3D Geometry 3 Question 39

39. Perpendicular are drawn from points on the line x+22=y+11=z3 to the plane x+y+z=3. The feet of perpendiculars lie on the line

(a) x5=y18=z213

(c) x4=y13=z27

(b) x2=y13=z25

(d) x2=y17=z25

(2013 Adv.)

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Answer:

Correct Answer: 39. (a)

Solution:

  1. Key Idea To find the foot of perpendiculars and find its locus. Formula used

Footof perpendicular from (x1,y1,z1) to

ax+by+cz+d=0 be (x2,y2,z2), then

x2x1a=y2y1b=z2z1c=(ax1+by1+cz1+d)a2+b2+c2

Any point on x+22=y+11=z3=λ

x=2λ2,y=λ1,z=3λ

Let foot of perpendicular from (2λ2,λ1,3λ)

to x+y+z=3 be (x2,y2,z2)

x2(2λ2)1=y2(λ1)1=z2(3λ)1=(2λ2λ1+3λ3)1+1+1

x22λ+2=y2+λ+1=z23λ=24λ3x2=2λ3,y2=17λ3,z2=2+5λ3λ=x202/3=y217/3=z225/3

Hence, foot of perpendicular lie on

x2/3=y17/3=z25/3x2=y17=z25



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