Let the foot of perpendicular from the point ( ,2,3) on the line (x-4)/(1)=(y-9)/(2)=(z-5)…

Question

Let the foot of perpendicular from the point $(\lambda ,2,3)$ on the line $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$ be the point ( $1,\mu ,2$ ). Then the distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $\frac{x-\lambda }{2}=\frac{y-\mu }{3}=\frac{z+5}{6}$ is equal to :

Accepted Answer

$\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$

$Q(1,\mu ,2)$ satisfies the line L

$	herefore \frac{1-4}{1}=\frac{\mu -9}{2}=\frac{2-5}{1}$

$\Rightarrow -3=\frac{\mu -9}{2}=-3\, \Rightarrow \mu =3$

D.R. of PQ : $\lambda -1,2-\mu ,1$

Since $PQ\perp L\Rightarrow (\lambda -1).1+2(2-\mu )+1.1=0$

$\Rightarrow \lambda -2\mu +4=0\, 	herefore \lambda =2$

Now lines : $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ \&

$\frac{x-2}{2}=\frac{y-3}{3}=\frac{z+5}{6}$

$\begin{gathered} \vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+{t}_{1}(2\hat{i}+3\hat{j}+6\hat{k}) \ \vec{r}=(2\hat{i}+3\hat{j}-5\hat{k})+{t}_{2}(2\hat{i}+3\hat{j}+6\hat{k}) \end{gathered}$

Distance

$\begin{gathered} =\frac{|((\hat{i}+2\hat{j}-4\hat{k})-(2\hat{i}+3\hat{j}-5\hat{k}))	imes (2\hat{i}+3\hat{j}+6\hat{k})|}{|2\hat{i}+3\hat{j}+6\hat{k}|} \ =\frac{|-9\hat{i}+8\hat{j}-\hat{k}|}{7}=\frac{\sqrt{146}}{7} \end{gathered}$

Solutions · 1