Question

If the projection of \( \mathbf{a} = \alpha \hat{i} + \hat{j} + 4 \hat{k} \) on \( \mathbf{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \) is 4 units, then \( \alpha \) is:

• A. \( -13 \)
• B. \( -5 \)
• C. \( 13 \)
• D. \( 5 \)

Hint:

The projection of vector \( \mathbf{a} \) on \( \mathbf{b} \) can be calculated using the formula \( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} \).

Answer 1

The Correct Option is D

The projection of a vector \( \mathbf{a} \) on a vector \( \mathbf{b} \) is given by: \[ \text{Projection of } \mathbf{a} \text{ on } \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} \] We are given that the projection is 4 units, so: \[ \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} = 4 \] First, calculate \( \mathbf{a} \cdot \mathbf{b} \): \[ \mathbf{a} \cdot \mathbf{b} = \alpha \cdot 2 + 1 \cdot 6 + 4 \cdot 3 = 2\alpha + 6 + 12 = 2\alpha + 18 \] Next, calculate \( |\mathbf{b}| \): \[ |\mathbf{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \] Now substitute into the projection formula: \[ \frac{2\alpha + 18}{7} = 4 \] Solving for \( \alpha \): \[ 2\alpha + 18 = 28 \quad \Rightarrow \quad 2\alpha = 10 \quad \Rightarrow \quad \alpha = 5 \]