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Question
If \[\vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} , \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k} \text{ and } \vec{c} = 3 \hat{i} + \hat{j}\] \[\vec{a} + \lambda \vec{b}\] is perpendicular to \[\vec{c}\] then find the value of λ.
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Solution
\[\text{ We have }\]
\[ \vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \vec{b} = - \hat{i} +2 \hat{j} + \hat{k} \]
\[\text{and}\]
\[ \vec{c} = 3 \hat{i} + \hat{j} \]
\[ \therefore \vec{a} + \lambda \vec{b} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda \left( - \hat{i} + 2 \hat{j} + \hat{k} \right) = \left( 2 - \lambda \right) \hat{i} + \left( 2 + 2\lambda \right) \hat{j} + \left( 3 + \lambda \right) \hat{k} \]
\[\text{ Given that } \vec{a} + \lambda \vec{b} \text{ is perpendicular to } \vec{c} .\]
\[ \Rightarrow \left( \vec{a} + \lambda \vec{b} \right) . \vec{c} = 0\]
\[ \Rightarrow \left[ \left( 2 - \lambda \right)\hat{i} + \left( 2 + 2\lambda \right) \hat{j} + \left( 3 + \lambda \right) \hat{k} \right] . \left( 3 \hat{i} + \hat{j} + 0 \hat{k} \right) = 0\]
\[ \Rightarrow 3 \left( 2 - \lambda \right) + 1 \left( 2 + 2\lambda \right) + 0 = 0\]
\[ \Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0\]
\[ \therefore \lambda = 8\]
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