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The point on the line `(x-2)/1=(y+3)/(-2)=(z+5)/(-2)`at a distance of 6 from the point `(2,-3,-5)`isa. `(3,-5,-3)`b. `(4,-7,-9)`c. `0,2,-1`d. none of theseA. `(3,-5,-3)`B. `(4,-7,-9)`C. `(0,1,-1)`D. `(-3,5,3)` |
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Answer» Correct Answer - B::C Direction cosines of the given line are `(1)/(3),-(2)/(3)-(2)/(3)` Hence, the equation of line can be a point in the form `(x-2)/((1)/(3))=(y+3)/((-2)/(3))=(z+5)/((-2)/(3))=r` Therefore, any point of the line is `(2+(r)/(3),-3-(2r)/(3),-5-(2r)/(3))` where `r=+-6` Points aer `(4,j-7,-9)` and `(0,1,-1)` |
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