InterviewSolution
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InterviewSolution
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The image of theline `(x-1)/3=(y-3)/1=(z-4)/(-5)`in the plane `2x-y+z+3=0`is the line(1) `(x+3)/3=(y-5)/1=(z-2)/(-5)`(2) `(x+3)/(-3)=(y-5)/(-1)=(z+2)/5`(3) `(x-3)/3=(y+5)/1=(z-2)/(-5)`(3) `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`A. `(x-3)/3=(y+5)/1=(z-2)/(-5)`B. `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`C. `(x+3)/3=(y-5)/1=(z-2)/(-5)`D. `(x+3)/(-3)=(y-5)/(-3)=(y-5)/(-1)=(z+2)/5` |
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Answer» Correct Answer - C Clearly, given line is parallel to the given plane. So, its image will also be parallel to the plane and so its direction ratios are proportional to 3,1,-5. Given line passesthrough (1,3,4). Let `(x_(1),y_(1),z_(1)` be its image in the given plane. Then `x_(1),y_(1),z_(1)` are given by `(x_(1)-1)/2=(y_(1)-3)/(-1)=(z_(1)-4)/1=-((2xx1-3+4+3))/(2^(2)+(-1)^(2)+1^(2))` `implies(x_(1)-1)/2=(y_(1)-3)/(-1)=(z_(1)-4)/1=-2` `impliesx_(1)=-3,y_(1)=5,z_(1)=2` So image of (1,3,4) in the given planeis (-3,5,2). Hence the equations of the required line are `(x+3)/3=(y-5)/1=(z-2)/(-5)`. |
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