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Question
Find a vector whose length is 3 and which is perpendicular to the vector \[\vec{a} = 3 \hat{ i } + \hat{ j } - 4 \hat{ k } \text{ and } \vec{b} = 6 \hat{ i } + 5 \hat{ j } - 2 \hat{ k } .\]
Solution
\[\text{ Given } : \]
\[ \vec{a} = 3 \hat{ i } + \hat{ j } - 4 \hat{ k } \]
\[ \vec{b} = 6 \hat{ i } + 5 \hat{ j } - 2 \hat{ k } \]
\[ \therefore a^\to \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat { k } \\ 3 & 1 & - 4 \\ 6 & 5 & - 2\end{vmatrix}\]
\[ = \left( - 2 + 20 \right) \hat{ i } - \left( - 6 + 24 \right) \hat{ j } + \left( 15 - 6 \right) \hat{ k } \]
\[ = 18 \hat{ i } - 18 \hat{ j } + 9 \hat{ k } \]
\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{{18}^2 + \left( - {18}^2 \right) + 9^2}\]
\[ = \sqrt{729}\]
\[ = 27\]
\[\text{ Required vector } = 3 \times \left\{ \frac{\vec{a} \times \vec{b}}{\left| \vec{a} \times \vec{b} \right|} \right\}\]
\[ = 3 \times \frac{18 \hat{ i } - 18 \hat{ j } + 9 \hat{ k } }{27}\]
\[ = \frac{3\left( 2 \hat{ i } - 2 \hat{ j } + \hat{ k } \right)}{3}\]
\[ = 2 \hat{ i } - 2 \hat{ j } + \hat{ k } \]
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