NCERT Solutions for Exercise 12.3 Class 11 Maths Chapter 12 - Introduction to Three Dimensional Geometry

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

Answer:

he line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB externally in the ratio 2:3.

Now, As we know by section formula , The coordinate of the point P which divides line segment $A(x_1,y_1,z_1)$ And $B(x_2,y_2,z_2)$ externally in ratio m:n is

$\left (\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n},\frac{mz_2-nz_1}{m-n} \right )$

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) externally in ratio 2:3 is

$\left (\frac{2(1)-3(-2)}{2-3},\frac{2(-4)-3(3)}{2-3},\frac{2(6)-3(5)}{2-3} \right )=\left ( -8,17,3\right )$

Hence required point is

. $( -8,17,3)$

Question:2 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:

Given Three points,

P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10)

Let point Q divides PR internally in the ratio $\lambda:1$

Now,

According to the section formula , The point Q in terms of P,Q and $\lambda$ is:

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

$\frac{9\lambda+3}{\lambda+1}=5$

${9\lambda+3}=5(\lambda+1)$

${9\lambda+3}=5\lambda+5$

$4\lambda =2$

$\lambda =\frac{1}{2}$

Hence, point Q divides PR in ratio 1:2.

Question:3 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Answer:

Given,

two points A(–2, 4, 7) and B(3, –5, 8)

Let Y-Z plane divides AB in $\lambda:1$

So, According to the section formula, the point which divides AB in $\lambda:1$ is

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

Since this point is in YZ plane, x coordinate of this point will be zero.

So,

$\frac{3\lambda-2}{\lambda+1}=0$

${3\lambda-2}=0$

$\lambda = \frac{2}{3}$

Hence YZ plane divides Line segment AB in a ratio 2:3.

Question:4 Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and $C \left ( 0 , 1/3 , 2 \right )$ are collinear.

Answer:

Given,

three points, A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2)

Let a point P divides Line segment AB in the ratio $\lambda:1$

SO, according to the section formula, the point P will be

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

Now, let's compare this point P with point C.

$\frac{-\lambda +2}{\lambda+1},=0$

$\lambda=2$

From here, we see that for $\lambda=2$, point C divides the line segment AB in ratio 2:1. Since point C divides the line segment AB, it lies in the line joining A and B and Hence they are colinear.

Question:5 Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).

Answer:

Given,

two points P (4, 2, – 6) and Q (10, –16, 6).

The point which trisects the line segment are the points which divide PQ in either 1:2 or 2:1

Let R (x,y,z) be the point which divides Line segment PR in ratio 1:2

Now, according to the section formula

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

Let S be the point which divides the Line segment PQ in ratio 2:1

So, The point S according to section formula is

$(x,y,z)=\left ( \frac{(2)10+(4)}{1+2},\frac{2(-16)+(2)}{1+2},\frac{(2)6-(6)}{1+2} \right )=(8,-10,2)$

Hence the points which trisect the line segment AB are (6,-4,-2) and (8,-10,2).