Q72 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

If $\theta$ is the angle between the vectors $4 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+3 \hat{j}-2 \hat{k}$ then $\sin 2 \theta=$

  • A
    $\sqrt{\frac{3}{95}}$
  • B
    $-\sqrt{\frac{3}{95}}$
  • Correct answer
    C
    $-\frac{\sqrt{285}}{49}$
  • D
    $\frac{\sqrt{285}}{49}$

Solution

$\cos \theta=\frac{(4 \hat{i}-\hat{j}+2 \hat{k}) \cdot(\hat{i}+3 \hat{j}-2 \hat{k})}{\sqrt{21} \times \sqrt{14}}=\frac{-3}{7 \sqrt{6}}$ $\sin \theta=\sqrt{1-\left(\frac{-3}{7 \sqrt{6}}\right)^2}=\sqrt{1-\frac{9}{49 \times 6}}=\frac{\sqrt{285}}{7 \sqrt{6}}$ $\begin{aligned} & \cos \theta \lt 0 \Rightarrow \theta \in\left(\frac{\pi}{2}, \pi\right) \Rightarrow \sin \theta\gt0 \\ & \sin 2 \theta=2 \cos \theta \cdot \sin \theta \\ & =2 \times\left(\frac{-3}{7 \sqrt{6}}\right) \times \frac{\sqrt{285}}{7 \sqrt{6}}=\frac{-\sqrt{285}}{49}\end{aligned}$

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

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