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Thedistance of the point (1, 0, 2) from the point of intersection of the line `(x-2)/3=(y+1)/4=(z-2)/(12)`and the plane x y + z = 16, is :(1) `2sqrt(14)`(2) 8 (3) `3sqrt(21)`(4) 27A. `2sqrt(14)`B. 8C. `3sqrt(21)`D. 13 |
Answer» Correct Answer - D Given equation of line is `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)=lambda" "["say"]…(i)` and equation of plane is `x-y+z=16" "…(ii)` Any point on the line (i) is, `(3lambda+2,4lambda-1,12lambda+2)` Let this point of intersection of the line and plane. `:.(3lambda+2)-(4lambda-1)+(12lambda+2)=16` `implies" "11lambda+5=16` `implies" "11lambda=11` `implies" "lambda=1` So, the point of intersection is (5, 3, 14) Now, distance between the points (1, 0, 2) and (5, 3, 14) `sqrt((5-1)^(2)+(3-0)^(2)+(14-2)^(2))` `=sqrt(16+9+144)=sqrt(169)=13` |
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