Let be distinct non-negative numbers. If the vectors , and lie in a plane, then is
- A.the arithmetic mean of and
- B.the geometric mean of and
- C.the harmonic mean of and
- D.equal to zero
Show answer and solution
Correct option: B
Solution
For the three vectors to be coplanar, the scalar triple product must be zero. Setting up the determinant:
|a a c|
|1 0 1| = 0
|c c b|
Expanding: a(0·b - 1·c) - a(1·b - 1·c) + c(1·c - 0·c) = 0
= a(-c) - a(b - c) + c(c) = 0
= -ac - ab + ac + c² = 0
= c² - ab = 0
So c² = ab, which means c is the geometric mean of a and b.
|a a c|
|1 0 1| = 0
|c c b|
Expanding: a(0·b - 1·c) - a(1·b - 1·c) + c(1·c - 0·c) = 0
= a(-c) - a(b - c) + c(c) = 0
= -ac - ab + ac + c² = 0
= c² - ab = 0
So c² = ab, which means c is the geometric mean of a and b.
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