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

Let \(\overrightarrow{ a }=\alpha \hat{ i }+2 \hat{ j }-\hat{ k }\) and \(\overrightarrow{ b }=-2 \hat{ i }+\alpha \hat{ j }+\hat{ k }\), where \(\alpha \in R\). If the area of the parallelogram whose adjacent sides are represented by the vectors \(\vec{a}\) and \(\vec{b}\) is \(\sqrt{15\left(\alpha^{2}+4\right)}\), then the value of \(2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}\) is equal to

  1. A \(10\)
  2. B \(7\)
  3. C \(9\)
  4. D \(14\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(14\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{ a }=\alpha \hat{ i }+2 \hat{ j }-\hat{ k }, \overrightarrow{ b }=-2\) \(\hat{ i }+\alpha \hat{ j }+\hat{ k }\) area of parallelogram \(=|\hat{a} \times \hat{b}|\) \(|\hat{a} \times \hat{b}|=\sqrt{(\alpha+2)^{2}+(\alpha-2)^{2}+\left(\alpha^{2}+4\right)^{2}}\)…