Let the line passing through the points (-1,2,1) and parallel to the line (x-1)/(2)=(y+1)/…

Question

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$

Accepted Answer

Equation of line through point $(-1,2,1)$ is $	o$

$\begin{gathered} \Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}-(2)=\lambda \ [[|\begin{array}{cc}x=2\lambda -1 \ y=3\lambda +2 \ z=4\lambda +1\end{array} \end{gathered}$

By $(1)	o \frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu ($ Let $)$

So, $[[|\begin{gathered} x=3\mu -2 \ y=2\mu +3 \ z=\mu +4 \end{gathered}$

For intersection point 'P'

$\begin{gathered} x=2\lambda -1=3\mu -2 \ y=3\lambda +2=2\mu +3\, [\lambda =1 \ \mu =1] \ z=4\lambda +1=\mu +4 \ P(x,y,z)=(1,5,5) \ \Q(4,-5,1) \ 	herefore PQ=\sqrt{9+100+16} \ =\sqrt{125}=5\sqrt{5} \end{gathered}$

Solutions · 1