Question

Let \(\vec{A} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} + \hat{j}\), where \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors. The projection of \(\vec{B}\) on \(\vec{A}\) is:

• A. \(\tfrac{1}{\sqrt{12}}\)
• B. \(\tfrac{1}{\sqrt{6}}\)
• C. \(\sqrt{6}\)
• D. \(\tfrac{1}{\sqrt{2}}\)

Hint:

Always remember: the scalar projection of \(\vec{B}\) on \(\vec{A}\) is \(\tfrac{\vec{A} \cdot \vec{B}}{|\vec{A}|}\). If asked for the vector projection, multiply this scalar by \(\tfrac{\vec{A}}{|\vec{A}|}\).

Answer 1

The Correct Option is B

Step 1: Recall projection formula.
The projection of \(\vec{B}\) on \(\vec{A}\) is: \[ \text{Proj}_{\vec{A}}(\vec{B}) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}. \] Step 2: Dot product.
\[ \vec{A} \cdot \vec{B} = (2)(1) + (-1)(1) + (1)(0) = 2 - 1 + 0 = 1. \] Step 3: Magnitude of \(\vec{A}\).
\[ |\vec{A}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}. \] Step 4: Projection value.
\[ \text{Projection} = \frac{1}{\sqrt{6}}. \] Final Answer: \[ \boxed{\tfrac{1}{\sqrt{6}}} \]