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On which of the following lines lies the point of intersection of the line, `(x-4)/(2)=(y-5)/(2)=(z-3)/(1)` and the plane, `x+y+z=2`?A. `(x-4)/(1)=(y-5)/(1)=(z-5)/(-1)`B. `(x+3)/(3)=(4-y)/(3)=(z+1)/(-2)`C. `(x-2)/(2)=(y-3)/(2)=(z+3)/(3)`D. `(x-1)/(1)=(y-3)/(2)=(z+4)/(-5)` |
Answer» Correct Answer - D Given equation of line is `(x-4)/(2)=(y-5)/(2)=(z-3)/(1)=r ("let")" "…(i)` `impliesx=2r+4,y=2r+5" and "z=r+3` `:."General point on the line (i) is "` `P(2r+4,2r+5,r+3)` So, the point of intersection of line (i) and plane `x+y+z=2` will be of the form `P(2r+4,2r+5,r+3)` for some `rinR.` `implies(2r+4)+(2r+5)+(r+3)=2` [`because` the point will lie on the plane] `implies5r=-10impliesr=-2` So, the point of intersection is P(0, 1, 1) [putting `r=-2` in `(2r+4,2r+5,r+3)]` Now, on checking the options, we get `(x-1)/(1)=(y-3)/(2)=(z+4)/(-5)` contain the point (0,1,1) |
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