Question

find the equation of plane​

Answer 1

Answer:

If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established. a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.

Step-by-step explanation:

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Answer 2

Answer:

In the first section of this chapter we saw a couple of equations of planes. However, none of those equations had three variables in them and were really extensions of graphs that we could look at in two dimensions. We would like a more general equation for planes.

So, let’s start by assuming that we know a point that is on the plane, P0=(x0,y0,z0)P0=(x0,y0,z0). Let’s also suppose that we have a vector that is orthogonal (perpendicular) to the plane, →n=⟨a,b,c⟩n→=⟨a,b,c⟩. This vector is called the normal vector. Now, assume thatP=(x,y,z)P=(x,y,z) is any point in the plane. Finally, since we are going to be working with vectors initially we’ll let →r0r0→and →rr→ be the position vectors for P0 and PPrespectively.

Here is a sketch of all these vectors.



Notice that we added in the vector →r−→r0r→−r0→which will lie completely in the plane. Also notice that we put the normal vector on the plane, but there is actually no reason to expect this to be the case. We put it here to illustrate the point. It is completely possible that the normal vector does not touch the plane in any way.

Now, because →nn→ is orthogonal to the plane, it’s also orthogonal to any vector that lies in the plane. In particular it’s orthogonal to →r−→r0r→−r0→. Recall from the Dot Product section that two orthogonal vectors will have a dot product of zero. In other words,

→n⋅(→r−→r0)=0⇒→n⋅→r=→n⋅→r0n→⋅(r→−r0→)=0⇒n→⋅r→=n→⋅r0→

This is called the vector equation of the plane.

A slightly more useful form of the equations is as follows. Start with the first form of the vector equation and write down a vector for the difference.

⟨a,b,c⟩⋅(⟨x,y,z⟩−⟨x0,y0,z0⟩)=0⟨a,b,c⟩⋅⟨x−x0,y−y0,z−z0⟩=0⟨a,b,c⟩⋅(⟨x,y,z⟩−⟨x0,y0,z0⟩)=0⟨a,b,c⟩⋅⟨x−x0,y−y0,z−z0⟩=0

Now, actually compute the dot product to get,

a(x−x0)+b(y−y0)+c(z−z0)=0a(x−x0)+b(y−y0)+c(z−z0)=0

Step-by-step explanation:

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