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# Vectors and Planes - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We’ve been talking about the equation of a plane. One really important equation is the vector equation of a plane. It’s a very simple equation but I want to make sure you understand before we talk about it any further. We have this normal vector n, and point A is the point at the base of that vector.

Let’s take P, any other point in the plane. Observe that, if n is a normal vector it’s going to be perpendicular to the vector AP. You know that when two vectors are perpendicular, their dot product is zero. I can say that n dot vector AP is 0. And this is vector AP right here.

Let me name these two vectors. This is the position vector for point A. Let’s call it r0, from the origin to point A. And the position vector for point P, this is just any arbitrary point on the plane is going to be r. Notice that r0 plus AP equals r. Let me write that down; r0 plus AP equals r. That means that AP is r minus r0.

Another way to write our vector equation for a plane is n.r minus r0, equals 0. The advantage of this is, you only need to know two things; the normal vector and a position vector for one point in the plane. That will be your given point A.

Let’s take a look at an example. This problem says find a vector and a rectangular equation for a plane with normal <1, -2, 2>. This is our normal vector n, which includes the point <3, 0, -5>. I'm going to make my initial point, this is my initial point a, and so I'm going to make position vector r0 equal to <3, 0, -5>. I have this equation that I've just shown you, n.r minus r0 equals zero. the dot product being zero means that these two vectors are perpendicular. n is <1, -2, 2> and r is just the position vector for any point on the plane. So it's going to have components <x, y, z>. minus r0. r0 is <3, 0, -5>.

Now I need this dot product to equal zero. Let me simplify in the parenthesis here, you know how to do vector subtraction. You just subtract component wise, you have x minus 3, y minus 0 which is y and z minus -5, z plus 5. So here's a simplified vector equation for the plane. This dot product equals zero. And that's all your teacher will want. If she wants a vector equation for the plane, this is a good answer.

Now let's say that you need a rectangular equation. You can very easily get it from this vector equation. Just perform the dot product. And remember the way that works is you multiply the components. 1 times x minus 3, is x minus 3, plus -2 times y, plus 2 times z, plus 5, equals 0. So you have x minus 3, I'm going to put the constants on the right side, so I'll add 3. You'll have minus 2y and you'll have 2z plus 10. When you subtract that 10, I'll have it over here. And so we have x minus 2y plus 2z equals -7. That's our plane. So that's a rectangular equation for the plane in the form ax plus by plus cz equals d. And this is a vector equation for the plane. Let me circle this. Those are the two basic forms for the equation of a plane.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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