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Question
Let \[\vec{a} = \hat{ i } + 4 \hat{ j } + 2 \hat{ k } , \vec{b} = 3 \hat{ i }- 2 \hat{ j } + 7 \hat{ k } \text{ and } \vec{c} = 2 \hat{ i } - \hat{ j } + 4 \hat{ k } .\] Find a vector \[\vec{d}\] which is perpendicular to both \[\vec{a} \text{ and } \vec{d}\] \[\text{ and } \vec{c} \cdot \vec{d} = 15 .\]
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Solution
\[\text{ Given } : \]
\[ \vec{a} = \hat{ i } + 4 \hat{ j } + 2 \hat{ k } \]
\[ \vec{b} = 3 \hat{ i } - 2 \hat{ j } + 7 \hat{ k } \]
\[ \vec{c} = 2 \hat{ i } - \hat{ j } + 4 \hat{ k } \]
\[\text{ Since d is perpendicular to both a and b, it is parallel to } \vec{a} \times \vec{b} . \]
\[ \text{ Suppose } d = \lambda\left( \vec{a} \times \vec{b} \right) \text{ for some scalar } \lambda . \]
\[d = \lambda \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 4 & 2 \\ 3 & - 2 & 7\end{vmatrix}\]
\[ = \lambda \left[ \left( 28 + 4 \right) \hat{ i } - \left( 7 - 6 \right) \hat{ j } + \left( - 2 - 12 \right) \hat{ k } \right]\]
\[ = \lambda \left[ 32 \hat{ i } - \hat{ j } - 14 \hat{ k } \right]\]
\[ \vec{c .} \vec{d} = 15 (\text{ Given } )\]
\[ \Rightarrow \left( 2 \hat{ i } - \hat{ j } + 4 \hat{ k } \right) . \lambda \left( 32 \hat{ i }- \hat{ j } - 14 \hat{ k } \right) = 15\]
\[ \Rightarrow \lambda\left( 64 + 1 - 56 \right) = 15\]
\[ \Rightarrow \lambda = \frac{5}{3}\]
\[ \therefore \vec{d} = \frac{5}{3}\left( 32 \hat{ i } - \hat{ j } - 14 \hat{ k } \right)\]
\[ \Rightarrow \vec{d} = \frac{1}{3}\left( 160 \hat{ i } - 5 \hat{ j } - 70 \hat{ k } \right)\]
Notes
The question should contain \["\text{ which is perpendicular to both } \vec{a} \text{ and } \vec{b} "\]
\[\text{ instead of } \]
\["\text { which is perpendicular to both } \vec{a} \text{ and } \vec{d} "\]
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