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, $m =x(2 \hat{ i }+\hat{ j })+y(\hat{ j }-\hat{ k })$
$m =2 x \hat{ i }+(x +y) \hat{ j }-y \hat{ k }$
$\therefore m$ is orthogonal to the vector
$\hat{ i }-\hat{ j }+\hat{ k }$, so $2 x-(x +y)-y=0$
$\Rightarrow x=2 y$ ...(i)
$\therefore |m |=1 $
$\Rightarrow 4 x^{2}+(x+ y)^{2}+y^{2}=1$
$\therefore 16 y^{2}+9 y^{2}+y^{2}=1$
$\Rightarrow 26 y^{2}=1$
$\Rightarrow y=\pm \frac{1}{\sqrt{26}} .$
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.