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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