MHT CET · Maths · Vector Algebra
If \(\bar{a}, \bar{b}, \bar{c}, \bar{d}\) are unit vectors such that \(\bar{a} \cdot \bar{b}=\frac{1}{2}, \bar{c} \cdot \bar{d}=\frac{1}{2}\) and the angle between \(\bar{a} \times \bar{b}\) and \(\bar{c} \times \overline{\mathrm{d}}\) is \(\frac{\pi}{6}\), then the value of \(|[\bar{a} \overline{\mathrm{~b}} \mathrm{~d}] \bar{c}-[\bar{a} \overline{\mathrm{~b}} \bar{c}] \mathrm{d}|=\)
- A \(\frac{3}{2}\)
- B \(\frac{3}{4}\)
- C \(\frac{3}{8}\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{8}\)
Step-by-step Solution
Detailed explanation
\(|[\bar{a} \overline{\mathrm{~b}} \mathrm{~d}] \bar{c}-[\bar{a} \overline{\mathrm{~b}} \bar{c}] \mathrm{d}| = |(\bar{a} \times \bar{b}) \times (\bar{c} \times \bar{d})|\) \(|\bar{a} \times \bar{b}| = |\bar{a}||\bar{b}|\sqrt{1-(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|})^2} = 1 \cdot 1 \cdot \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}\)
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