Question:
The equation of the line through the point $(0,1,2)$ and perpendicular to the line
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is :
Correct Option: , 4
Solution:
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$
$\Rightarrow \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(2 \mathrm{r}+1,3 \mathrm{r}-1,-2 \mathrm{r}+1)$
Since, $\overline{\mathrm{QP}} \perp(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
$\Rightarrow 4 r+2+9 r-6+4 r+2=0$
$\Rightarrow r=\frac{2}{17}$
$\Rightarrow P\left(\frac{21}{17}, \frac{-11}{17}, \frac{13}{17}\right)$
$\Rightarrow \overline{\mathrm{PQ}}=\frac{21 \hat{\mathrm{i}}-28 \hat{\mathrm{j}}-21 \hat{\mathrm{k}}}{17}$
So, $\overline{\mathrm{QP}}: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-1}{4}=\frac{\mathrm{z}-2}{3}$