Question

Find the distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to line x-1/2 , y-3/3 , z+2/-6

Answer 1

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Answer 2

The distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to line x-1/2 , y-3/3 , z+2/-6 is  $\frac{\sqrt{881} }{7}$

Step-by-step explanation:

Given:

The equation of the given line is

$\frac{x-1}{2}=\frac{y-3}{3} =\frac{z+2}{-6}$           [1]

The direction ratios of the above line are proportional to 2, 3, 7.Let P(1, −2, 3) be the given point.The equation of the given plane is, x − y + z = 5.Let PQ be the perpendicular distance of the point P(1, −2, 3) from the given plane.Now, the equation of PQ, passing through P(1, −2, 3) and parallel to the line (1) is given by,

$\frac{x-1-1}{2}=\frac{y-3-(-2)}{3}=\frac{z+2-3)}{-6}$

$\frac{x-2}{2}=\frac{y-1}{3}=\frac{z-1)}{-6}=K$

x-2=2K, y-1=3K, z-1=-6K

x=2K+2, y=3K+1, z=-6K+1          [say]

Now, the coordinates of any point Q on this line are Q(2K+2, 3K+1, -6K+1).If point Q lies on the given plane, then it must satisfy the equation of the plane. So, we have,

x − y + z = 5.

(2K+2)-(3K+1)+(-6K+1)=5

2K+2-3K-1-6K+1=5

-7K+2=5

-7K=5-2

-7K=3

$K=-\frac{3}{7}$

Threrfore,

$x=2(\frac{-3}{7}) +2, y=3(\frac{-3}{7} )+1, z=-6(\frac{-3}{7} )+1$

$x=(\frac{-6}{7}) +2, y=(\frac{-9}{7} )+1, z=(\frac{-18}{7} )+1$

$x=(\frac{-6+14}{7}), y=(\frac{-9+7}{7} ), z=(\frac{-18+7}{7} )$

$x=(\frac{8}{7}), y=(\frac{-2}{7} ), z=(\frac{-11}{7} )$

So, the coordinates of Q are

$Q(\frac{8}{7},\frac{-2}{7},\frac{-11}{7} )$ and we have P(1, −2, 3).

Now, required distance PQ = $\sqrt{(1-\frac{8}{7})^2+(-2-(\frac{-2}{7} ))^2+(3-(\frac{-11}{7} )) ^2}$

PQ = $\sqrt{(\frac{7-8}{7})^2+(\frac{-14+2}{7} )^2+(\frac{21+11}{7} ) ^2}$

PQ = $\sqrt{(\frac{-1}{7})^2+(\frac{-12}{7} )^2+(\frac{32}{7} ) ^2}$

PQ = $\sqrt{\frac{1}{49}+\frac{-144}{49}+\frac{1024}{49}}$

PQ = $\sqrt{\frac{1-144+1024}{49} }$

PQ = $\sqrt{\frac{881}{49} }$

PQ = $\frac{\sqrt{881} }{7}$