Q87 / 200 · Manipal (MET) 2013 · Mathematics

A unit vectors coplanar with $\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{i}+2 \mathbf{j}+\mathbf{k}$ and perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$ is

  • Correct answer
    A
    $\frac{j-k}{\sqrt{2}}$
  • B
    $\frac{i+j+k}{\sqrt{3}}$
  • C
    $\frac{i+j+2 k}{\sqrt{6}}$
  • D
    $\frac{-j+2 k}{\sqrt{5}}$

Solution

$\begin{aligned} \text { Let } \mathbf{a} & =t(\mathbf{i}+\mathbf{j}+2 \mathbf{k})+s(\mathbf{i}+2 \mathbf{j}+\mathbf{k}) \\ & =(t+3 s) \mathbf{i}+(t+2 s) \mathbf{j}+(2 t+s) \mathbf{k}\end{aligned}$ given, $\mathbf{a} \cdot(\mathbf{i}+\mathbf{j}+\mathbf{k})=0$ $\therefore t+s+t+2 s+2 t+s=0$ $\begin{array}{rrr}\Rightarrow & 4(t+s)=0 \\ \Rightarrow & t+s=0\end{array}$ $\therefore \quad \mathbf{a}=t(\mathbf{1}+\mathbf{j}+2 \mathbf{k})-t(\mathbf{i}+2 \mathbf{j}+\mathbf{k})$ $=t(-1+\mathbf{k})$ $\begin{aligned} & \text { But }|\mathbf{a}|=1 \\ & \therefore \quad 1=2 t^2\end{aligned}$ $\Rightarrow \quad t= \pm \frac{1}{\sqrt{2}}$ $\therefore \quad a= \pm\left(\frac{k-j}{\sqrt{2}}\right)$

Hence, the correct answer is option: 'A'.

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