Let the line passing through the points (-1,2,1) and parallel to the line x-12=y+13=z4 int…

Question

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ intersect the line $\dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is:

Accepted Answer

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

$$\begin{aligned}
&\Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}-(2)=\lambda\
&\text { So, }\left[\[\begin{gathered} x=2 \lambda-1 \ y=3 \lambda+2 \ z=4 \lambda+1 \end{gathered}\end{aligned}$$\right.
\]

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

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

For intersection point 'P'

$$\begin{aligned}
& \mathrm{x}=2 \lambda-1=3 \mu-2 \
& \mathrm{y}=3 \lambda+2=2 \mu+3 \quad\left[\[\begin{gathered} \lambda=1 \ \mu=1 \end{gathered}\end{aligned}$$\right] \
\& \mathrm{z}=4 \lambda+1=\mu+4 \
\& \text { So, point } \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5) \
\& \& \mathrm{Q}(4,-5,1) \
\& \therefore \mathrm{PQ}=\sqrt{9+100+16} \
\& =\sqrt{125}=5 \sqrt{5}
\]

Solutions · 1