JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\theta\) be the angle between the vectors \(\vec{a}\) and \(\vec{b}\), where \(|\vec{a}|=4,|\vec{b}|=3 \quad \theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\). Then \(|(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})|^{2}+4(\vec{a} \cdot \vec{b})^{2}\) is equal to
- A \(576\)
- B \(489\)
- C \(578\)
- D \(598\)
Answer & Solution
Correct Answer
(A) \(576\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}|=4,|\vec{b}|=3 \quad \theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\) \(|(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})|^{2}+4(\vec{a} \cdot \vec{b})^{2}\) \(|\vec{a} \times \vec{b}-\vec{b} \times \vec{a}|^{2}+4 a^{2} b^{2} \cos ^{2} \theta\)…
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