Question

Let $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{w} = \hat{i} + 3\hat{k}$. If $\vec{u}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \,\, \vec{v} \,\, \vec{w}]$ is

• A. 6
• B. 10
• C. 13
• D. $\sqrt{89}$

Hint:

The scalar triple product $[\vec{u}\,\,\vec{v}\,\,\vec{w}]$ geometrically represents the volume of a parallelepiped. For two fixed base vectors $\vec{v}$ and $\vec{w}$, the volume is maximized when the vector $\vec{u}$ stands perfectly perpendicular to the base plane, which means it points in the exact same direction as $\vec{v} \times \vec{w}$. Under these conditions, the maximum value simplifies simply to the magnitude $|\vec{v} \times \vec{w}|$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given two constant vectors $\vec{v}$ and $\vec{w}$, and an unspecified unit vector $\vec{u}$. We need to find the maximum possible scalar value that the scalar triple product $[\vec{u} \,\, \vec{v} \,\, \vec{w}]$ can achieve.

Step 2: Key Formula or Approach:
By definition, the scalar triple product can be written using a dot product and a cross product: $$[\vec{u} \,\, \vec{v} \,\, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$$ Using the geometric definition of the vector dot product: $$\vec{u} \cdot (\vec{v} \times \vec{w}) = |\vec{u}| \cdot |\vec{v} \times \vec{w}| \cdot \cos \theta$$ where $\theta$ is the angle between vector $\vec{u}$ and the vector cross product $(\vec{v} \times \vec{w})$. To maximize this value, $\cos \theta$ must take its maximum possible value of 1 (which occurs when $\vec{u}$ is aligned parallel to $\vec{v} \times \vec{w}$). Since $\vec{u}$ is a unit vector, $|\vec{u}| = 1$. Thus: $$\text{Maximum Value} = 1 \times |\vec{v} \times \vec{w}| \times 1 = |\vec{v} \times \vec{w}|$$

Step 3: Detailed Explanation:
1. First, calculate the cross product vector $\vec{v} \times \vec{w}$ using a determinant format: $$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 1 & 0 & 3 \end{vmatrix}$$ 2. Expand the determinant along the first row: $$\vec{v} \times \vec{w} = \hat{i} \left[(2)(3) - (-1)(0)\right] - \hat{j} \left[(2)(3) - (-1)(1)\right] + \hat{k} \left[(2)(0) - (2)(1)\right]$$ $$\vec{v} \times \vec{w} = \hat{i} [6 - 0] - \hat{j} [6 + 1] + \hat{k} [0 - 2]$$ $$\vec{v} \times \vec{w} = 6\hat{i} - 7\hat{j} - 2\hat{k}$$ 3. Now, calculate the absolute vector magnitude of this cross product: $$|\vec{v} \times \vec{w}| = \sqrt{(6)^2 + (-7)^2 + (-2)^2}$$ $$|\vec{v} \times \vec{w}| = \sqrt{36 + 49 + 4} = \sqrt{89}$$ Therefore, the maximum value of the scalar triple product is $\sqrt{89}$.

Step 4: Final Answer:
The maximum value of the scalar triple product is $\sqrt{89}$, which corresponds to option (D).