The value of 'a' so that the volume of parallelopiped formed by i+aj+k, j+ak and ai+k becomes minimu

Q: Wait, why do we use the determinant to find the volume?

Great question! The scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ geometrically represents the volume of the parallelepiped formed by the vectors. Since the cross product $\vec{v} \times \vec{w}$ gives a vector whose magnitude is the area of the base and whose direction is perpendicular to the base, the dot product with $\vec{u}$ effectively calculates the base area multiplied by the height, which is exactly the volume. The determinant is just the most efficient algebraic way to compute this product.

Q: Why do we need the second derivative test here?

When we set the first derivative $\frac{dV}{da} = 0$, we find the 'stationary points' where the slope of the volume function is zero. This could be a local minimum, a local maximum, or even a point of inflection. The second derivative test tells us the concavity of the function at those points. If the second derivative is positive, the function is concave up, meaning we have found a local minimum. It is a crucial step to ensure we haven't accidentally found the maximum!

Q: Is the volume always positive?

Yes, volume is a physical quantity and must be non-negative. The scalar triple product can result in a negative value depending on the orientation (or 'handedness') of the vectors. That is why the formal definition of volume is the *absolute value* of the scalar triple product, $V = |[\vec{u}, \vec{v}, \vec{w}]|$. In this specific problem, we are analyzing the function $V(a) = a^3 - a + 1$ to find its local minimum, which is a valid approach for optimization.

Question

The value of 'a' so that the volume of parallelopiped formed by $\hat{i}+a\hat{j}+\hat{k}$, $\hat{j}+a\hat{k}$ and $a\hat{i}+\hat{k}$ becomes minimum is

Options:

(A) $-3$

(B) $3$

(C) $1/\sqrt{3}$

(D) $\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system. You have three vectors, $\vec{u} = \hat{i} + a\hat{j} + \hat{k}$, $\vec{v} = \hat{j} + a\hat{k}$, and $\vec{w} = a\hat{i} + \hat{k}$, stretching out from the origin like the edges of a crystal.

These vectors define a parallelepiped—a slanted, three-dimensional box. As you vary the parameter $a$, you are essentially pulling and pushing on these edges, causing the box to stretch, shrink, and warp.

Our mission is to find the precise value of $a$ that compresses this box into its smallest possible volume.

### The Engine of Volume: The Scalar Triple Product

To solve this, we need the right tool. In vector algebra, the volume $V$ of a parallelepiped formed by coterminous edges $\vec{u}$, $\vec{v}$, and $\vec{w}$ is given by the absolute value of their scalar triple product: $V = |[\vec{u}, \vec{v}, \vec{w}]|$.

This product is elegantly computed as the determinant of a $3 	imes 3$ matrix where the vectors form the rows. Let's set up our matrix:

$$V = \begin{vmatrix} 1 & a & 1 \ 0 & 1 & a \ a & 0 & 1 \end{vmatrix}$$

This matrix is the DNA of our parallelepiped. Every entry tells us how the shape is oriented in space. If we make a mistake here, the entire geometry collapses.

Let's expand this determinant along the first row with precision:

$$V = 1 \cdot \begin{vmatrix} 1 & a \ 0 & 1 \end{vmatrix} - a \cdot \begin{vmatrix} 0 & a \ a & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \ a & 0 \end{vmatrix}$$

### The Calculus of Optimization

Now, let's simplify. The first minor is $(1 \cdot 1 - a \cdot 0) = 1$. The second minor is $(0 \cdot 1 - a \cdot a) = -a^2$. The third minor is $(0 \cdot 0 - 1 \cdot a) = -a$.

Putting it all together, we get the volume function: $V(a) = 1(1) - a(-a^2) + 1(-a) = 1 + a^3 - a$. This cubic polynomial, $V(a) = a^3 - a + 1$, is our map.

To find the minimum, we turn to the power of calculus. We need to find where the rate of change of volume with respect to $a$ is zero:

$$\frac{dV}{da} = \frac{d}{da}(a^3 - a + 1) = 3a^2 - 1$$

Setting $\frac{dV}{da} = 0$, we get $3a^2 - 1 = 0$, which leads us to $a^2 = \frac{1}{3}$, or $a = \pm \frac{1}{\sqrt{3}}$.

### The Final Verdict

We have two candidates, but which one is the minimum? We use the second derivative test: $\frac{d^2V}{da^2} = 6a$.

At $a = \frac{1}{\sqrt{3}}$, the second derivative is $6(\frac{1}{\sqrt{3}}) = 2\sqrt{3}$, which is positive, confirming a local minimum. At $a = -\frac{1}{\sqrt{3}}$, the second derivative is negative, indicating a local maximum.

Thus, the volume is minimized when **$a = \frac{1}{\sqrt{3}}$**. You have successfully navigated the geometry, the algebra, and the calculus to find the answer.

Animated Solution for Mathematics - Vector Algebra: The value of 'a' so that the volume of parallelopiped formed by $\hat{i} + a \hat{j} + \hat{k}$ , $\hat{j} + a \hat{k}$ and $a \hat{i} + \hat{k}$ becomes minimum is

Select Answer:

Visualized Solution

Defining the Vectors

The Volume Formula

Setting up the Determinant

Expanding the Determinant

Simplifying the Volume Function

Condition for Extrema

Differentiating the Function

Solving for Critical Points

Second Derivative Test

Final Answer and Summary

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Engine of Volume

The Calculus of Optimization

The Final Verdict

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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; } The value of 'a' so that the volume of parallelopiped formed by i^+aj^​+k^, j^​+ak^ and ai^+k^ becomes minimum is