Q50 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics

For $\lambda>0$, let $\theta$ 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 $\theta)^2$ is equal to

  • A
    50
  • B
    40
  • Correct answer
    C
    25
  • D
    20

Solution

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

Hence, the correct answer is option: 'C'.

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