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Question 31

Mathematics Scalar and Vector Products Hard

Let&nbsp;<span class="math-tex">\(\vec a, \vec b\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\vec c\)</span>&nbsp;be three vector having magnitudes 1, 1 and 2 respectively, If&nbsp;<span class="math-tex">\(\vec a \times (\vec a \times \vec c) - \vec b = 0\)</span>&nbsp;then the acute angle between&nbsp;<span class="math-tex">\(\vec a\)</span>&nbsp;and&nbsp;<span class="math-tex">\(\vec c\)</span>&nbsp;is

(A) &pi; / 4
(B) &pi; / 6
(C) &pi; / 3
(D) None
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\[ \mathbf{a} \times (\mathbf{a} \times \mathbf{c}) + \mathbf{b} = 0 \] \[ \Rightarrow (\mathbf{a} \cdot \mathbf{c}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{a}) \mathbf{c} = -\mathbf{b} \] \[ \Rightarrow \mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{a} = \mathbf{b} \quad (\because \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 1) \] \[ \Rightarrow |\mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{a}|^2 = |\mathbf{b}|^2 \] \[ \Rightarrow |\mathbf{c}|^2 + |\mathbf{a} \cdot \mathbf{c}|^2 |\mathbf{a}|^2 - 2 (\mathbf{a} \cdot \mathbf{c}) (\mathbf{a} \cdot \mathbf{c}) = |\mathbf{b}|^2 \] If \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{c}\), we get: \[ 4 + (2 \cos \theta)^2 - 2 (2 \cos \theta)^2 = 1 \] \[ \Rightarrow 4 \cos^2 \theta = 3 \Rightarrow \cos \theta = \frac{\sqrt{3}}{2} \quad (\because \theta \text{ is acute}) \] \[ \Rightarrow \theta = \frac{\pi}{6} \]

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