Find the distance between the straight line  and the plane 2x - 4y + 4z - 7 = 0 ?

CONCEPT:

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d and a point whose coordinate is, (x1, y1, z1)

Now, distance between the point and the plane = 

CALCULATION:

Given: Equation of line  and equation of plane is 2x - 4y + 4z - 7 = 0

First let's find out if they given line is parallel to the plane or not.

As we can see that, the direction ratios of the given line are: (0, 1, 1)

Similarly, the direction ratios of the normal to the given plane are: (2, -4, 4)

⇒ 2 × 0 - 1 × 4 + 1 × 4 = 0

So, the given line is parallel to the given plane.

Let's find out a point on the given line

As we can see that, point Q (3, - 1, 2) lies on the line.

So, finding the distance between the point Q and the given plane is same as finding the distance between the given line and plane.

As we know that, distance between a point and a plane is given by: 

Here, x1 = 3, y1 = - 1 and z1 = 2

⇒ 

Hence, option C is the correct answer.