Vector Algebra
Scalar Triple Product and Vector Perpendicularity
Grade 12
Question:
<p>Let <strong>a</strong>, <strong>b</strong>, <strong>c</strong> be three non-coplanar vectors and <strong>d</strong> be a non-zero vector, which is perpendicular to <strong>a</strong> + <strong>b</strong> + <strong>c</strong>. If <strong>d</strong> = \(\sin x\)(\<strong>a</strong> × <strong>b</strong>) + \(\cos y\)(<strong>b</strong> × <strong>c</strong>) + 2(<strong>c</strong> × <strong>a</strong>), then the minimum value of \(x^2 + y^2\) is</p>
<p>(a) \(\pi^2\)</p>
<p>(b) \(\frac{\pi^2}{2}\)</p>
<p>(c) \(\frac{\pi^2}{4}\)</p>
<p>(d) \(\frac{5\pi^2}{4}\)</p>
Step-by-Step Solution
Key Concept: Use the perpendicularity condition to establish a constraint on sin x and cos y, then find the minimum value using the bounds on sine and cosine.
Step 1: Given \(\mathbf{d} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0\) Step 2: Computing dot products with a , b , c : \(\mathbf{a} \cdot \mathbf{d} = \cos y [\mathbf{a} \mathbf{b} \mathbf{c}]\) \(\mathbf{b} \cdot \mathbf{d} = 2[\mathbf{b} \mathbf{c} \mathbf{a}]\) \(\mathbf{c} \cdot \mathbf{d} = \sin x [\mathbf{a} \mathbf{b} \mathbf{c}]\) Step 3: Adding these equations and using \(\mathbf{d} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0\): \(\sin x + \cos y + 2 = 0\) \(\sin x + \cos y = -2\) Step 4: Since \(-1 \leq \sin x \leq 1\) and \(-1 \leq \cos y \leq 1\), we must have: \(\sin x = -1\) and \(\cos y = -1\) Step 5: This gives \(x = -\frac{\pi}{2}\) and \(y = \pi\) \(x^2 + y^2 = \frac{\pi^2}{4} + \pi^2 = \frac{5\pi^2}{4}\) ∴ Answer is (d).
Correct Answer: D