Let the plane P contain the line 2x+y−z−3=0=5x−3y+4z+9 and be parallel to the line 2x+2=−43−y=5z−7. Then the distance of the point A(8,−1,−19) from the plane P measured parallel to the line −3x=4y−5=−122−z is equal to
Answer: 26
Solution
The plane P is of the form P1+λP2=0:
(2x+y−z−3)+λ(5x−3y+4z+9)=0
Normal vector: n=(5λ+2,1−3λ,4λ−1)
Since P is parallel to b=(2,4,5): n⋅b=0
2(5λ+2)+4(1−3λ)+5(4λ−1)=0 ⟹ λ=−61
Plane equation: 7x+9y−10z−27=0
Line through A parallel to (−3,4,12):
−3x−8=4y+1=12z+19=t
Point B on plane: (8−3t,−1+4t,−19+12t)
Substitute into plane: 7(8−3t)+9(4t−1)−10(12t−19)=27 ⟹ t=2