For $\lambda0$, let $ heta$ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of (14 cos $ heta)^2$ is equal to

Question

For $\lambda>0$, let $	heta$ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of (14 cos $	heta)^2$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0, \lambda>0 \ & |\vec{a}|^2-|\vec{b}|^2=0 ightarrow 1+\lambda^2+9=9+1+4 \ & 	herefore \lambda=2, \cos 	heta=\frac{\vec{a}-\vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{3-\lambda-6}{\sqrt{14} \cdot \sqrt{14}} \ & 14 \cos 	heta=3-8=-5 \ & 	herefore(14 \cos 	heta)^2=25\end{aligned}$

For $\lambda>0$, let $\theta$ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$…

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