Find the vector of magnitude 5 which are perpendicular to the vectors $\vec{a}$ = $2\hat{i} + \hat{j} – 3\hat{k}$ and $\vec{b}$ = $\hat{i} – 2\hat{j} + \hat{k}$

Solution :

Unit vectors perpendicular to $\vec{a}$ & $\vec{b}$ = $\pm$$\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|$

$\therefore$  $\vec{a}\times\vec{b}$ = $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}$ = $-5\hat{i} – 5\hat{j} – 5\hat{k}$

$\therefore$ Unit Vectors = $\pm$ $-5\hat{i} – 5\hat{j} – 5\hat{k}\over 5\sqrt{3}$

Hence the required vectors are $\pm$ $5\sqrt{3}\over 3$($\hat{i} + \hat{j} + \hat{k}$)

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