Find the distance

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