Let a = i + 2 j + k and b = 2 i + j - k. Let c be a unit vector in the plane of the vectors a and b and be perpendicular to a. Then such a vector c is:
- 1 2(- i + k)
- 1 5( j - 2 k)
- 1 3( i - j + k)
- 1 3(- i + j - k)
Answer
(1) 1 2(- i + k)
Solution
c lies in the plane of a and b: c = a + b
a a = 1 + 4 + 1 = 6, a b = 2 + 2 - 1 = 3
c a = 0 6 + 3 = 0 = -2
c a - 2 b = (1 - 4) i + (2 - 2) j + (1 + 2) k = -3 i + 3 k
c = 1 2(- i + k)
Answer: (1)