First slide

Planes in 3D

Question

Given $\vec{α} = 3 \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{β} = \hat{i} - 2 \hat{j} - 4 \hat{k}$ are the position vectors of the points A and B. Then the distance of the point $- \hat{i} + \hat{j} + \hat{k}$ form the plane passing through B and perpendicular to AB is

Moderate

A

5
B

10
C

15
D

20

Solution

$\vec{A B} = \vec{β} - \vec{α} = - 2 \hat{i} - 3 \hat{j} - 6 \hat{k}$
Equation of the plane passing through B and perpendicular to AB is
$\begin{array}{l} (\vec{r} - \vec{OB}) \cdot \vec{AB} = 0 \\ \vec{r} \cdot (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) + 28 = 0 \end{array}$
Hence, the required distance from
$\begin{array}{l} \vec{r} = - \hat{i} + \hat{j} + \hat{k} \\ = |\frac{(- \hat{i} + \hat{j} + \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) + 28}{| 2 \hat{i} + 3 \hat{j} + 6 \hat{k} |}| \\ = |\frac{- 2 + 3 + 6 + 28}{7}| \\ = 5 units \end{array}$

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