Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>Let \(\vec{r} = (\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + (\vec{c} \times \vec{a})\), where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are non-zero non-coplanar vectors. If \(\vec{r}\) is orthogonal to \(3\vec{a} + 5\vec{b} + 2\vec{c}\), then the value of \(\sec^2 y + \cosec^2 x + \sec y \cosec x\) is</p>
<p>(a) 3</p>
<p>(b) 4</p>
<p>(c) 5</p>
<p>(d) 6</p>
Step-by-Step Solution
Key Concept: Use orthogonality condition with scalar triple products to establish constraints on trigonometric functions, then apply range limitations.
Step 1: Given \(\vec{r} \cdot (3\vec{a} + 5\vec{b} + 2\vec{c}) = 0\) Step 2: Expanding using scalar triple product properties: \[\vec{a} \cdot (\vec{b} \times \vec{c})[2\sin x + 3\cos y + 5] = 0\] Step 3: Since \(\vec{a} \cdot (\vec{b} \times \vec{c}) \neq 0\): \[2\sin x + 3\cos y + 5 = 0\] \[2\sin x + 3\cos y = -5\] Step 4: Since \(-1 \leq \sin x \leq 1\) and \(-1 \leq \cos y \leq 1\), the only solution is: \[\sin x = -1, \cos y = -1\] Step 5: Therefore: \(\cosec x = -1\), \(\sec y = -1\) Step 6: \(\sec^2 y + \cosec^2 x + \sec y \cosec x = 1 + 1 + (-1)(-1) = 1 + 1 + 1 = 3\) ∴ Answer is (a) 3
Correct Answer: A