# Finding a plane based off 2 points and the z axis

1. Jul 13, 2014

### hagobarcos

1. The problem statement, all variables and given/known data

Find an equation of the plane that passes through the points (4,2,1) and (-2, 9, 6), and is parallel to the z axis.

2. Relevant equations

a(X-Xo) + b(Y-Yo) + c(Z-Zo) = 0

3. The attempt at a solution

Okay so for this one, I first tried to make a vector out of the two given points.
If P(4,2,1) and Q(-2,9,6),

then PQ = <-6,7,5>.

Next I fumbled around a bit trying to figure out how to either a) find another vector to form a cross product with, or b) somehow figure out what the <a,b,c> numbers should look like in order to be parallel to z axis.

When I tried approach a) I came up with another point on the z axis, (0,0,1), used it as a vector of its own, crossed it with PQ, picked up the resultant, and plugged in the direction numbers to the plane equation.

For part b) I tried making a vector out of (0,0,1) and P, crossing it with PQ and using the resultant direction numbers.

However, both approaches seem to be rather wrong.

#### Attached Files:

• ###### Cal Hw 11.5.jpg
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2. Jul 13, 2014

### SammyS

Staff Emeritus
Why pick a point on the z-axis?

Instead, use a vector which is parallel to the z-axis .

3. Jul 14, 2014

### hagobarcos

Does the unit vector z count? <0,0,1>

4. Jul 14, 2014

### SammyS

Staff Emeritus
Why not?

Beginning or ending at either of the given points, where does this vector take you ?

5. Jul 14, 2014

### hagobarcos

Well, since the unit vector always points up by 1 unit, for point P(4,2,1), the vector says to go up one to (4, 2, 2).

For point Q(-2,9,6), it says go up one unit in the z axis, to (-2,9,8).

With these new points, the cross product of the resulting vectors should yield the direction numbers <a,b,c> of the normal vector to the plane that I am trying to describe, no?

6. Jul 14, 2014

### hagobarcos

Maybe this is still wrong, but I have a resultant of 7i + 6j + 0k from crossing <0,0,1> with <-6,7,5>, which I then used to produce 7(x-4) + 6(y-2) = 0, which turns out to

7x + 6y - 40 = 0.

However, upon asking Wolfram Alpha to plot it, I got a line!

7. Jul 14, 2014

### hagobarcos

Well actually, after some meddling, got this:

#### Attached Files:

• ###### 11.5 plane.jpg
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8. Jul 14, 2014

### SammyS

Staff Emeritus
That plane does contain the two points, and is parallel to (or contains) <0, 0, 1>, right? Since the plane doesn't pass through the origin, it's parallel to the z-axis.

9. Jul 15, 2014

### hagobarcos

Excellent. Thank you for the assistance :D