Question

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

• A. $\frac{-1}{\sqrt{3$
• B. $\frac{1}{\sqrt{3$
• C. $-\sqrt{3}$
• D. $\sqrt{3}$

Hint:

Logic Tip: When looking for a maximum, the second derivative must be negative (concave down). By simply looking at the second derivative $V''(\alpha) = 6\alpha$, it is immediately obvious that we need the negative root of $\alpha$ to satisfy $6\alpha<0$.

Answer 1

The Correct Option is A

Concept:
The volume $V$ of a parallelepiped formed by three coterminous vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is: \[ V = |[\vec{a}\ \vec{b}\ \vec{c}]| \] i.e., the absolute value of their scalar triple product.
Step 1: Express $V$ as a function of $\alpha$.
\[ \vec{a} = \hat{i} + \alpha\hat{j} + \hat{k}, \quad \vec{b} = \hat{j} + \alpha\hat{k}, \quad \vec{c} = \alpha\hat{i} + \hat{k} \] \[ V(\alpha) = \begin{vmatrix} 1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1 \end{vmatrix} \] Expanding: \[ V(\alpha) = 1(1\cdot1 - \alpha\cdot0) - \alpha(0\cdot1 - \alpha^2) + 1(0\cdot0 - 1\cdot\alpha) \] \[ V(\alpha) = 1 + \alpha^3 - \alpha \] 
Step 2: Find critical points.
\[ \frac{dV}{d\alpha} = 3\alpha^2 - 1 \] \[ 3\alpha^2 - 1 = 0 \Rightarrow \alpha^2 = \frac{1}{3} \] \[ \alpha = \pm \frac{1}{\sqrt{3}} \] 
Step 3: Second derivative test.
\[ \frac{d^2V}{d\alpha^2} = 6\alpha \] For $\alpha = \frac{1}{\sqrt{3}}$: \[ 6\left(\frac{1}{\sqrt{3}}\right) > 0 \Rightarrow \text{Minimum} \] For $\alpha = -\frac{1}{\sqrt{3}}$: \[ 6\left(-\frac{1}{\sqrt{3}}\right) < 0 \Rightarrow \text{Maximum} \] 
Final Answer:
\[ \boxed{\alpha = -\frac{1}{\sqrt{3}}} \]