Question

Let \[ \vec{a} = (\sin^2\alpha)\hat{i} + (\cos 2\alpha)\hat{j} + (\cos^2\alpha)\hat{k}, \qquad 0\leq\alpha\leq\frac{\pi}{2}, \] and \[ \vec{b} = \hat{i} - 2\hat{j} + \hat{k}. \] If \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other, then the value of \(\alpha\) is equal to:

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{2}\)
• E. 0

Hint:

Recognizing that \(\sin^2 \alpha + \cos^2 \alpha = 1\) immediately simplifies the dot product equation. Always look for fundamental identities in trigonometric vector problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Two vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular if and only if their dot product is zero (\(\vec{a} \cdot \vec{b} = 0\)).

Step 2: Key Formula or Approach:
1. \(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z = 0\).
2. Trigonometric identity: \(\cos 2\alpha = 1 - 2\sin^2 \alpha\).
3. \(\sin^2 \alpha + \cos^2 \alpha = 1\).

Step 3: Detailed Explanation:
Calculate the dot product: \[ (\sin^2 \alpha)(1) + (\cos 2\alpha)(-2) + (\cos^2 \alpha)(1) = 0 \]
Combine the squared terms: \[ (\sin^2 \alpha + \cos^2 \alpha) - 2\cos 2\alpha = 0 \]
Substitute the fundamental identity \(\sin^2 \alpha + \cos^2 \alpha = 1\): \[ 1 - 2\cos 2\alpha = 0 \]
\[ \cos 2\alpha = \frac{1}{2} \]
For the range \(0 \leq \alpha \leq \pi/2\), the double angle \(2\alpha\) ranges from \(0\) to \(\pi\). \[ 2\alpha = \frac{\pi}{3} \implies \alpha = \frac{\pi}{6} \]

Step 4: Final Answer:
The value of \(\alpha\) is \(\pi/6\).