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MHT CET · Maths · Vector Algebra

If \(\theta\) is angle between the vectors \(\bar{a}\) and \(\bar{b}\) where \(|\bar{a}|=4, \quad|\bar{b}|=3 \quad\) and \(\theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\), then \(|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2\) has the value

  1. A 576
  2. B 24
  3. C 144
  4. D 12
Verified Solution

Answer & Solution

Correct Answer

(A) 576

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & |(\overline{\mathrm{a}}-\overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+\overline{\mathrm{b}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2 \\ & =|(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-(\overline{\mathrm{b}} \times \overline{\mathrm{a}})-(\overline{\mathrm{b}} \times \overline{\mathrm{b}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2 \\ & =|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-(\overline{\mathrm{b}} \times \overline{\mathrm{a}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2 \\ & =|2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2 \ldots[(\overline{\mathrm{a}} \times \overline{\mathrm{b}})=-(\overline{\mathrm{b}} \times \overline{\mathrm{a}})] \\ & =4|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2 \\ & =4\left[|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2+(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2\right] \\ & =4|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2 \\ & =4(4)^2(3)^2 \\ & =576\end{aligned}\)