The distance of the point $(2i+j-k)$ from the plane ${r} \cdot(i-2j+4k)=9$ is

Question

The distance of the point $(2 \mathbf{i}+\mathbf{j}-\mathbf{k})$ from the plane $\mathbf{r} \cdot(\mathbf{i}-2 \mathbf{j}+4 \mathbf{k})=9$ is

(a) $\frac{13}{\sqrt{21}}$

(b) $\frac{3}{\sqrt{21}}$

(c) $\frac{13}{21}$

(d) $\frac{13}{3 \sqrt{21}}$

Answer 1

We know, that the perpendicular distance of a point $P$ with position vector a from the plane $\mathbf{r} \cdot \mathbf{n}=d$ is given by $\frac{|\mathbf{a} \cdot \mathbf{n}-d|}{|\mathbf{n}|}$

Here, $\quad \mathbf{a}=2 \mathbf{i}+\mathbf{j}-\mathbf{k}, \mathbf{n}=\mathbf{i}-2 \mathbf{j}+4 \mathbf{k}$ and $d=9$

$\begin{aligned}\therefore \quad \text { Distance } & =\frac{|(2 \mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-2 \mathbf{j}+4 \mathbf{k})-9|}{\sqrt{1+4+16}} \\& =\frac{|2-2-4-9|}{\sqrt{21}}=\frac{13}{\sqrt{21}}\end{aligned}$

So, The correct option of this question will be (A).