Question

If the components of \( \vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) along and perpendicular to \( \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \) respectively are \( \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) \) and \( \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}) \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

• A. 18
• B. 26
• C. 23
• D. 16

Hint:

For vector components, sum the parallel and perpendicular components to find the resultant vector.

Answer 1

The Correct Option is B

Step 1: Given Components of \( \vec{a} \) 

We are given the components of \( \vec{a} \) along and perpendicular to \( \vec{b} \): \[ \vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp} \] Where: \[ \vec{a}_{\parallel} = \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) \quad \text{and} \quad \vec{a}_{\perp} = \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}) \]

Step 2: Combine the Components

Combining the vectors: \[ \vec{a} = 4\hat{i} + \hat{j} - 3\hat{k} \]

Step 3: Compute \( \alpha^2 + \beta^2 + \gamma^2 \)

Calculating the sum of squares of the components: \[ \alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 \] \[ = 16 + 1 + 9 = 26 \]