Question

Find the length of the perpendicular from (2, -3, 1) to the line $\frac{x+1}{2}= \frac{y-3}{3}= \frac{z+2}{-1}$.

Answer 1

Answer:

$\frac{\sqrt{7434}}{14}\:\:\:units$

Step-by-step explanation:

We know that standard equationof line in cartesian form is

$\frac{x-x_{1}}{a} =\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$

on comparison with given equation ,we get that (2,3,-1) are the direction ratio's of the line.

If from  P(2,-3,1) we join that point to extended till the line ,to meet at Q(x,y,z)

thus Q also exist on the line

so Line AB

$\frac{x+1}{2}= \frac{y-3}{3}= \frac{z+2}{-1} = \lambda\\\\x=2\lambda-1\\\\y=3\lambda+3\\\\z=-\lambda-2$

Thus Q( 2λ-1,3λ+3,-λ-2)

Now we can find direction ratio of Perpendicular

Direction Ratio of perpendicular PQ(2λ-3,3λ+6,-λ-3)

So line AB and PQ are perpendicular ,

2(2λ-3)+3(3λ+6)-1(-λ-3)=0

4λ-6+9λ+18+λ+3=0

14λ=-15

λ=-15/14

so point Q[2(-15/14)-1,3(-15/14)+3,15/14-2]

[-22/7,-3/14,-13/14]

Now distance between two points

$PQ=\sqrt{({2+\frac{22}{7})^2+({-3+\frac{3}{14}})^2+({1+\frac{13}{14}})^2}$

$=\sqrt{\frac{5184+1521+729}{196}}$

$=\frac{\sqrt{7434}}{14}$