# 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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