Name: Introduction to Section Formula in 3D
Uploaded: 2019-07-19
Duration: 10 min 4 s
Description: We've already learnt about section formula in 2D. Let's learn the section formula in 3D and find the coordinates of a point dividing a line segment in a spec...

Question

Show that the points

P (- 3, - 2, 4), Q (- 9, - 8, 10)

and

R (- 5, - 4, 6)

are collinear and

R

divides

P Q

in the ratio

1 : 2

.

Accepted Answer

The given points are $P (- 3, - 2, 4), Q (- 9, - 8, 10)$ and $R (- 5, - 4, 6)$ .

We know three points are collinear if the sum any two sides is equal to the third.

So we will find the $P Q$ , $Q R$ , $P R$ and compare to find out whether the given points are collinear or not.

We know the distance between two points $A & P$ is given by the formula,

$A P = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}}$

Substituting the corresponding values for $P$ and $Q$ points, we get

$P Q = \sqrt{(- 3 - (- 9))^{2} + (- 2 - (- 8))^{2} + (4 - 10)^{2}}$

$\Rightarrow P Q = \sqrt{(6)^{2} + (6)^{2} + (- 6)^{2}}$

$\Rightarrow P Q = \sqrt{3 \times 36}$

$\Rightarrow P Q = 6 \sqrt{3} \dots (i)$

Substituting the corresponding values for $Q$ and $R$ points, we get

$Q R = \sqrt{(- 9 - (- 5))^{2} + (- 8 - (- 4))^{2} + (10 - 6)^{2}}$

$\Rightarrow Q R = \sqrt{(- 4)^{2} + (- 4)^{2} + (4)^{2}}$

$\Rightarrow Q R = \sqrt{3 \times 16}$

$\Rightarrow Q R = 4 \sqrt{3} \dots (ii)$

Substituting the corresponding values for $P$ and $R$ , we get

$P R = \sqrt{(- 3 - (- 5))^{2} + (- 2 - (- 4))^{2} + (4 - 6)^{2}}$

$\Rightarrow P R = \sqrt{2^{2} + 2^{2} + (- 2)^{2}}$

$\Rightarrow P R = \sqrt{3 \times 4}$

$\Rightarrow P R = 2 \sqrt{3} \dots (iii)$

Now consider the condition that need to be satisfied for the points to be collinear,

$P Q = Q R + P R$

Substituting values from equation $(i), (ii) & (iii)$ , we get

$6 \sqrt{3} = 4 \sqrt{3} + 2 \sqrt{3}$

$\Rightarrow 6 \sqrt{3} = 6 \sqrt{3}$

So, the given points are satisfying the required condition.

Hence, the given points are collinear in nature.

Now we need to show that $R$ divides $P Q$ ,

i.e., $\frac{P R}{Q R} = \frac{2 \sqrt{3}}{4 \sqrt{3}}$

$\Rightarrow \frac{P R}{Q R} = \frac{1}{2}$

So, $R$ divides $P Q$ in the ratio $1 : 2$ .

Show that the points $P (- 3, - 2, 4), Q (- 9, - 8, 10)$ and $R (- 5, - 4, 6)$ are collinear and $R$ divides $P Q$ in the ratio $1 : 2$ .

Important Questions on Introduction to Three Dimensional Geometry

Learn and Prepare for any exam you want !

Show that the points P(-3,-2,4), Q(-9,-8,10) and R(-5,-4,6) are collinear and R divides PQ in the ratio 1:2.

Important Questions on Introduction to Three Dimensional Geometry

Learn and Prepare for any exam you want !

Show that the points $P (- 3, - 2, 4), Q (- 9, - 8, 10)$ and $R (- 5, - 4, 6)$ are collinear and $R$ divides $P Q$ in the ratio $1 : 2$ .