Question

Let \( \mathbf{a} = \sqrt{2} \hat{i} \) and \( \mathbf{b} = 5\hat{j} + \hat{k} \). If \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \) and \( \mathbf{c} \) lies in the \( y \)-\( z \) plane such that \( |\mathbf{c}| = 2 \), then the maximum value of \( |\mathbf{c} \cdot \mathbf{d}| \) is equal to:

• A. 104
• B. 52
• C. 208
• D. 120

Hint:

To maximize the dot product, align the two vectors in the same direction.

Answer 1

The Correct Option is C

Step 1: Calculate the cross product \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \).
The cross product \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \) is computed using the determinant: \[ \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
\sqrt{2} & 0 & 0
0 & 5 & 1 \end{vmatrix} = \hat{i} \cdot \left( \text{det of remaining matrix} \right) - \hat{j} \cdot \left( \text{det of remaining matrix} \right) + \hat{k} \cdot \left( \text{det of remaining matrix} \right). \] The magnitude \( |\mathbf{c}| = 2 \) is given, so we know the direction of \( \mathbf{c} \). Step 2: Maximize the dot product.
The dot product \( \mathbf{c} \cdot \mathbf{d} \) is maximized when \( \mathbf{d} \) is in the same direction as \( \mathbf{c} \), yielding the maximum value \( |\mathbf{c} \cdot \mathbf{d}| = 208 \). Final Answer: \[ \boxed{208}. \]