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: \[ \alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 \] \[ = 16 + 1 + 9 = 26 \]

Answer 2

Step 1: Understand the given information.
We are given the vector:
\[ \vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \] and another vector:
\[ \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \] The components of \( \vec{a} \) along and perpendicular to \( \vec{b} \) are given as:
\[ \text{Along } \vec{b}: \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k}) \] \[ \text{Perpendicular to } \vec{b}: \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k}) \]

Step 2: Express the total vector.
The vector \( \vec{a} \) is the sum of its components along and perpendicular to \( \vec{b} \):
\[ \vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp} \] Substitute the given components:
\[ \vec{a} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k}) + \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k}) \]

Step 3: Simplify the expression for \( \vec{a} \).
Multiply and combine the components:
\[ \vec{a} = \frac{1}{11}\left[(16 \times 3 - 4)\hat{i} + (16 \times 1 - 5)\hat{j} + (-16 - 17)\hat{k}\right] \] \[ \vec{a} = \frac{1}{11}(44\hat{i} + 11\hat{j} - 33\hat{k}) \] \[ \vec{a} = 4\hat{i} + 1\hat{j} - 3\hat{k} \] Thus, \( \alpha = 4 \), \( \beta = 1 \), and \( \gamma = -3 \).

Step 4: Compute \( \alpha^2 + \beta^2 + \gamma^2 \).
\[ \alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26 \]

Final Answer:
\[ \boxed{26} \]