Write the number of vectors of unit length perpendicular to both the vectors $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k} and \vec{b} = \hat{j} + \hat{k}$ .

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Solution

Unit vectors perpendicular to $\vec{a}$ and $\vec{b}$ are $\pm (\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|})$ .
$\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{matrix}| = - \hat{i} - 2 \hat{j} + 2 \hat{k}$

∴ Unit vectors perpendicular to $\vec{a}$ and $\vec{b}$ are $\pm \frac{- \hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt{{(- 1)}^{2} + {(- 2)}^{2} + {(2)}^{2}}} = \pm (- \frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k})$

Thus, there are two unit vectors perpendicular to the given vectors.

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4 \hat{i} - \hat{j} + 3 \hat{k} and - 2 \hat{i} + \hat{j} - 2 \hat{k} .

\vec{a} = 2 \hat{i} + \hat{j} + \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} .

\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k} and \vec{b} = \hat{j} + \hat{k} is

\to a = 2^i +^j + 2^k

\to b =^j +^k

\sqrt{2}

3 i - j - k

i + j - 2 k

2 i + 2 j + k

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