The value of , so that the volume of parallelopiped formed by , and become minimum, is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The volume equals the absolute value of the scalar triple product, which is the determinant:
|1 a 1|
|0 1 a|
|a 0 1|
= 1(1·1 - a·0) - a(0·1 - a·a) + 1(0·0 - 1·a)
= 1 + a³ - a
So V(a) = |a³ - a + 1|. To minimize, differentiate f(a) = a³ - a + 1:
f'(a) = 3a² - 1 = 0 → a = ±1/√3
Check f''(a) = 6a > 0 at a = 1/√3, confirming a minimum there (at a = -1/√3, f'' < 0, it's a local max of f, not minimum of |f|).
At a = 1/√3: f(1/√3) = 1/(3√3) - 1/√3 + 1 = 1 - 2/(3√3) > 0, so |f| is minimized at a = 1/√3.
|1 a 1|
|0 1 a|
|a 0 1|
= 1(1·1 - a·0) - a(0·1 - a·a) + 1(0·0 - 1·a)
= 1 + a³ - a
So V(a) = |a³ - a + 1|. To minimize, differentiate f(a) = a³ - a + 1:
f'(a) = 3a² - 1 = 0 → a = ±1/√3
Check f''(a) = 6a > 0 at a = 1/√3, confirming a minimum there (at a = -1/√3, f'' < 0, it's a local max of f, not minimum of |f|).
At a = 1/√3: f(1/√3) = 1/(3√3) - 1/√3 + 1 = 1 - 2/(3√3) > 0, so |f| is minimized at a = 1/√3.
[AIEEE 2003]