Let the foot of perpendicular from the point ( , 2, 3) on the line x-41= y-92= z-51 be the point (1, , 2). Then the distance between the lines x-12= y-23= z+46 and x- 2= y- 3= z+56 is equal to :
- 127
- 1457
- 1467
- 1437
Answer
(3) 1467
Solution
Any point of the line: (4+r, 9+2r, 5+r); foot (1, , 2) r=-3 =9-6=3
Foot N(1, 3, 2), P( , 2, 3): NP=( -1, -1, 1) (1, 2, 1)
( -1)-2+1=0 =2
Lines: x-12= y-23= z+46 and x-22= y-33= z+56 are parallel, b=2 i+3 j+6 k, | b|=7
A(1, 2, -4), B(2, 3, -5): AB= i+ j- k
AB b= vmatrix i & j & k\ 1 & 1 & -1\ 2 & 3 & 6 vmatrix=9 i-8 j+ k
d= | AB b || b|= 81+64+17= 1467
Answer: (3)