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Find the image of the point (1,6,3) in t...

Find the image of the point `(1,6,3)` in the line `x/1=(y-1)/2=(z-2)/3`

Text Solution

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The equations of the given line are
`(x)/(1) = (y-1)/(2) =(z-2)/(3) =r` (say) …(i)
The general point on this line is
`(r , 2r +1, 3r+2)`

Let N be the foot of the perpendicular drawn from the point P(1,6,3) on the given line.
Then this point is N(r,2r +1,3r +2) for some fixed value fo r
D.r's of PN are (r-1,2r-5,3r -1)
D.r's of the given line are 1,2,3
Since PN is perpendicular to the given line (i) we have
`1.(r -1) +2(2r-4)+3(3r -1)=0rArr 14r =14 rArr r=1`
So, the point N is given by N(1,3,5)
Let Q`(alpha, beta ,gamma)` be the image of P(1,6,3) in the given line.
Then N is the midpoint of PQ.
`:. (alpha+1)/(2)=1 (beta+6)/(2)=3 (gamma+3)/(2) 5 rArr alpha =1 , beta =0 " and " gamma =7`
Hence the image of the point P(1,6,3) in the given line is Q(1,0,7)
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