1
Question
The coordinates of the foot of perpendicular from the point (1,0,0) to the line x−12=y+1−3=z+108 are?
The coordinates of the foot of perpendicular from the point (1,0,0) to the line x−12=y+1−3=z+108 are?
Open in App
Solution
The correct option is C (3,−4,−2)
R.E.F image Given point = (1,0,0)Given line x−12=y+1−3=2+108=λx=2λ+1;y=−3λ−1 z=8λ−10(2λ+1,−3λ−1,8λ−10)Q=(x,y,z)from line PQ=(2λ+1−1,−3λ−1+0,8λ−10+0)PQ=(2λ,−3λ−1,8λ−10)=0from (2,−3,8) to PQ 2(2λ)−3(−3λ−1)+8(8λ−10)=04λ+9λ+3+64λ−80=077λ−77=0λ=7777=1∴ coordinate of foot of ⊥erQ=(3,−4,−2)2λ+12(1)+1=3
−3λ−1−3(1)−1=−48λ−10
8(1)=10=−2
R.E.F image
Given point = (1,0,0)
Given line x−12=y+1−3=2+108=λ
x=2λ+1;y=−3λ−1 z=8λ−10
(2λ+1,−3λ−1,8λ−10)Q=(x,y,z)
from line PQ=(2λ+1−1,−3λ−1+0,8λ−10+0)
PQ=(2λ,−3λ−1,8λ−10)=0
from (2,−3,8) to PQ
2(2λ)−3(−3λ−1)+8(8λ−10)=0
4λ+9λ+3+64λ−80=0
77λ−77=0
λ=7777=1
∴ coordinate of foot of ⊥erQ=(3,−4,−2)
2λ+1
2(1)+1
=3
−3λ−1
−3(1)−1
=−4
8λ−10
8(1)=10
=−2
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
