InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Perpendiculars are drawn from points on the line `(x+2)/2=(y+1)/(-1)=z/3` to the plane `x + y + z=3` The feet of perpendiculars lie on the line (a) `x/5=(y-1)/8=(z-2)/(-13)` (b) `x/2=(y-1)/3=(z-2)/(-5)` (c) `x/4=(y-1)/3=(z-2)/(-7)` (d) `x/2=(y-1)/(-7)=(z-2)/5`A. `(x)/(5)=(y-1)/(8)=(z-2)/(-13)`B. `(x)/(2)=(y-1)/(3)=(z-2)/(-5)`C. `(x)/(4)=(y-1)/(3)=(z-2)/(-7)`D. `(x)/(2)=(y-1)/(-7)=(z-2)/(5)` |
|
Answer» Correct Answer - D To find the foot of perpendiculars and find its locus. Fromula used Footof perpendicular from `(x_(1),y_(1),z_(1))`to `ax+by+cz+d=0`be`(x_(2),y_(2),z_(2))`, then `(x_(2)-x_(1))/(a)=(y_(2)-y_(1))/(b)=(z_(2)-z_(1))/(c)=(-(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))` Any point on `(x+2)/(2)=(y+1)/(1)=(z)/(3)=lambda` `implies" "x=2lambda-2,y=-lambda-1,z=3lambda` Let foot of perpendicular from `(2lambda-2,-lambda-1,3lambda)` to `x+y+z=3` be `(x_(2),y_(2),z_(2)).` `:.(x_(2)-(2lambda-2))/(1)=(y_(2)-(-lambda-1))/(1)=(z_(2)-(3lambda))/(1)` `=-((2lambda-2-lambda-1+3lambda-3))/(1+1+1)` `implies" "x_(2)-2lambda+2=y_(2)+lambda+1=z_(2)-3lambda=2-(4lambda)/(3)` `:." "x_(2)=(2lambda)/(3),y_(2)=1-(7lambda)/(3),z_(2)=2+(5lambda)/(3)` `implies" "lambda=(x_(2)-0)/(2//3)=(y_(2)-1)/(-7//3)=(z_(2)-2)/(5//3)` Hence, foot of perpendicular lie on `(x)/(2//3)=(y-1)/(-7//3)=(z-2)/(5//3)implies(x)/(2)=(y-1)/(-7)=(z-2)/(5)` |
|