Q79 / 155 · WBJEE 2025 · Mathematics

If $\vec{\alpha}=3 \vec{i}-\vec{k},|\vec{\beta}|=\sqrt{5}$ and $\vec{\alpha} \cdot \vec{\beta}=3$, then the area of the parallelogram for which $\vec{\alpha}$ and $\vec{\beta}$ are adjacent sides is

  • A
    $\sqrt{17}$
  • B
    $\sqrt{14}$
  • C
    $\sqrt{7}$
  • Correct answer
    D
    $\sqrt{41}$

Solution

$\begin{aligned}|\vec{\alpha} \times \vec{\beta}| & =|\vec{\alpha} \| \vec{\beta}| \sin \theta, \quad \cos \theta=\frac{3}{\sqrt{50}} \\ & =\sqrt{50} \times \frac{\sqrt{41}}{\sqrt{50}}=\sqrt{41}\end{aligned}$

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

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