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WBJEE · Maths · Vector Algebra

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

  1. A \(\sqrt{17}\)
  2. B \(\sqrt{14}\)
  3. C \(\sqrt{7}\)
  4. D \(\sqrt{41}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt{41}\)

Step-by-step Solution

Detailed explanation

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