22+y-12+z-22=0-32+y-22+z+124+y2-2y+1+z2-4z+4=9+y2-4y+4+z2+2z+1-2y-4z+9=-4y+2z+14-2y+4y-4z-2z=14-92y-6z=5y-3z=52 y=52+3z                                  2Putting the value of y in equation 1:  y-z=7452+3z-z=742z=74-522z=7-1042z=-34 z=-38Putting the value of z in equation 2:  y=52+3zy=52+3-38y=52-98y=20-98 y=118
Hence, the required point is 0,118,-38.

(iii)  We know that the y-coordinate of every point on the zx-plane is zero.
So, let P (x, 0, z) be a point on the zx-plane such that PA = PB = PC
      Now, PA = PB
  x-12+0+12+z-02=x-22+0-12+x-22
x2+1-2x+1+z2=x2-4x+4+1+z2-4z+4-2x+2=-4x-4z+9-2x+4x-4z=72x-4z=7x-2z=72                          ...1     PB=PCPB2=PC2x-22+0-12+z-22=x-32+0-22+z+12x2-4x+4+1+z2-4z+4=x2-6x+9+4+z2+2z+1-4x-4z+9=-6x+2z+14-4x+6x-4z-2z=14-92x-6z=5x-3z=52 x=52+3z                              ...2Putting the value of x in equation 1:   x-2z=7252+3z-2z=7252+z=72z=72-52z=7-52z=22 z=1Putting the value of z in equation 2:      x=52+3zx=52+31x=52+3x=5+62 x=112
Hence, the required point is 112, 0, 1.
      

Page No 28.9:

Question 5:

Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).

Answer:

Let M be the point on the z-axis.
Then, the coordinates of M will be 0, 0, z.

Let M be equidistant from the points A 1, 5, 7 and B 5, 1, -4

AM = 0-12+0-52+z-72
      =-12+-52+z-72=1+25+z2-14z+49=z2-14z+75

BM = 0-52+0-12+z+42
      =-52+-12+z2+8z+16=25+1+z2+8z+16=z2+8z+42

Then, AM = BM

 z2-14z+75=z2+8z+42z2-14z+75=z2+8z+42-14z-8z=42-75-22z=-33z=3322z=32
Thus, the coordinates of M are 0, 0, 32.

Page No 28.9:

Question 6:

Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).

Answer:

Let the point on the y-axis be Y 0, y, 0 which is equidistant from the points P 3, 1, 2 and Q 5, 5, 2.
Then ,PY=QY
Now,  0-32+y-12+0-22=0-52+y-52+0-22

-32+y2+2y+1+-22=-52+y-52+-229+y2+2y+1+4=25+y2+10y+25+4y2+2y+14=y2+10y+54
y2+2y+14=y2+10y+54

2y-10y=54-14-8y=40-y=408y=-5

Thus, the required point on the y-axis is (0, −5, 0).

Page No 28.9:

Question 7:

Find the points on z-axis which are at a distance 21 from the point (1, 2, 3).

Answer:

Let the point be A (0, 0, z).Then,

AP = 21
0-12+0-22+z-32=21-12+-22+z-32=211+4+z-32=21z-32=21-5z-32=16z-3=±4z-3=4   or z-3=-4z=7 or z=-1
Hence, the coordinates of the required point are (0, 0, 7)  and (0, 0, −1).

Page No 28.9:

Question 8:

Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.

Answer:

Let A (1, 2, 3) , B (2, 3, 1) and C (3, 1, 2) are the coordinates of the triangle ABC
AB = 2-12+3-22+1-32
      =12+12+-22=1+1+4=6

 BC = 3-22+1-32+2-12
       =12+-22+12=1+4+1=6

 AC = 3-12+1-22+2-32
      =22+-12+-12=4+1+1=6
Now, AB = BC = AC

Therefore, it is an equilateral triangle.



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