Question

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

Answer 1

The cartesian equation of the line is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$..........(i)

the given point is (1,6,3)

so, find te 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

so, let A,B,C be the direction cosines of PR. since PR is $\perp$

$a\times 1+b\times 2+c\times 3=0$..............(2)

$a+2b+3c$=0

step 2;-

Now $\frac{x-1}{a}=\frac{y-6}{b}=\frac{z-3}{c}=k$--------(iii)

anypoint on the line (ii) is $(ak+1,bk+6,ck+3)$

let the point is Q.

but Q is also lie on line (i)

therefore ,

$\frac{ak+1}{1}=\frac{bk+6-1}{2}=\frac{ck+3-2}{3}$

$=\frac{ak+1}{1}=\frac{bk+5}{2}=\frac{ck+1}{3}$

$=\frac{1(ak+1)+2(bk+5)+3(k+1)}{1\times 1+2\times 2+3\times 3}$

$\Rightarrow \frac{14+(a+2b+3c)k}{14}=1$

$=ak=0,bk=-3\&ck=2$

$Q=(0+1,-3+6,2+3)$

$=(1,3,5)$

step3;- therefore Q is the midpoint of PR

$\frac{1+x^{\prime }}{2}=1$ and $\frac{6+y^{\prime }}{2}=3$ and $\frac{3+z^{\prime }}{2}=5$

$x^{\prime }=1,y^{\prime }=0$ and $z^{\prime }=7$

hence (1,0,7) is the image of P in line (i)