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Question
If \[\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}\] \[\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\] \[\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}\] find λ such that \[\vec{a}\] is perpendicular to \[\lambda \vec{b} + \vec{c}\]
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Solution
The given vectors are \[\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}\] \[\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\] and \[\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}\]
Now,
\[\lambda \vec{b} + \vec{c} = \lambda\left( \hat{i} + \hat{j} - 2 \hat{k} \right) + \left( \hat{i} + 3 \hat{j} - \hat{k} \right) = \left( \lambda + 1 \right) \hat{i} + \left( \lambda + 3 \right) \hat{j} - \left( 2\lambda + 1 \right) \hat{k}\] It is given that
\[\vec{a} \perp \left( \lambda \vec{b} + \vec{c} \right)\]
\[ \Rightarrow \vec{a} . \left( \lambda \vec{b} + \vec{c} \right) = 0\]
\[ \Rightarrow \left( 2 \hat{i} - \hat{j} + \hat{k} \right) . \left[ \left( \lambda + 1 \right) \hat{i} + \left( \lambda + 3 \right) \hat{j} - \left( 2\lambda + 1 \right) \hat{k} \right] = 0\]
\[ \Rightarrow 2\left( \lambda + 1 \right) - \left( \lambda + 3 \right) - \left( 2\lambda + 1 \right) = 0\]
\[ \Rightarrow 2\lambda + 2 - \lambda - 3 - 2\lambda - 1 = 0\]
\[ \Rightarrow \lambda = - 2\]
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