Find tthe equation of the plane
Answers
Step-by-step explanation:
the y-z-plane has standard equation x = 0 and the x-z-plane has standard equation y = 0. A plane parallel to the x-y-plane must have a standard equation z = d for some d, since it has normal vector k. A plane parallel to the y-z-plane has equation x = d, and one parallel to the x-z-plane has equation y = d.
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HERE IS YOUR ANS.
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.
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So, let’s start by assuming that we know a point that is on the plane,
P0=(x0,y0,z0)
. Let’s also suppose that we have a vector that is orthogonal (perpendicular) to the plane,
→n=⟨a,b,c⟩
. This vector is called the normal vector. Now, assume that
P=(x,y,z)
is any point in the plane. Finally, since we are going to be working with vectors initially we’ll let
→r0
and
→r
be the position vectors for P0 and P
respectively.
This graph has a standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. On the graph is the graph of a plane that is above the xy-plane. The point $P_{0}$ is on the plane and to the right of the point P. The vector $\vec{r}-\vec{r}){0}$ is on the line starting at $P_{0}$ and ending at P is on the plane. The vector $\vec{n}$ is also shown on the plane and normal to the plane. The position vectors $\vec{r}_{0}$ and $\vec{r}$ are also shown on the graph as well as the vector $\vec{v}$ which is shown starting at the origin and parallel to the line.
Notice that we added in the vector
→r−→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
→n
is orthogonal to the plane, it’s also orthogonal to any vector that lies in the plane. In particular it’s orthogonal to
→r−→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⋅→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
Now, actually compute the dot product to get,
a(x−x0)+b(y−y0)+c(z−z0)=0
This is called the scalar equation of plane. Often this will be written as,
ax+by+cz=d
where d=ax0+by0+cz0.
This second form is often how we are given equations of planes. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. A normal vector is,
→n=⟨a,b,c⟩
hope it helps.
AKASH KUMAR OR ITEMHEAVEN.