Question

The value of ‘a’ so that the volume of parallelopiped formed by $\mathbf { i } + a \mathbf { j } + \mathbf { k }$ ; $\mathbf { j } + a \mathbf { k }$ and $a \mathbf { i } + \mathbf { k }$ becomes minimum is

• A. – 3
• B. 3
• C. $1 / \sqrt { 3 }$
• D. $\sqrt { 3 }$

Answer 1

Answer: (C)

$1 / \sqrt { 3 }$

Answer 2

Volume of the parallelepiped

V = = $( \mathbf { i } + a \mathbf { j } + \mathbf { k } ) \cdot \{ ( \mathbf { j } + a \mathbf { k } ) \times ( a \mathbf { i } + \mathbf { k } ) \}$

= $\left. ( \mathbf { i } + a \mathbf { j } + \mathbf { k } ) \cdot \left\{ \mathbf { i } + a ^ { 2 } \mathbf { j } - a \mathbf { k } \right) \right\}$ = $1 + a ^ { 3 } - a$

$\frac { d V } { d a }$ = $3 a ^ { 2 } - 1$ ; $\frac { d ^ { 2 } V } { d a ^ { 2 } } = 6 a$ ; $\frac { d V } { d a } = 0 \Rightarrow 3 a ^ { 2 } - 1 = 0 \Rightarrow a = \pm \frac { 1 } { \sqrt { 3 } }$At $a = \frac { 1 } { \sqrt { 3 } } , \frac { d ^ { 2 } V } { d a ^ { 2 } } = \frac { 6 } { \sqrt { 3 } } > 0$

$\therefore$ V is minimum at $a = \frac { 1 } { \sqrt { 3 } }$