Let the line L pass through the point (0,1,2), intersect the line (x-1)/(2)=(y-2)/(3)=(z-3…

Question

Let the line $L$ pass through the point $(0,1,2)$, intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is :

Accepted Answer

Given line, $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{1}=\lambda $ .........(i)

$	herefore $ Any point on the line $A(2\lambda +1,3\lambda +2,\lambda +3)$

Since, line $L$ pass through the point $B(0,1,2)$ and intersects line (i)

$	herefore $ Direction ratio of $AB$ are

$\begin{gathered} 0-2\lambda -1,1-3\lambda -2,2-\lambda -3 \ -2\lambda -1,-3\lambda -1,-\lambda -1 \end{gathered}$

Line $L$ parallel to the plane $2x+y-3z=4$

$\begin{gathered} 	herefore 2(-2\lambda -1)+1(-3\lambda -1)-3(-\lambda -1)=0 \ \Rightarrow -4\lambda -2-3\lambda -1+3\lambda +3=0 \ \Rightarrow -4\lambda =0 \ \Rightarrow \lambda =0 \ 	herefore AB=-1,-1,-1 \ 1,1,1 \end{gathered}$

$	herefore $ Equation of line $L$ is $\frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu $

$\Rightarrow x=\mu ,y=\mu +1,z=\mu +2$

$	herefore $ Direction ratios of $PQ=\mu -1,\mu +10,\mu $

Since, $PQ\perp L$

$\begin{gathered} 	herefore 1(\mu -1)+1(\mu +10)+\mu (1)=0 \ \Rightarrow \mu -1+\mu +10+\mu =0 \ \Rightarrow 3\mu +9=0 \ \Rightarrow \mu =-3 \ 	herefore Q(-3,-2,-1) \end{gathered}$

$	herefore $ Distance of the point $P(1,-9,2)$ from the line $L$ is

$\begin{gathered} =\sqrt{(-3-1{)}^{2}+(-2+9{)}^{2}+(-1-2{)}^{2}} \ =\sqrt{16+49+9}=\sqrt{74} \end{gathered}$

Solutions · 1