For a positive real number $\lambda$, if the vector $\hat{\mathrm{a}}=\lambda \hat{\mathrm{i}}-5…

For a positive real number $\lambda$, if the vector $\hat{\mathrm{a}}=\lambda \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ satisfies the equation $[\hat{\mathrm{i}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}})+\hat{\mathrm{j}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}})+\hat{\mathrm{k}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}})]^2=440$, then $\lambda=$
  1. 3
  2. 4
  3. 7
  4. 11

Solution

$\because[\hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})]^2=440$ $\begin{aligned} & \Rightarrow[(\hat{i} \hat{i}) \vec{a}-(\hat{i} \cdot \vec{a}) \hat{i}+(\hat{j} \cdot \hat{j}) \vec{a}-(\hat{j} \cdot \vec{a}) \hat{j}+(\hat{k} \cdot \hat{k}) \vec{a}-(\hat{k} \cdot \vec{a}) \hat{k}]^2=440 \\ & \Rightarrow[1 \cdot \vec{a}-\lambda \hat{i}+1 \cdot \vec{a}-(-5) \hat{j}+1 \cdot \vec{a}-(6) \hat{k}]=440 \\ & \Rightarrow[3 \vec{a}-\vec{a}]^2=440 \Rightarrow 4[\vec{a}]^2=440 \Rightarrow|\vec{a}|^2=110 \\ & \Rightarrow \lambda^2+25+36=110 \Rightarrow \lambda=7\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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