InterviewSolution
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InterviewSolution
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The equation of the line passing through `(-4, 3, 1),` parallel to the plane `x +2y-z-5=0` and intersecting the line `(x+1)/-3=(y-3)/2=(z-2)/-1`A. `(x+4)/(3)=(y-3)/(-1)=(z-1)/(1)`B. `(x+4)/(-1)=(y-3)/(1)=(z-1)/(1)`C. `(x+4)/(1)=(y-3)/(1)=(z-1)/(3)`D. `(x-4)/(2)=(y+3)/(1)=(z+1)/(4)` |
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Answer» Correct Answer - A Any point on the line`(x+1)/(-3)=(y-3)/(2)=(z-2)/(-1)` is of the form `(-3lambda-1,2lambda+3,-lambda+2)` `["take"(x+1)/(-3)=(y-3)/(2)=(z-2)/(-1)=lambdaimpliesx=-3lambda-1,` `y=2lambda+3" and "z=-lambda+2`] So, the coordinates of point of intersection of two lines will be `(-3lambda-1,2lambda+3,-lambda+2)` for some `lambdainR` Let the point `A=(-3lambda-1,2lambda+3,-lambda+2)` and `B-=(-4, 3, 1)` Then, `AB=OB-OA` `=(-4hati+3hatj+hatk)-{(-3lambda-1)hati+(2lambda+3)hatj+(-lambda+2)hatk}` `=(3lambda-3)hati-2lambdahatj+(lambda-1)hatk` Now, as the line is parallel to the given plane and so AB will be parallel to the given plane and so AB will be perpendicular to the normal of plane. `impliesAB.lambda=0," where" n=hati+2hatj-hatk` is normal to the plane. `implies((3llambda-3)hati-2lambdahatj+(lambda-1)hatk).(hati+2hatj-hatk)=0` `implies3(lambda-1)-4lambda+(-1)(lambda-1)=0` `[because` If `a=a_(1)hatja_(2)hatj+a_(3)hatk" and "b=b_(1)hati+b_(2)hatj+b_(3)hatk,` then `a.b=a_(1)b_(1)+a_(2)b_(2)+a_(3)b_(3)`] `implies3lambda-3-4lambda-lambda+1=0` `implies" "-2lambda=2implieslambda=-1` Now, the required equation is the equation of line joining A(2, 1 , 3) and B(-4, 3, 1), which is `(x-(-4))/(2-(-4))=(y-3)/(1-3)=(z-1)/(3-1)` [`because` Equation of line joining `(x_(1),y_(1),z_(1))` and `(x_(2),y_(2),z_(2))` is `(x-x_(1))/(x_(2)-x_(1))=(y-y_(1))/(y_(2)-y_(1))=(z-z_(1))/(z_(2)-a_(1))`] `implies" "(c+4)/(6)=(y-3)/(-2)=(z-1)/(2)` or`" "(x+4)/(3)=(y-3)/(-1)=(z-1)/(1)" "["multiplying by "2]` |
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