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# Defining a plane in R3 with a point and normal vector

Determining the equation for a plane in R3 using a point on the plane and a normal vector. Created by Sal Khan.

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• Could we just say that we can find a vector on the plane with the two points (x, y, z) and (x0, y0, z0)? •  If you only gave me two points, there are an infinite number of planes that can pass through it. Think of a piece of paper passing through two points. You can rotate the piece of paper however you want and it can still hold those two points.

The difference is that we knew the vector PERPENDICULAR to the plane, too. Now look at your piece of paper. How many ways can you fit a piece of paper that goes through two points and also has a vector (which you can represent with a pencil) going straight up in the air? One.
• At Sal starts talking about "naught" or "not" in relation to X0. He never defines what not/naught means. Does it just mean that a variable is perpendicular to it's non-not/naught counterpart? (Meaning X is perpendicular to X0) • At , why is this method of defining a plane in R^3 preferred by computer game programmers? • The reason this is a useful solution as it uses less memory. For example, you could define a plane using 3 points contained on the plane. This would use 9 double values at 4 bytes each. Using a point and a vector (or just two points one of which is off the plane) takes up 6 doubles.

Its also useful to have the perpendicular vector for the plane handy. For example, if you want to find the minimum distance of a point to a plane, you will need the plane vector.
• What is he doing at ? Why is he adding the two vectors ? Vector n and the position vector x0 ? • That's just the name that he's giving to this method of defining a plane. In other words, Sal is calling it the "Normal vector and point definition of a plane". Then, as the video progresses, he shows how this definition of a plane can, via basic algebra, be rewritten in the form Ax+By+Cz = D (another equivalent definition).

Thus, just as there is more than one way to define a line (for instance, there is "Slope-Intercept form" and "Point-Slope form" as well as others) there is also more than one way to define a plane--that's the main point of the video, I believe.

Hope this helps!
• Looking at the Plane Equation (AX + BY + CZ = D): (A, B, C) are components in the Normal to a plane and (X, Y, Z) are components in a vector that lies on the same plane. What does 'D' represent? I don't see what D tangibly stands for, because the plane equation almost always ends up being set equal to 0 as the math collapses. • If A^2 + B^2 + C^2 = 1 then D is the distance of the closest point on the plane to the origin. In other words D is the scalar projection of every point on the plane onto the vector <A, B, C>. In every case, even when A^2 + B^2 + C^2 is not one, D is the dot product of every vector with endpoint on the plane with the vector <A, B, C> because the plane equation really means:
<A, B, C> . <x, y, z> = D. D = 0 only if the plane passes thru the origin.
• Having a little trouble understanding this intuitively. So the equation of a plane is Ax + By + Cz = D. Taking the dot product between a vector ON the plane and a vector perpendicular to the plane gives us an equation in a similar form. But WHY does this have to be the equation of the plane? How does taking the dot product between these two vectors end up as the equation of the plane? • The idea is that we take the dot product between the normal vector and every vector (specifically, the difference between every position x and a fixed point on the plane x0). Note that x contains variables x, y and z. Then we solve for when that dot product is equal to zero, because this will give us every vector which is parallel to the plane. Visualise this, and you will realise that this set of vectors defines an entire plane parallel to our original plane (because every vector which is normal to the normal vector must lie on the plane).
The solution to our dot product equation will then give us an expression for x, y and z for which we obtain x - x0 perpendicular to n, which gives us the equation for the plane in position space.
• At , the perpendicular line should be n-vector rite? (nt n-cap) • Yes. For this example, Sal has decided to use direction vector n that can be any (non-zero) length to indicate the direction of the plane and position vector x_0 to indicate a point on the plane. In practice, you would want to make n a unit vector because it makes your calculations easier and any non-zero length vector n is acceptable, so why not make it have a length of 1 to make your life easier? That's why he keeps slipping up writing n-hat or n-cap even though he's trying to point out that n doesn't have to be unit length. Good habits are hard to break.
• Sorry for this question is not a math question. I am not a native English speaker, but I am understanding this video quite well, but I just wanna make sure how it is pronounced either X0, N0, or whatever.
Would it be "X not"? • I have a question and it may be a silly one.
The coordinates of a vector specify its end position correct? If so what specifies where the vector is going to "start" from? What makes the normal vector start from the plane instead of origin? We obviously don't want it to start from origin. • A vector can start and finish at any point where it's defined, we can move it around and it will still hold the same values.

Usually, we let the vector start at the origin, but it will still be the same vector if we let it start from somewhere else (again, where it is defined).

The normal vector has similar properties, we might have a normal n=(1,1,1), and this vector n can also be moved around, as long as its tail is on the plane. If the origin is a part of the plane, we might aswell let the normal start from the origin or move it around on the plane however we want.

When we move a vector, we have to preserve the length and direction properties. So we cant rotate it or try to make it longer or smaller when wemove it around.
• Are there any videos of Sal explaining the "regular linear equation" for planes?
I find it difficult to understand how Ax+By+Cz = d could define a plane. • The closest match to a good Khan Academy video on your question (not helpful in this case):

An oversimplified answer to your question would just be to compare the equation for a line in R2 to the equation for a plane in R3.

A line is a continuous set of line segments that goes on infinitely in 2 dimensions, defined by a slope (rate of change) and a point. Line segments come from R1, the next lowest dimension, and they're made up of an infinite number of points, which come from R0.
The equation for a line in R2 is Ax + By = C; it involves the two dimensions that make up R2.

Remember, when you go from R2 to R3, you add a new directionality at a 90 degree angle to the other underlying dimensions. If you already have length and width, it's like adding height. It's like the difference between a square and a cube. In linear algebra, it's like going from just the unit vectors i and j to i, j, and k.

So, going back to your question about how Ax + By + Cz = D could possibly define a plane in R3, the only difference between that equation and the equation for a line is the variable of z, the new directionality you get when you go from R2 to R3. Comparing the equation for a plane in R3 to the equation for a line in R2 makes this clear.

A plane could be viewed as a continuous set of lines (although this isn't really true because it's actually made up of points), defined by a vector normal to the plane (slope) and a point. Lines come from R2, the next lowest dimension, and lines are made up of line segments, which come from R1, and line segments are made up of an infinite number of points, which come from R0.
The equation for a plane in R3 is Ax + By + Cz = D; it involves the three dimensions that make up R3.