JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let and be two vectors such that and . Then is equal to .
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Correct answer: 36
- We use the identity relating dot product and cross product:
|\vec a ^2 |\vec b ^2 = (\vec a \cdot \vec b)^2 + |\vec a \times \vec b ^2
Equivalently,
(\vec a \cdot \vec b)^2 = |\vec a ^2 |\vec b ^2 - |\vec a \times \vec b ^2
- Substitute the given values:
|\vec a| = \sqrt{14} \Rightarrow |\vec a|^2 = 14 |\vec b| = \sqrt{6} \Rightarrow |\vec b|^2 = 6 |\vec a \times \vec b| = \sqrt{48} \Rightarrow |\vec a \times \vec b|^2 = 48
So,
- Compute:
Hence,
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