First slide

Planes in 3D

Question

The distance between the line $\vec{r} = 2 \hat{i} - 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - \hat{j} + 4 \hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5 \hat{j} + \hat{k}) = 5$ is

Moderate

A

$\frac{10}{3 \sqrt{3}}$
B

$\frac{10}{9}$
C

$\frac{10}{3}$
D

$\frac{3}{10}$

Solution

It is obvious that the given line and plane are parallel.
Given point on the line is A (2, - 2,3). B (0,0, 5) is a point on the plane.
Therefore, $\vec{A B} = (2 - 0) \hat{i} + (- 2 - 0) \hat{j} + (3 - 5) \hat{k}$
Then distance of ,B from the plane - projection of
$\vec{AB} on vector \hat{i} + 5 \hat{j} + \hat{k}$
$\begin{aligned} p = |\frac{∣ (2 \hat{i} - 2 \hat{j} - 2 \hat{k}) \cdot (\hat{i} + 5 \hat{j} + \hat{k})}{\sqrt{1 + 25 + 1}}| \\ = |\frac{2 - 10 - 2}{\sqrt{27}}| = \frac{10}{3 \sqrt{3}} \end{aligned}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App