JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(|\vec{a}|=2,|\vec{b}|=3\) and the angle between the vectors \(\vec{a}\) and \(\vec{b}\) be \(\frac{\pi}{4}\). Then \(|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2\) is equal to
- A \(482\)
- B \(441\)
- C \(841\)
- D \(882\)
Answer & Solution
Correct Answer
(D) \(882\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}|=2,|\vec{b}|=3\) \(|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2\) \(|-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2\) \(|-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2\) \(|-7 \vec{a} \times \vec{b}|^2\)…
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