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JEE Mains · Maths · STD 12 - 10. vector algebra

माना \(\theta\) सदिश \(\overrightarrow{ a }\) तथा \(\overrightarrow{ b }\) के मध्य कोण है जहां \(|\overrightarrow{ a }|=4,|\overrightarrow{ b }|=3 \theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\) है। तब \(|(\overrightarrow{ a }-\overrightarrow{ b }) \times(\overrightarrow{ a }+\overrightarrow{ b })|^2+4(\overrightarrow{ a } \cdot \overrightarrow{ b })^2\) बराबर होगा

  1. A \(576\)
  2. B \(489\)
  3. C \(578\)
  4. D \(598\)
Verified Solution

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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