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Let \(\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k},\ \vec{b}=3 \hat{i}-5 \hat{j}+\hat{k}\) and \(\vec{c}\) be a vector such that \(\vec{a} \times \vec{c}=\vec{c} \times \vec{b}\) and \((\vec{a}+\vec{c}) \cdot(\vec{b}+\vec{c})=168.\) Then the maximum value of \(|\vec{c}|^{2}\) is:

(1) 77

(2) 462

(3) 308

(4) 154

1 Answer

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Best answer

Correct option is (3) 308  

\(\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}\)

\(\vec{\mathrm{b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)

\(\vec{\mathrm{a}} \times \vec{\mathrm{c}}=\vec{\mathrm{c}} \times \vec{\mathrm{b}}\)

\(\vec{\mathrm{a}} \times \vec{\mathrm{c}}+\vec{\mathrm{b}} \times \vec{\mathrm{c}}=0\)

\((\vec{a}+\vec{b}) \times \vec{c}=0\)

\(\Rightarrow \vec{\mathrm{c}}=\lambda(\vec{\mathrm{a}}+\vec{\mathrm{b}})\)

\(\vec{\mathrm{c}}=\lambda(5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\)    ....(1)

\(|\vec{\mathrm{c}}|^{2}=\lambda^{2}(25+36+16)\)

\(|\vec{\mathrm{c}}|^{2}=77 \lambda^{2}\)

\((\vec{a}+\vec{c}) \cdot(\vec{b}+\vec{c})=168\)

\(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{c} \cdot \vec{b}+|\vec{c}|^{2}=168\)

\(14+\vec{c} \cdot(\vec{a}+\vec{b})+77 \lambda^{2}=168\)

using equation (1)

\(\lambda|5 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|^{2}+77 \lambda^{2}=154\)

\(77 \lambda+77 \lambda^{2}-154=0\)

\(\lambda^{2}+\lambda-2=0\)

\(\lambda=-2,1\)

\(\therefore\) Maximum value of \(|\overrightarrow{\mathrm{c}}|^{2}\) occurs when \(\lambda=-2\)

\(|\vec{\mathrm{c}}|^{2}=77 \lambda^{2}\)

\(=77 \times 4\)

= 308

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