Question

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

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

Hint:

When calculating projections, always start with the dot product and the magnitudes of the vectors involved. Ensure you understand the direction and magnitude of the resultant projection.

Answer 1

The Correct Option is B

The projection of vector \( \vec{B} \) on vector \( \vec{A} \) is given by the formula: \[ {Proj}_{\vec{A}} \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|} \hat{A} \] where \( \hat{A} \) is the unit vector in the direction of \( \vec{A} \), and \( \vec{A} \cdot \vec{B} \) is the dot product of \( \vec{A} \) and \( \vec{B} \).
Let’s first compute the dot product \( \vec{A} \cdot \vec{B} \). We have: \[ \vec{A} = 2\ell - \mathbf{j} + \mathbf{k}, \quad \vec{B} = \ell + \mathbf{f} \] Since \( \ell, \mathbf{f}, \) and \( \mathbf{k} \) are unit vectors, the dot product is: \[ \vec{A} \cdot \vec{B} = (2\ell) \cdot \ell + (-\mathbf{j}) \cdot \mathbf{f} + \mathbf{k} \cdot \mathbf{f} \] Using the fact that the unit vectors are orthogonal (i.e., \( \ell \cdot \mathbf{f} = 0 \), \( \mathbf{j} \cdot \mathbf{k} = 0 \), etc.), we get: \[ \vec{A} \cdot \vec{B} = 2 \times 1 + 0 + 0 = 2 \] Now, we compute the magnitude of \( \vec{A} \): \[ |\vec{A}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \] Thus, the projection of \( \vec{B} \) on \( \vec{A} \) is: \[ {Proj}_{\vec{A}} \vec{B} = \frac{2}{\sqrt{6}} \hat{A} \] Therefore, the magnitude of the projection is: \[ \frac{2}{\sqrt{6}} \] which simplifies to \( \frac{1}{\sqrt{6}} \) when considering the unit vector in the direction of \( \vec{A} \).
Thus, the correct option is \( \frac{1}{\sqrt{6}} \), which corresponds to option (B).