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

Question

Let the line $$\mathrm{L}$$ 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 $2 x+y-3 z=4$. Then the distance of the point $$\mathrm{P}(1,-9,2)$$ from the line $$\mathrm{L}$$ is :

Accepted Answer

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

$\therefore$ 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)

$\therefore$ Direction ratio of $A B$ are

$$
\[\begin{gathered} 0-2 \lambda-1,1-3 \lambda-2,2-\lambda-3 \ \text { i.e., }  -2 \lambda-1,-3 \lambda-1,-\lambda-1 \end{gathered}$$
\]

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

$$
\begin{aligned}
& \therefore 2(-2 \lambda-1)+1(-3 \lambda-1)-3(-\lambda-1)=0 \\
& \Rightarrow-4 \lambda-2-3 \lambda-1+3 \lambda+3=0 \\
& \Rightarrow -4 \lambda=0 \\
& \Rightarrow  \lambda=0 \\
& \therefore \text { Direction ratio of } A B=-1,-1,-1 \\
& \text { i.e., } 1,1,1
\end{aligned}
$$

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

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

$\therefore$ Direction ratios of $P Q=\mu-1, \mu+10, \mu$

Since, $P Q \perp L$

$$
\begin{aligned}
& \therefore  1(\mu-1)+1(\mu+10)+\mu(1)=0 \\
& \Rightarrow  \mu-1+\mu+10+\mu=0 \\
& \Rightarrow 3 \mu+9=0 \\
& \Rightarrow  \mu=-3 \\
& \therefore Q \text { is }(-3,-2,-1)
\end{aligned}
$$

$\therefore$ 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