Question

A unit vector in the plane of the vectors $2i+j+k,i-j+k$ and orthogonal to $5i+2j+6k$ is

• A. $\frac{6i-5k}{\sqrt{61}}$
• B. $\frac{3j-k}{\sqrt{10}}$
• C. $\frac{2i-5j}{\sqrt{29}}$
• D. $\frac{2i+j-2k}{3}$

Answer 1

Answer: (B)

$\frac{3j-k}{\sqrt{10}}$

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