If the volume of parallelopiped formed by the vectors i + λj + k, j + λk and λi + k is minimum, then

Q: Wait, why do we use the absolute value for the volume?

Great question! The scalar triple product can yield a negative value depending on the orientation of the vectors (the 'handedness' of the basis). However, volume is a physical quantity and must always be non-negative. Thus, we take the modulus to ensure our result is physically meaningful.

Q: Why do we reject the negative root in the final step?

We don't just reject it blindly! We use the second derivative test. When we found $\lambda = \pm \frac{1}{\sqrt{3}}$, we plugged them into $f''(\lambda) = 6\lambda$. The positive root gave a positive second derivative (indicating a local minimum), while the negative root gave a negative second derivative (indicating a local maximum). Since the problem asks for the minimum, we select the positive root.

Q: Can I just use the cross product and dot product separately instead of a determinant?

You absolutely can! The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is defined as $\vec{a} \cdot (\vec{b} \times \vec{c})$. Using the determinant is simply a more efficient, organized way to compute this. It reduces the risk of algebraic errors, especially with signs, which is crucial in JEE Advanced.

Question

If the volume of parallelopiped formed by the vectors $\hat{i} + \lambda\hat{j} + \hat{k}$, $\hat{j} + \lambda\hat{k}$ and $\lambda\hat{i} + \hat{k}$ is minimum, then $\lambda$ is equal to :

Options:

(A) $\sqrt{3}$

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

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

(D) $-\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Space: Unlocking the Parallelepiped

Imagine you are standing in a vast, three-dimensional void. Before you float three vectors: $\vec{a} = \hat{i} + \lambda\hat{j} + \hat{k}$, $\vec{b} = \hat{j} + \lambda\hat{k}$, and $\vec{c} = \lambda\hat{i} + \hat{k}$.

These vectors are the structural ribs of a parallelepiped, a slanted, elegant box suspended in space. Our mission is to find the value of $\lambda$ that shrinks this box to its smallest possible volume.

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

To find the volume of this parallelepiped, we reach for one of the most powerful tools in our vector toolkit: the scalar triple product. Geometrically, the volume $V$ is the magnitude of the scalar triple product of the three vectors that define its adjacent edges.

Mathematically, we express this as $V = |[\vec{a} \vec{b} \vec{c}]|$. Think of this as a 'volume machine'—you feed in the vectors, and it spits out the space they enclose.

### The Determinant: Organizing the Chaos

To compute this product, we organize our vectors into a $3 	imes 3$ determinant. This is where precision is paramount. We place the components of $\vec{a}$, $\vec{b}$, and $\vec{c}$ into the rows of our matrix:

$$V = \left| \begin{vmatrix} 1 & \lambda & 1 \ 0 & 1 & \lambda \ \lambda & 0 & 1 \end{vmatrix} ight|$$

Expanding this determinant along the first row, we carefully compute the minors: $1(1 - 0) - \lambda(0 - \lambda^2) + 1(0 - \lambda)$. Simplifying this, we arrive at the expression $V = |1 + \lambda^3 - \lambda|$.

Let us define the function inside the modulus as $f(\lambda) = \lambda^3 - \lambda + 1$. Our goal is now clear: minimize the magnitude of this cubic function.

### The Calculus of Optimization

Now, we transition from geometry to calculus. To find the minimum of $f(\lambda)$, we must find its critical points. We take the derivative with respect to $\lambda$:

$$f'(\lambda) = \frac{d}{d\lambda}(\lambda^3 - \lambda + 1) = 3\lambda^2 - 1$$

Setting $f'(\lambda) = 0$, we find $3\lambda^2 = 1$, which leads us to $\lambda^2 = \frac{1}{3}$, or $\lambda = \pm \frac{1}{\sqrt{3}}$. We have two candidates for our minimum.

Which one is the winner? We turn to the second derivative test to decide. Differentiating again, we get $f''(\lambda) = 6\lambda$.

When we test $\lambda = \frac{1}{\sqrt{3}}$, we find $f''(\frac{1}{\sqrt{3}}) = \frac{6}{\sqrt{3}} > 0$. A positive second derivative confirms that this point is a local minimum.

Conversely, testing $\lambda = -\frac{1}{\sqrt{3}}$ yields a negative second derivative, identifying it as a local maximum.

### The Elegant Conclusion

We have navigated the geometry, mastered the determinant, and utilized the power of calculus to isolate the exact value of $\lambda$.

The volume of our parallelepiped is minimized when **$\lambda = \frac{1}{\sqrt{3}}$**.

This result is not just a number; it is the precise configuration where the vectors align to enclose the smallest possible space. Keep this logical flow in your mind—it is the same path you will take for any optimization problem in your JEE journey.

Animated Solution for Mathematics - Vector Algebra: If the volume of parallelopiped formed by the vectors $\hat{i} + λ \hat{j} + \hat{k}$ , $\hat{j} + λ \hat{k}$ and $λ \hat{i} + \hat{k}$ is minimum, then $λ$ is equal to :

Select Answer:

Visualized Solution

Identify the Vectors

The Parallelepiped

Volume Formula

Determinant Setup

Expanding the Determinant

Simplifying the Volume Expression

Differentiation for Extrema of $f (λ)$

First Derivative $f^{'} (λ)$

Finding Critical Points for $λ$

Second Derivative Test $f^{''} (λ)$

Testing the Critical Points

Final Conclusion for $λ$

The Sigma Insight: Scalar Triple Product

The Geometry of Space

The Engine of Volume

The Determinant

The Calculus of Optimization

The Elegant Conclusion

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