Question

$\overline{V}=2\overline{i}+\overline{j}-\overline{k}$ and $\overline{W}=\overline{i}+3\overline{k}$. If $\overline{U}$ is a unit vector, then the maximum value of the scalar triple product $[\overline{U}\overline{V}\overline{W}]$ is

• A. \(-1 \)
• B. \(\sqrt{10}+\sqrt{6} \)
• C. \(\sqrt{59} \)
• D. \(\sqrt{60} \)

Hint:

For any two vectors $\overline{V}$ and $\overline{W}$, the scalar triple product $[\overline{U}\overline{V}\overline{W}]$ with a unit vector $\overline{U}$ is maximized when $\overline{U}$ is collinear with the vector $\overline{V}\times\overline{W}$.

Answer 1

The Correct Option is C

Concept: The scalar triple product $[\overline{U}\overline{V}\overline{W}] = \overline{U}\cdot(\overline{V}\times\overline{W})$. Since $\overline{U}$ is a unit vector, the maximum value of $\overline{U}\cdot(\overline{V}\times\overline{W})$ is equal to the magnitude $|\overline{V}\times\overline{W}|$.

Step 1: Calculate the cross product \(\overline{V} \times \overline{W}\). \[ \overline{V} \times \overline{W} = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} \] \[ = \overline{i}(3 - 0) - \overline{j}(6 - (-1)) + \overline{k}(0 - 1) \] \[ = 3\overline{i} - 7\overline{j} - 1\overline{k} \]

Step 2: Evaluate the magnitude of the resulting vector \(\overline{V} \times \overline{W}\). \[ |\overline{V} \times \overline{W}| = \sqrt{(3)^2 + (-7)^2 + (-1)^2} \] \[ = \sqrt{9 + 49 + 1} = \sqrt{59} \]

Step 3: Determine the maximum value of the scalar triple product. Since $\overline{U}$ is a unit vector, the maximum value of $\overline{U} \cdot (\overline{V} \times \overline{W})$ is $|\overline{V} \times \overline{W}| \cdot |\overline{U}| \cos(0^{\circ}) = \sqrt{59} \cdot 1 = \sqrt{59}$. center minipage0.3

Maximum Value = $\sqrt{59}$ minipage center