Question

Let \( \mathbf{a} = 3\hat{i} - \hat{j} + 2\hat{k} \), \( \mathbf{b} = \mathbf{a} \times (\hat{i} - 2\hat{k}) \) and \( \mathbf{c} = \mathbf{b} \times \hat{k} \). Then the projection of \( \mathbf{c} - 2\hat{j} \) on \( \mathbf{a} \) is:

• A. \( 2\sqrt{14} \)
• B. \( 2\sqrt{7} \)
• C. \( 3\sqrt{7} \)
• D. \( \sqrt{14} \)

Hint:

To find the projection of a vector on another, use the formula \( \text{Proj}_{\mathbf{a}} \mathbf{v} = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{a}|} \) and remember to compute cross products when needed.

Answer 1

The Correct Option is A

To find the projection of \( \mathbf{c} - 2\hat{j} \) on \( \mathbf{a} \), first compute the vectors \( \mathbf{b} \) and \( \mathbf{c} \) using the given cross products. Then, use the projection formula: \[ \text{Proj}_{\mathbf{a}} \mathbf{v} = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{a}|}. \] Substitute \( \mathbf{c} - 2\hat{j} \) and \( \mathbf{a} \) into the formula. 
Final Answer: \( 2\sqrt{14} \).

Answer 2

Step 1: Define the vectors.
\[ \mathbf{a} = 3\hat{i} - \hat{j} + 2\hat{k} \] \[ \mathbf{u} = \hat{i} - 2\hat{k} \] Given,
\[ \mathbf{b} = \mathbf{a} \times \mathbf{u} \] \[ \mathbf{c} = \mathbf{b} \times \hat{k} \]

Step 2: Calculate \(\mathbf{b} = \mathbf{a} \times \mathbf{u}\).
\[ \mathbf{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix} = \hat{i}\begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix} - \hat{j}\begin{vmatrix} 3 & 2 \\ 1 & -2 \end{vmatrix} + \hat{k}\begin{vmatrix} 3 & -1 \\ 1 & 0 \end{vmatrix} \] Calculate minors:
\[ \hat{i}((-1)(-2) - (0)(2)) - \hat{j}(3 \times -2 - 1 \times 2) + \hat{k}(3 \times 0 - 1 \times -1) \] \[ = \hat{i}(2) - \hat{j}(-6 - 2) + \hat{k}(0 + 1) = 2\hat{i} + 8\hat{j} + \hat{k} \]
So, \[ \mathbf{b} = 2\hat{i} + 8\hat{j} + \hat{k} \]

Step 3: Calculate \(\mathbf{c} = \mathbf{b} \times \hat{k}\).
\[ \mathbf{b} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{vmatrix} = \hat{i}\begin{vmatrix} 8 & 1 \\ 0 & 1 \end{vmatrix} - \hat{j}\begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} + \hat{k}\begin{vmatrix} 2 & 8 \\ 0 & 0 \end{vmatrix} \] Calculate minors:
\[ \hat{i}(8 \times 1 - 0) - \hat{j}(2 \times 1 - 0) + \hat{k}(2 \times 0 - 0) = 8\hat{i} - 2\hat{j} + 0\hat{k} \] \[ \mathbf{c} = 8\hat{i} - 2\hat{j} \]

Step 4: Calculate \(\mathbf{c} - 2\hat{j}\).
\[ \mathbf{c} - 2\hat{j} = 8\hat{i} - 2\hat{j} - 2\hat{j} = 8\hat{i} - 4\hat{j} \]

Step 5: Calculate the projection of \(\mathbf{c} - 2\hat{j}\) on \(\mathbf{a}\).
Projection formula:
\[ \text{proj}_{\mathbf{a}} (\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{a}\|} \] Compute dot product:
\[ (8\hat{i} - 4\hat{j}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) = 8 \times 3 + (-4) \times (-1) + 0 = 24 + 4 = 28 \] Compute magnitude of \(\mathbf{a}\):
\[ \|\mathbf{a}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 +1 +4} = \sqrt{14} \] Therefore, projection length:
\[ \frac{28}{\sqrt{14}} = 2 \sqrt{14} \]

Final answer:
\[ \boxed{2 \sqrt{14}} \]