Let a unit vector u=x i+y j+z k make angles 2, 3 and 2 3 with the vectors 1 2 i+ 1 2 k, 1 2 j+ 1 2 k and 1 2 i+ 1 2 j respectively. If v= 1 2 i+ 1 2 j+ 1 2 k then | u- v|2 is equal to
- 112
- 52
- 7
- 9
Answer
(2) 52
Solution
∵ All three given vectors are unit vectors and | u|=1 ⇒ u p= (angle)
x+z 2= 2=0 x+z=0
y+z 2= 3= 12 y+z= 1 2
x+y 2= 2 3=- 12 x+y=- 1 2
Adding: x+y+z=0 y=0, z= 1 2, x=- 1 2
u- v=- 2 2 i- 1 2 j+0 k
| u- v|2=2+ 12= 52
Answer: (2)