Find the image of the point (2, -1, 5) in the line
Find the image of the point (2, -1, 5) in the line

Find the image of the point in the line

$\stackrel{\to }{\mathrm{r}}=\left(11\stackrel{^}{i}–2\stackrel{^}{j}–8\stackrel{^}{\mathrm{k}}\right)+\lambda \left(10\stackrel{^}{i}–4\stackrel{^}{j}–11\stackrel{^}{k}\right)$
\overrightarrow{\mathrm{r}}=(11 \hat{i}-2 \hat{j}-8 \hat{\mathrm{k}})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})

Given: Point Equation of line The equation of line can be re-arranged as The general point on this line is

$\left(10r+11,–4r–2,–11r–8\right)$
(10 r+11,-4 r-2,-11 r-8)

Let be the foot of the perpendicular drawn from the point on the given line. Then, this point is for some fixed value of .
D.r.’s of PN are D.r.’s of the given line is Since, PN is perpendicular to the given line, we have,

$10\left(10r+9\right)–4\left(–4r–3\right)–11\left(–11r–3\right)=0$
10(10 r+9)-4(-4 r-3)-11(-11 r-3)=0
$\begin{array}{l}100r+90+16r+12+121r+33=0\\ 237r=135\\ r=\frac{135}{237}\end{array}$
\begin{array}{l}
100 r+90+16 r+12+121 r+33=0 \\
237 r=135 \\
r=\frac{135}{237}
\end{array}

Then, the image of the point is

$\frac{\alpha –11}{–1.1}=0,\frac{\beta +2}{\overline{I}}=1,\frac{\gamma +8}{9}=1$
\frac{\alpha-11}{-1.1}=0, \frac{\beta+2}{\bar{I}}=1, \frac{\gamma+8}{9}=1

Therefore, the image is .