Find the distance of point 
from the line 
measured parallel to the plane: x – y +2z – 3 = 0
Let the point be A = (-2, 3, -4)
The line is 
Compare line with 
Where 
is point on the line and 
is the direction of line
and 
The point on line is (1, 2, -1) and direction is <1, 3, -9>
Rewriting the line equation in cartesian from
Let it be equal to k so that we can write the general point on the line
⇒ x – 1 = k and y – 2 = 3k and z + 1 = -9k
⇒ x = k + 1 and y = 3k + 2 and z = -9k – 1
So, any general point on the line say B has coordinates (k + 1, 3k + 2, -9k – 1)
We have to measure the distance from point A to the line parallel to the plane x – y + 2z – 3 = 0
Comparing x – y + 2z – 3 = 0 with general plane equation ax + by + cz + d = 0 where <a, b, c> is the normal to the plane
let the normal is 
Say we have to find the distance AB such that AB is parallel to plane
AB parallel to plane which means AB is perpendicular to the normal of plane that is CD
As 
is perpendicular to 
their dot product is zero because the right hand side will have cos90° which is 0
⇒ k + 3 + (3k – 1) (-1) + (-9k + 3)2 = 0
⇒ k + 3 – 3k + 1 – 18k + 6 = 0
⇒ -20k + 10 = 0
⇒ 20k = 10
⇒ k = 1/2
Putting k = 1/2 in B we will get the point B
⇒ B = (1/2 + 1, 3(1/2) + 2, -9(1/2) – 1)
Now the distance AB
Couldn't generate an explanation.
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