Question

find the foot of the perpendicular and image of the point (2,4,-1) to the line x+5/1=y+3/4=z-6/-9. Also find perpendicular distance from the given point to the line​

Answer 1

Answer:

Let P(2,4,−1)P(2,4,−1) be the given point and the Line L→x−51=y−34=z−6−9=λL→x−51=y−34=z−6−9=λ

⇒d.r.⇒d.r. of the line L=(1,4,−9)L=(1,4,−9)

Any point Q(x,y,z)Q(x,y,z) on the line L is given by Q(λ+5,4λ+3,−9λ+6)Q(λ+5,4λ+3,−9λ+6)

Let this point QQ be the foot of ⊥⊥ of the point P on the line LL.

d.r of

PQ

the ⊥⊥

=(λ+5−2,4λ+3−4,−9λ+6+1)=(λ+5−2,4λ+3−4,−9λ+6+1)

=(λ+3,4λ−1,−9λ+7)

∴

PQ

(1,4,−9)=0∴PQ¯.(1,4,−9)=0

(λ+3,4λ−1,−9λ+7).(1,4,−9)=0(λ+3,4λ−1,−9λ+7).(1,4,−9)=0

⇒(λ+3)1+(4λ−1)4−9(−9λ+7)=0⇒(λ+3)1+(4λ−1)4−9(−9λ+7)=0

⇒λ+3+16λ−4+81λ−63=0⇒λ+3+16λ−4+81λ−63=0

⇒98λ=64⇒98λ=64

⇒λ=3249

PQ

is ⊥⊥⊥ to the line L

Substituting λλ in Q we get the foot of ⊥Q⊥Q as

Q(3249Q(3249+5,12849+5,12849+3,−28849+3,−28849+6)+6)

⊥⊥ distance≡

PQ

=7

thank you