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Find the image of the point (1,6,3) in the line $ \large\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $

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  • To get distance between a point and line get foot of $ \perp$ of the point on the line and get the distance.
Step 1:
The cartesian equation of the line is $\large\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$----(1)
The given point is (1,6,3)
To find the image of P(1,6,3) in the line draw a line PR $\perp$ to the line .
Let R be the image of P and Q is the mid point of PR
Let $a,b,c$ be the direction cosines of PR.Since PR is $\perp$ to the line,apply the condition of perpendicularity.
$a\times 1+b\times 2+c\times 3=0$-------(2)
Step 2:
Now let $\large\frac{x-1}{a}=\frac{y-6}{b}=\frac{z-3}{c}=$$k$(say)------(3)
Any point on the line (2) is $(ak+1,bk+6,ck+3)$
Let the point be Q
But Q also lies on line (1)
$\therefore \large\frac{ak+1}{1}=\frac{bk+6-1}{2}=\frac{ck+3-2}{3}$
$\Rightarrow \large\frac{ak+1}{1}=\frac{bk+5}{2}=\frac{ck+1}{3}$
$\Rightarrow \large\frac{1(ak+1)+2(bk+5)+3(k+1)}{1\times 1+2\times 2+3\times 3}$
$\Rightarrow \large\frac{14+(a+2b+3c)k}{14}$$=1$
$\Rightarrow ak=0,bk=-3$ and $ck=2$
Step 3:
Since $Q$ is the midpoint of PR
$\large\frac{1+x'}{2}$$=1$ and $\large\frac{6+y'}{2}$$=3$ and $\large\frac{3+z'}{2}$$=5$
$\Rightarrow x'=1,y'=0$ and $z'=7$
Hence (1,0,7) is the image of P in line (1)
answered Nov 7, 2013 by sreemathi.v

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