Find the distance between the point and the plane.

Point: P(0, 0, 0)

Line: 2x + 3y + z = 12

My Work

Let n = normal vector

n = <2, 3, 1>

I now need a second point. Call it Point Q.

No, you need to find

**the** point at which this line intersects the given plane.

Let y = 0 and z = 0 in the given line.

2x + 3(0) + 0 = 12

2x + 0 = 12

2x = 12

x = 12/2

x = 6

Point Q is (6, 0, 0).

I also need vector PQ.

PQ = <6-0, 0-0, 0-0>

PQ = <6, 0, 0>

To find the distance, we use the distance formula below.

D = |PQ*n|/||n||

After plugging the information above into the formula, my answer is D = root{14}/7.

Okay, you have found the distance from P to this arbitrary point Q. What does that have to do with the distance from P to the plane?

The book's answer is

[6*root{14}]/7.

Why am I wrong?

Why should you be right? You find an arbitrary second point on the line and determine the distance from (0, 0, 0) to that. That has

**nothing** to do with the plane! You need to find

**the** point at which the line through (0, 0, 0), perpendicular to the plane,

**intersects the plane**!

Any line perpendicular to this plane has "direction" given by the vector <2, 3, 1>. Such a line, passing through (0, 0, 0), has parametric equation x= 2t, y= 3t, z= t.

To determine where that line intersects the plane, replace (x, y, z) in the equation of the plane and solve for t.

The point, Q, you found has nothing to do with the plane. It is just another point on the line. You

**could** find a second, arbitrary point on the line in order to use the "two point" form for the line. But in three dimensions, that involves first finding the "direction vector" of the line- and that

**is** the normal vector to the plane so you would be doing the same work two different ways.