Question

A vector \(\vec{a}\) makes equal acute angles on the coordinate axis.Then the projection of vector \(\vec{b}=5\hat{i}+7\hat{j}-\hat{k}\) on \(\vec{a}\) is

• A. \(\frac{11}{15}\)
• B. \(\frac{11}{\sqrt3}\)
• C. \(\frac{4}{5}\)
• D. \(\frac{3}{5\sqrt3}\)

Answer 1

The Correct Option is B

Since vector \( \vec{a} \) makes equal acute angles with the coordinate axes, we can conclude that its direction ratios are all equal. Thus, we assume that: \[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \] The projection of vector \( \vec{b} \) on \( \vec{a} \) is given by the formula: \[ \text{Projection of } \vec{b} \text{ on } \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \] First, we calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (1)(5) + (1)(7) + (1)(-1) = 5 + 7 - 1 = 11 \] Next, we calculate the magnitude of \( \vec{a} \): \[ |\vec{a}| = \sqrt{(1^2 + 1^2 + 1^2)} = \sqrt{3} \] Now, we can find the projection: \[ \text{Projection of } \vec{b} \text{ on } \vec{a} = \frac{11}{\sqrt{3}} \] Thus, the projection of \( \vec{b} \) on \( \vec{a} \) is \( \frac{11}{\sqrt{3}} \).

Answer 2

Given: 

Vector \( \vec{b} = 5\hat{i} + 7\hat{j} - \hat{k} \) 
Vector \( \vec{a} \) makes equal acute angles with the coordinate axes.

Step 1: Direction Ratios of \( \vec{a} \)

If \( \vec{a} \) makes equal acute angles with the coordinate axes, then let:
\[ \vec{a} = a(\hat{i} + \hat{j} + \hat{k}) \] Normalizing: \[ \vec{a} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]

Step 2: Projection of \( \vec{b} \) on \( \vec{a} \)

Projection formula: \[ \text{Projection} = \vec{b} \cdot \hat{a} \]

\[ \vec{b} \cdot \vec{a} = (5\hat{i} + 7\hat{j} - \hat{k}) \cdot \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \] \[ = \frac{1}{\sqrt{3}}(5 + 7 - 1) = \frac{11}{\sqrt{3}} \]

Final Answer: \( \frac{11}{\sqrt{3}} \)