Question

Find the distance of the point (1,-2,3) from the plane x-y+z = 5 measured parallel to the line x+12=y+32=z+1−6

Answer 1

Answer:

Step-by-step explanation:

Given: Point ( 1, -2, 3 ) , Equation of Plane : x - y + z = 5 and Equation of Line : $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z+1}{-6}$

To find Distance of Point from plane measured along line

Let say point A ( 1, -2, 3 )

Given Equation of Plane, x - y + z = 5

Given Equation of Line l, $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z+1}{-6}$

First we find equation of line AB which is parallel to line l,

Slope of line l ( 2, 2, -6 )

using this slope and point A,

Equation of line AB , given by

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

from this equation we find genral formula for a point on line AB,

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

$\frac{x-1}{2}=k\:\:,\:\:\frac{y+2}{2}=k\:\:,\:\:\frac{z-3}{-6}=k$

$x-1=2k\:\:,\:\:y+2=2k\:\:,\:\:z-3=-6k$

$x=2k+1\:\:,\:\:y=2k-2\:\:,\:\:z=-6k+3$

So, Point B ( 2k+1, 2k-2, -6k+3 )

Point B lies on Plane also.

Now we coordinates of point B in equation of plane,

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

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

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

-6k + 6 = 5

-6k = 5 - 6

-6k = -1

$k=\frac{-1}{-6}$

$k=\frac{1}{6}$

⇒ Coordinates of Point B ( $2\times\frac{1}{6}+1\:,\:2\times\frac{1}{6}-2\:,\:-6\times\frac{1}{6}+3$ )

⇒ Coordinates of Point B ( $\frac{1}{3}+1\:,\:\frac{1}{3}-2\:,\:-1+3$ )

⇒ Coordinates of Point B ( $\frac{1+3}{3}\:,\:\frac{1-6}{3}\:,\:2$ )

⇒ Coordinates of Point B ( $\frac{4}{3}\:,\:\frac{-5}{3}\:,\:2$ )

Finally By using distance formula we get the distance between point A and Point B

Distance = $\sqrt{(1-\frac{4}{3})^2+(-2+\frac{-5}{3})^2+(3+2)^2}$

               = $\sqrt{(\frac{3-4}{3})^2+(\frac{-6-5}{3})^2+(5)^2}$

               = $\sqrt{(\frac{-1}{3})^2+(\frac{-11}{3})^2+25}$

               = $\sqrt{\frac{1}{9}+\frac{121}{9}+25}$

               = $\sqrt{\frac{1}{9}+\frac{121}{9}+\frac{225}{9}}$

               = $\sqrt{\frac{1+121+225}{9}}$

               = $\sqrt{\frac{347}{9}}$

               = $\frac{\sqrt{347}}{3}$

Therefore, Distance = $\frac{\sqrt{347}}{3}$   units