Question

Let $m$ be the unit vector orthogonal to the vector $\hat{i}-\hat{j}+\hat{k}$ and coplanar with the vectors $2\hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. If $a=\hat{i}-\hat{k}$, then the length of the perpendicular from the origin to the plane $r\cdot m=a\cdot m$ is

• A. $\frac{1}{\sqrt{26}}$
• B. $\frac{1}{\sqrt{5}}$
• C. $\frac{5}{\sqrt{26}}$
• D. 1

Answer 1

Answer: (C)

The vector $m$ is coplane with $(2\hat{i}+\hat{j})$ and $(\hat{j}-\hat{k})$
So, $m=x(2\hat{i}+\hat{j})+y(\hat{j}-\hat{k})$
$m=2x\hat{i}+(x+y)\hat{j}-y\hat{k}$
$\therefore m$ is orthogonal to the vector
$\hat{i}-\hat{j}+\hat{k}$, so $2x-(x+y)-y=0$
$\Rightarrow x=2y$ ...(i)
$\therefore |m|=1$
$\Rightarrow 4x^2+(x+y{)}^2+y^2=1$
$\therefore 16y^2+9y^2+y^2=1$
$\Rightarrow 26y^2=1$
$\Rightarrow y=\pm \frac{1}{\sqrt{26}}.$
So, $x=\pm \frac{2}{\sqrt{26}}$
$\therefore m=\pm \left(\frac{4}{\sqrt{26}}\hat{i}+\frac{3}{\sqrt{26}}\hat{j}-\frac{1}{\sqrt{26}}\hat{k}\right)$
The length of the perpendicular from the origin to the plane $r\cdot m=a\cdot m$ is
So, $|a.m|=\left|(\hat{i}-\hat{k})\cdot \left[\frac{4}{\sqrt{26}}\hat{i}+\frac{3}{\sqrt{26}}\hat{j}-\frac{1}{\sqrt{26}}\hat{k}\right]\right|$
$=\frac{4}{\sqrt{26}}+\frac{1}{\sqrt{26}}=\frac{5}{\sqrt{26}}$ units.