NCERT Solutions for Class 11 Maths Chapter 11 Introduction to 3D Geometry

Class 11 Math Ex 12.3 NCERT Solution

Q1. Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.

A.1. i.Let P(x, y, z) be the point which divides line segment joining (–2, 3, 5) and (1, –4, 6) internally in the ratio 2 : 3

Therefore,

x =  $\frac{2\left(1\right) + 3\left(-2\right)}{2 + 3}$ =  $\frac{2 - 6}{5}$ =  $\frac{-4}{5}$

y =  $\frac{2\left(-4\right) + 3\left(3\right)}{2 + 3}$ =  $\frac{-8 + 9}{5}$ =  $\frac{1}{5}$

z =  $\frac{2\left(6\right) + 3\left(5\right)}{2 + 3}$ =  $\frac{12 + 15}{5}$ =  $\frac{27}{5}$

Thus, the required points are  $\left(\frac{-4}{5},\frac{1}{5},\frac{27}{5}\right)$

ii.

Let P(x, y, z) be the point which divides line segment joining (–2, 3, 5) and (1, –4, 6) externally in the ratio 2 : 3

Therefore,

x =  $\frac{2\left(1\right) - 3\left(-2\right)}{2 - 3}$ =  $\frac{2 + 6}{-1}$ = –8

y =  $\frac{2\left(-4\right) - 3\left(3\right)}{2 - 3}$ =  $\frac{-8 - 9}{-1}$ = 17

z =  $\frac{2\left(6\right) - 3\left(5\right)}{2 - 3}$ =  $\frac{12 - 15}{-1}$ = 3

Thus, the required points are (–8, 17, 3).

A.2. Let point Q divides PR in the k : 1. Then co-ordinate of Q will be

$(\frac{k(9)+1(3)}{k+1},\frac{k(8)+1(2)}{k+1},\frac{k(-10)+1(-4)}{k+1})$

=> (5, 4, –6) =  $·\left(\frac{9k+3}{k+1}·,·\frac{8k+2}{k+1}·,·\frac{-10k-4}{k+1}·\right)$

Equating the co-ordinates we get,

$\frac{9k+3}{k + 1}$ = 5

=>  $9k+3=5\left(k+1\right)$

=>  $9k+3=5k+5$

=>  $9k-5k=5-3$

=>  $4k=2$

=>  $k=\frac{2}{4}$

=>  $k=\frac{1}{2}$

Putting  $k=\frac{1}{2}$ in y-coordinate and z-coordinate

$\frac{8k + 2}{k + 1}$

=  $\frac{\left(8 x\frac{1}{2}\right) + 2}{\frac{1}{2} + 1}$

= (4 + 2) ÷  $\frac{3}{2}$

= 6 x  $\frac{2}{3}$

= 4

And

$\frac{\left(-1\mathrm{0x}\frac{1}{2} \right) - 4}{\frac{1}{2} + 1}$

= (–5 – 4) ÷  $\frac{3}{2}$

= – 9 x  $\frac{2}{3}$

= – 6

Which is matching with the given co-ordinates of Q.

Hence, the ration in which Q divides PR is k : 1

=  $\frac{1}{2}$ : 1

= 1 : 2

A.3. Let YZ-plane divides the line segment joining A(–2, 4, 7) and B (3, –5, 8) at point P (x, y, z) in the ratio k : 1.

Then the co-ordinates of P are.

$\left(\frac{k(3)+1(-2)}{k+1},\frac{k(-5)+1(4)}{k+1},\frac{k(8)+1(7)}{k+1}\right)$

$\left(\frac{3k-2}{k+1},\frac{-5k+4}{k+1},\frac{8k+7}{k+1}\right)$

As P lies on YZ-plane its x-coordinate is zero.

i.e.  $\frac{3k-2}{k + 1}=0$

=>  $3k-2=0$

=>  $k=\frac{2}{3}$

Hence the YZ-plane divides AB internally in ratio.

$\frac{2}{3}$ : 1 = 2 : 3

A.4. Let P divides AB in ratio k : 1. Then co-ordinates of point P are

$(\frac{k(-1)+1(2)}{k+1},\frac{k(2)+1(-3)}{k+1},\frac{k(1)+1(4)}{k+1})$

=  $\left(\frac{-k+2}{k+1},\frac{2k-3}{k+1},\frac{k+4}{k+1}\right)$

Let us examine whether the value of k, the point P coincides with point C

Putting  $\frac{-k+2}{k + 1}=0$

=>  $-k+2=0$

=>  $k=2$

Put  $k=2$ in

$\frac{2k - 3}{k + 1}$

=  $\frac{2\left(2\right) - 3}{2 + 1}$

=  $\frac{4 - 3}{3}$

=  $\frac{1}{3}$

And put  $k=2$ in

$\frac{k + 4}{k + 1}$

=  $\frac{2 + 4}{2 + 1}$

=  $\frac{6}{3}$

= 2

Therefore, C  $\left(0,\frac{1}{3},2\right)$ is a point which divides AB internally in ratio 2 : 1 and is same as P. Hence A, B and C are collinear.

A.5. Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) trisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6).

