Question

The unit vector in the plane of the vectors $2\hat{i}+\hat{j}+\hat{k},\hat{i}-\hat{j}+\hat{k}$ and orthogonal to $5\hat{i}+2\hat{j}+6\hat{k}$ is

• A. $\frac{-3}{\sqrt{10}}\hat{j}-\frac{1}{\sqrt{10}}\hat{k}$
• B. $\frac{3}{\sqrt{10}}\hat{j}+\frac{1}{\sqrt{10}}\hat{k}$
• C. $\frac{-3}{\sqrt{10}}\hat{j}+\frac{1}{\sqrt{10}}\hat{k}$
• D. $\frac{3}{\sqrt{10}}\hat{j}-\frac{1}{\sqrt{10}}\hat{k}$

Answer 1

Answer: (C), (D)

$\frac{3}{\sqrt{10}}\hat{j}-\frac{1}{\sqrt{10}}\hat{k}$
$\frac{-3}{\sqrt{10}}\hat{j}+\frac{1}{\sqrt{10}}\hat{k}$

Let a unit vector in the plane of $2\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ be $\hat{a}=\alpha (2\times \hat{i}+\hat{j}+\hat{k})+\beta (\hat{i}-\hat{j}+\hat{k})$
$=(2\times \alpha +\beta )\hat{i}+(\alpha -\beta )\hat{j}+(\alpha +\beta )\hat{k}$
As $\hat{a}$ is unit vector,we have
$={(2\times \alpha +\beta )}^2+{(\alpha -\beta )}^2+{(\alpha +\beta )}^2=1$
$\Rightarrow 6\times {\alpha }^2+4\times \alpha \beta +3\times {\beta }^2=1$ on simplification
As $\hat{a}$ is orthogonal to $5\hat{i}+2\hat{j}+6\hat{k}$,
we get $5\times (2\times \alpha +\beta )+2\times (\alpha -\beta )+6\times (\alpha +\beta )=0$
$\Rightarrow 18\times \alpha +9\times \beta =0$
$\Rightarrow \beta =-2\times \alpha$
From $\left(i\right)$, we get $6\times {\alpha }^2-8\times {\alpha }^2+12\times {\alpha }^2=1$
$\Rightarrow \alpha =\pm \frac{1}{\sqrt{10}}$
$\Rightarrow \beta =\pm \frac{2}{\sqrt{10}}$
Thus,$\hat{a}=\pm (\frac{3}{\sqrt{10}}\hat{j}-\frac{1}{\sqrt{10}}\hat{k})$