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

Step 1: Understand what we are maximising.
The scalar triple product is $[\vec U\,\vec V\,\vec W]=\vec U\cdot(\vec V\times\vec W)$. Since $\vec U$ is a unit vector, its dot with any fixed vector is largest when $\vec U$ points the same way.
Step 2: State the key bound.
The maximum of $\vec U\cdot(\vec V\times\vec W)$ over all unit $\vec U$ is simply $|\vec V\times\vec W|$, achieved when $\vec U$ is parallel to $\vec V\times\vec W$.
Step 3: Compute the cross product.
\[ \vec V\times\vec W=\begin{vmatrix} \vec i & \vec j & \vec k\\ 2 & 1 & -1\\ 1 & 0 & 3\end{vmatrix}. \]
Step 4: Expand each component.
$\vec i$ part $=1\cdot3-(-1)\cdot0=3$; $\vec j$ part $=-(2\cdot3-(-1)\cdot1)=-7$; $\vec k$ part $=2\cdot0-1\cdot1=-1$. So $\vec V\times\vec W=3\vec i-7\vec j-\vec k$.
Step 5: Take its magnitude.
\[ |\vec V\times\vec W|=\sqrt{3^2+(-7)^2+(-1)^2}=\sqrt{9+49+1}=\sqrt{59}. \]
Step 6: Conclude.
The maximum value of the triple product is therefore $\sqrt{59}$, which is option 3.
\[ \boxed{[\vec U\,\vec V\,\vec W]_{\max}=\sqrt{59}} \]