Since A divides PQ internally in ratio 1 : 2. Then co-ordinates of A

=  $\left(\frac{1(10)+2(4)}{1+2},\frac{1(-16)+2(2)}{1+2},\frac{1(6)+2(-6)}{1+2}\right)$

=  $\left(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3}\right)$

=  $\left(\frac{18}{3},\frac{-12}{3},\frac{-6}{3}\right)$

= (6, –4, –2)

Similarly B divides PQ internally in ratio 2 : 1. Then co-ordinates of B

=  $\left(\frac{2(10)+1(4)}{2+1},\frac{2(-16)+1(2)}{2+1},\frac{2(6)+1(-6)}{2+1}\right)$

=  $\left(\frac{20+4}{3}·,·\frac{-32+2}{3}·,·\frac{12-6}{3}\right)$

=  $\left(\frac{24}{3},\frac{-30}{3},\frac{6}{3}\right)$

= (8, –10, 2)

Hence the points which trisects the line segment joining the points P(4, 2, –6) and Q(10, –16, 6) are (6, –4, –2) and (8, –1

A.1. Let D(x, y, z) be the fourth vertex of the parallelogram ABCD.

            In a parallelogram, the diagonal AC and BD bisects each other at point say O.

=>  $\left(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\right)=\left(\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2}\right)$

=> (1, 0, 2) =  $\left(\frac{1+x}{2},\frac{2+y}{2},\frac{-4+z}{2}\right)$

Equating the coordinates we get,

$\frac{1+x}{2}$ = 1

=>  $1+x=2$

=>  $x=1$

And

$\frac{2 + y}{2}$ = 0

=>  $y=-2$

And

$\frac{- 4 + z }{2}$ = 2

=>  $-4+z=4$

=>  $z=4+4$

=>  $z=8$

So, coordinates of fourth vertex is (1, –2, 8)

Answer

In a triangle ABC, the medians are the line segment that joins a vertex to the mid-point of the side that is opposite to that vertex. So, AE, BF and CG are the three medians.

 

Q3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

A.3. We know that, the centroid of a triangle with vertices (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is

$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$

So, $\left(\frac{2a+(-4)+8}{3},\frac{2+3b+14}{3},\frac{6+(-10)+2c}{3}\right)$  = (0, 0, 0)

Equating the coordinates we get,

$\frac{2a+\left(-4\right)+8}{3}$ = 0

=> $2a-4+8=0$

=> $2a+4=0$

=>  $2a=-4$

=>  $a=\frac{-4}{2}$

=>  $a=-2$

And,

$\frac{2+3b+14}{3}=0$

=>  $2+3b+14=0$

=>  $3b+16=0$

=>  $3b=-16$

=>  $b=\frac{-16}{3}$

And,

$\frac{6+\left(-10\right)+2c}{3}=0$

=>  $6-10+2c=0$

=>  $-4+2c=0$

=>  $2c=4$

=>  $c=\frac{4}{2}$

=>  $c=2$

Q4. Find the coordinates of a point on y-axis which are at a distance of   from the point P (3, –2, 5).

A.4.        Let Q be the point on y-axis which are at a distance  from point P. As Q is on y-axis it has the coordinates of form (0, y, 0).

=> ${\left(0-3\right)}^2+{\left(y+2\right)}^2+{\left(0-5\right)}^2$  = 25 x 2

=> 9 +  $y^2+2^2+2.y.2+25=50$

=> $y^2+4y+34+4-50=0$

=> $y^2+4y-12=0$

=> $y^2+6y-2y-12=0$

=> $y\left(y+6\right)-2\left(y+6\right)=0$

=> $\left(y+6\right)\left(y-2\right)=0$

=>  $y=-6, y=2$

So the coordinates Q are (0, 2, 0) and (0, –6, 0).

Q5. A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

[Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by  $\left(\frac{8k+2}{k+1},\frac{-3}{k+1},\frac{10k+4}{k+1}\right)].$

A.5. Given, x-coordinate of R = 4

Let R divides line segment joining points P(2, –3, 4) and Q(8, 0,10) internally in the ratio k : 1. Then coordinate of R is

$\left(\frac{k(8)+1(2)}{k+1},\frac{k(0)+1(-3)}{k+1},\frac{k(10)+1(4)}{k+1}\right)$

=  $\left(\frac{8k+2}{k+1},\frac{-3}{k+1},\frac{10k+4}{k+1}\right)$

Then,

$\frac{8k+2}{k + 1}$ = 4

=>  $8k+2=4\left(k+1\right)$

=>  $8k+2=4k+4$

=>  $8k-4k=4-2$

=>  $4k=2$

=>  $k=\frac{2}{4}$

=>  $k=\frac{1}{2}$

Hence,  $y=$  $\frac{-3}{k + 1}$

=  $\frac{-3}{\frac{1}{2} + 1}$

=  $-3÷\frac{1+2}{2}$

=  $-\mathrm{3\times } \frac{2}{3}$

=  $-2$

And,

z =  $\frac{10k + 4}{k + 1}$

=  $\frac{\left(10\frac{1}{2}\right) + 4}{\frac{1}{2} + 1}$