The magnitude of the projection of the vector $2 \hat{i}+\hat{j}+\widehat{k}$, on the vector perpendicular…
The magnitude of the projection of the vector $2 \hat{i}+\hat{j}+\widehat{k}$, on the vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2 \hat{j}+3 \hat{k}$, is
$\frac{5}{\sqrt{6}}$ units
$\frac{1}{\sqrt{6}}$ units
$\sqrt{6}$ units
$\frac{2}{\sqrt{6}}$ units
Solution
The perpendicular vector is $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right|=\hat{i}-2 \hat{j}+\hat{k}$
the required projection
$=\frac{(2 \hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k})}{|\hat{i}-2 \hat{j}+\hat{k}|}=\frac{2-2+1}{\sqrt{1^2+(-2)^2+1^2}}=\frac{1}{\sqrt{6}}$