NIMCET 2008 — Mathematics PYQ

The volume $V$ of a parallelopiped formed by three vectors $\vec{a},\vec{b},$ and $\vec{c}$ is given by the magnitude of their Scalar Triple Product:

Step 2: Set up the Scalar Triple Product

Let the given vectors be:

• $\vec{a}=1\hat{i}+\lambda \hat{j}+1\hat{k}$

• $\vec{b}=0\hat{i}+1\hat{j}+\lambda \hat{k}$

• $\vec{c}=\lambda \hat{i}+0\hat{j}+1\hat{k}$

The volume is the determinant of the matrix formed by these components:

Step 3: Expand the determinant

Expanding along the first row:

Step 4: Find the critical points for minimum volume

To find the minimum value, we take the first derivative of $V$ with respect to $\lambda$ and set it to zero:

Setting $\frac{dV}{d\lambda }=0$:

Step 5: Use the second derivative test

Find the second derivative to check for minima:

• At $\lambda =\frac{1}{\sqrt{3}}$, \frac{d^2V}{d\lambda^2} = \frac{6}{\sqrt{3}} > 0 (This indicates a local minimum).

• At $\lambda =-\frac{1}{\sqrt{3}}$, \frac{d^2V}{d\lambda^2} = -\frac{6}{\sqrt{3}} < 0 (This indicates a local maximum).

Conclusion:

The volume of the parallelopiped is minimum when $\lambda =\frac{1}{\sqrt{3}}$.

Correct Option: (c)