The distance of the point Q(0,2,-2) from the line passing through the point P(5,-4,3) and perpendicular to the lines r=(-3 i+2 k)+ (2 i+3 j+5 k), R and r=( i-2 j+ k)+ (- i+3 j+2 k), R is :
- 74
- 86
- 54
- 20
Answer
(1) 74
Solution
The required line is perpendicular to both given lines, so its direction is the cross product of their directions:
(2 i + 3 j + 5 k) (- i + 3 j + 2 k) = -9 i - 9 j + 9 k, i.e. d = i + j - k
A general point on the line through P is M(5 + , -4 + , 3 - ).
[figure]
QM = (5 + ) i + ( - 6) j + (5 - ) k
M is the foot of the perpendicular from Q when QM d = 0: (5 + ) + ( - 6) - (5 - ) = 0, so 3 - 6 = 0 and = 2
Hence M(7, -2, 1) and QM = 7 i - 4 j + 3 k
QM = 49 + 16 + 9 = 74
Answer: (1)