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

Using section formula, show that the following three points are collinear:

(- 2, 3, 5), (1, 2, 3), (7, 0, - 1)

.

Accepted Answer

Given points are $P (- 2, 3, 5), Q (1, 2, 3)$ and $R (7, 0, - 1)$ .

Let $Q$ divides $P R$ internally in the ratio $k : 1$ , then

Applying the section formula, we get

$Q = (\frac{m (x_{2}) + n (x_{1})}{m + n}, \frac{m (y_{2}) + n (y_{1})}{m + n}, \frac{m (z_{2}) + n (z_{1})}{m + n})$

$\Rightarrow Q = (\frac{k (7) + 1 (- 2)}{k + 1}, \frac{k (0) + 1 (3)}{k + 1}, \frac{k (- 1) + 1 (5)}{k + 1})$

$\Rightarrow Q = (\frac{7 k - 2}{k + 1}, \frac{3}{k + 1}, \frac{- k + 5}{k + 1})$

Equating this with the values of $Q (1, 2, 3)$ , we get

$\frac{7 k - 2}{k + 1} = 1$

$\Rightarrow 7 k - 2 = k + 1$

$\Rightarrow 6 k = 3$

$\Rightarrow k = \frac{1}{2}$

For this value of $k$

$y = \frac{3}{k + 1}$

$\Rightarrow y = \frac{3}{\frac{1}{2} + 1}$

$\Rightarrow y = \frac{3}{\frac{3}{2}} = 2$

$z = \frac{- k + 5}{k + 1}$

$\Rightarrow z = \frac{- \frac{1}{2} + 5}{\frac{1}{2} + 1}$

$\Rightarrow z = \frac{\frac{9}{2}}{\frac{3}{2}} = 3$

Since $k > 0$ and $k = \frac{1}{2}$ , therefore $Q$ divides $P R$ internally in the ratio $1 : \frac{1}{2}$ or $2 : 1$ .

And the given points are collinear in nature.

Using section formula, show that the following three points are collinear: $(- 2, 3, 5), (1, 2, 3), (7, 0, - 1)$ .

Important Questions on Introduction to Three Dimensional Geometry

Learn and Prepare for any exam you want !

Using section formula, show that the following three points are collinear: (-2,3,5), (1,2,3), (7,0,-1).

Important Questions on Introduction to Three Dimensional Geometry

Learn and Prepare for any exam you want !

Using section formula, show that the following three points are collinear: $(- 2, 3, 5), (1, 2, 3), (7, 0, - 1)$ .