1.

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. `3sqrt(21)`B. `13`C. `2sqrt(14)`D. 8

Answer» Correct Answer - B
The coordinates of an arbitrary point on the line
`(x-2)/3=(y+1)/4=(z-2)/12` are given by
`(x-2)/3=(y+1)/4=(z-2)/12=lamda` where `lamda` is a parameter
`implies x=3lamda+2,y=4lamda-1,z=12lamda+2`
Suppose the given line intersects the plane `x-y+z=16` at `(3lamda+2,4lamda-1,12lamda+2)`. Then
`(3lamda+2)-(4lamda-1)+(12lamda+2)=16implies11lamda=11implieslamda=1`
so the coordinate of the point of intersection are (5,3,14).
Hence required distance `=sqrt((5-1)^(2)+(3-0)^(2)+(14-2)^(2))=13`


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