Question
The unit vector perpendicular to both vectors $\hat{i}+\hat{k}$ and $\hat{i}-\hat{k}$ is:

Answer

Let the required vector be $x \hat{i}+y \hat{j}+z \hat{k}$.
Then, $x^2+y^2+z^2=1$ .............(i)
Also, $(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{k})=0$
$\Rightarrow \quad x+z=0 ........(ii)$
And $(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-\hat{k})=0$
$\Rightarrow x - z =0 .........(iii)$
Solving (ii) and (iii), we get $x=z=0$
$\therefore \quad$ From (i), $y^2=1 \Rightarrow y= \pm 1$
So, required vector is $\pm \hat{j}$.

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