WBJEE · Maths · Vector Algebra
Let \(\vec{\alpha}, \vec{\beta}, \vec{\gamma}\) be the three unit vectors such that \(\vec{\alpha} \cdot \vec{\beta}=\vec{\alpha} \cdot \vec{\gamma}=0\) and the angle between \(\vec{\beta}\) and \(\vec{\gamma}\)
is \(30^{\circ} .\) Then \(\vec{\alpha}\) is
- A \(2(\vec{\beta} \times \vec{\gamma})\)
- B \(-2(\vec{\beta} \times \vec{y})\)
- C \(\pm 2(\vec{\beta} \times \vec{\gamma})\)
- D \((\vec{\beta} \times \vec{y})\)
Answer & Solution
Correct Answer
(C) \(\pm 2(\vec{\beta} \times \vec{\gamma})\)
Step-by-step Solution
Detailed explanation
\(\vec{\alpha}=\lambda(\vec{\beta} \times \vec{\gamma})=\lambda\left(|\vec{\beta}||\vec{\gamma}|\left|\sin 30^{\circ}\right|\right.\) \(\Rightarrow \quad|\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)\)…
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