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Question 12

Mathematics Scalar and Vector Products Hard

If&nbsp;<span class="math-tex">\(\vec a\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\vec b\)</span>&nbsp;are vectors in space, given by&nbsp;<span class="math-tex">\(\vec a = \frac {\hat i - 2\hat j}{\sqrt 5}\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\vec b = \frac {2\hat i + \hat j + 3\hat k}{\sqrt {14}}\)</span>&nbsp;then the value of&nbsp;<span class="math-tex">\(\left(2\vec a + \vec b\right)\left[\left(\vec a \times \vec b\right)\times \left(\vec a - 2\vec b\right)\right]\)</span>&nbsp;is

(A) 3
(B) 4
(C) 5
(D) 6
✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\(|\vec{a}| = 1\), \(|\vec{b}| = 1\), and \(\vec{a} \cdot \vec{b} = \frac{2 - 2 + 0}{\sqrt{70}} = 0\).

Use \((\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}\):

  • \((\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{b} \cdot \vec{a})\vec{a} = \vec{b}\)
  • \((\vec{a} \times \vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a} = -\vec{a}\)

So \((\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b}) = \vec{b} + 2\vec{a}\).

\((2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b}) = 4|\vec{a}|^2 + |\vec{b}|^2 + 4\vec{a} \cdot \vec{b} = 4 + 1 + 0 = 5\)

Answer: (C) 5

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