The distance of the point (7,-2,11) from the line x-61= y-40= z-83 along the line x-52= y-1-3= z-56, is :
- 12
- 18
- 21
- 14
Answer
(4) 14
Solution
Through P(7,-2,11) parallel to (2,-3,6): Q=(7+2 , -2-3 , 11+6 )
[figure]
Q lies on L, where y-40 means y=4
-2-3 =4 =-2 Q=(3,4,-1) (also satisfies x-61= z-83=-3)
PQ=| | 22+32+62=2 7=14
Answer: (4)