Let V = 2i + j - k and W = i + 3k. If U is a unit vector, then the maximum value of the scalar tripl

Q: Wait, why do we reject the negative root here? Can't the scalar triple product be negative?

That is a fantastic observation! The scalar triple product $[\vec{U} \vec{V} \vec{W}]$ can indeed be negative, depending on the orientation of the vectors. However, the question asks for the *maximum* value. Since we can choose the direction of the unit vector $\vec{U}$ freely, we can always align it to make the dot product positive. Thus, we focus on the magnitude, which represents the maximum possible volume.

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm for physics! However, this is a pure 3D geometry problem. While you could think of this in terms of projections, the most robust tool here is the vector algebra of the scalar triple product. It is the most direct path to the solution.

Q: Is the scalar triple product always equal to the volume?

Yes, geometrically, the absolute value of the scalar triple product is the volume of the parallelepiped formed by the three vectors. If the vectors are coplanar, the volume is zero, which is a great way to check if your vectors are linearly dependent!

Question

Let $\vec{V} = 2\hat{i} + \hat{j} - \hat{k}$ 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

Options:

(A) $-1$

(B) $\sqrt{10} + \sqrt{6}$

(C) $\sqrt{59}$

(D) $\sqrt{60}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have two fixed vectors, $\vec{V} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{W} = \hat{i} + 3\hat{k}$, anchored at the origin.

They define a plane, and together with a third, variable unit vector $\vec{U}$, they form a parallelepiped. Our mission is to find the maximum volume this shape can occupy.

### The Power of the Scalar Triple Product

The volume of a parallelepiped defined by vectors $\vec{U}$, $\vec{V}$, and $\vec{W}$ is given by the scalar triple product, denoted as $[\vec{U} \vec{V} \vec{W}]$. Algebraically, this is defined as $\vec{U} \cdot (\vec{V} 	imes \vec{W})$.

This formula is our key. It separates the fixed geometry of $\vec{V}$ and $\vec{W}$ from the variable direction of $\vec{U}$. We must first conquer the cross product $\vec{V} 	imes \vec{W}$.

Using the determinant method, we set up:

$$
\vec{V} 	imes \vec{W} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & 0 & 3 \end{vmatrix}
$$

Expanding this, we get $\hat{i}(1 \cdot 3 - (-1) \cdot 0) - \hat{j}(2 \cdot 3 - (-1) \cdot 1) + \hat{k}(2 \cdot 0 - 1 \cdot 1)$. This simplifies beautifully to $3\hat{i} - 7\hat{j} - \hat{k}$.

This resulting vector is perpendicular to the plane formed by $\vec{V}$ and $\vec{W}$.

### The Art of Maximization

Now, we have the expression $[\vec{U} \vec{V} \vec{W}] = \vec{U} \cdot (3\hat{i} - 7\hat{j} - \hat{k})$. Let $\vec{A} = 3\hat{i} - 7\hat{j} - \hat{k}$.

We want to maximize $\vec{U} \cdot \vec{A}$. We know that $\vec{U} \cdot \vec{A} = |\vec{U}| |\vec{A}| \cos 	heta$, where $	heta$ is the angle between $\vec{U}$ and $\vec{A}$.

Since $\vec{U}$ is a unit vector, $|\vec{U}| = 1$. The expression becomes $|\vec{A}| \cos 	heta$.

To maximize this, we simply need $\cos 	heta$ to be at its maximum, which is $1$. This occurs when $	heta = 0^\circ$, meaning $\vec{U}$ must be perfectly parallel to $\vec{A}$.

The maximum value is then just the magnitude of $\vec{A}$.

### The Final Triumph

We calculate the magnitude of our cross product vector:

$$
|\vec{A}| = |3\hat{i} - 7\hat{j} - \hat{k}| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}
$$

And there it is! The **maximum volume of the parallelepiped is $\sqrt{59}$**.

It is a testament to the elegance of vector algebra that such a complex geometric problem collapses into a simple magnitude calculation. You have navigated the space, aligned the vectors, and found the peak.

Animated Solution for Mathematics - Vector Algebra: Let $V = 2 \hat{i} + \hat{j} - \hat{k}$ and $W = \hat{i} + 3 \hat{k}$ . If $U$ is a unit vector, then the maximum value of the scalar triple product $[U V W]$ is

Select Answer:

Visualized Solution

Visualizing the Vectors $V$ and $W$

The Scalar Triple Product Formula

Setting up $V \times W$

Calculating $V \times W$

Maximizing the Dot Product

Applying the Constraints

Aligning Vector $U$

Final Calculation of Magnitude

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Power of the Scalar Triple Product

The Art of Maximization

The Final Triumph

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let V=2i^+j^​−k^ and W=i^+3k^. If U is a unit vector, then the maximum value of the scalar triple product [UVW] is