Find the image of the point (1, 6, 3) with respect to the line x1=y−12=z−23 is
Let P (1, 6, 3) be the given point and let L be the foot of the perpendicular from P to the given line.
The coordinates fo a general point on the given line are
x−01=y−12=z−23i.e.x=λ,y=2λ+1,z=3λ+2
If the coordinates of L are(λ,2λ+1,3λ+2) then the direction ratios of PL areλ−1,2λ−5,3λ−1.
But the direction ratios of given line which is perpendicular to PL are 1, 2, 3 .Therefore (λ−1)1+(2λ−5)2+(3λ−1)3=0, which gives λ=1. Hence coordinates of L are (1,3,5).
Let Q(x1,y1,z1) be the image fo P (1, 6, 3) in the given line then L is the md -point of PQ. Therefore,
x1+12=1,y1+62=3,z1+32=5⇒x1=1,y1=0,z1=7