# Equations of planes and the meaning of the d value

1. Apr 27, 2014

equations of planes and the meaning of the "d" value

1. The problem statement, all variables and given/known data
Find the equations of the following plane

Through (2, 0, 2), (0, 1, 2) and (2, 1, -1)

2. Relevant equations
standard form of equation of a plane:
$ax + by + cz = d$

3. The attempt at a solution
hello
I'm trying to conceptually understand how to derive the value for "d".

I understand to find the a b and c coefficients you find the vectors which make the outside of the plane, take the cross product of the two to find the normal vector and then using that result. However, in my solutions it tells me that
"since (2; 0; 2) is in the plane d = 10"

But I have no idea on how that works. I've searched online and had answers such as "d is the distance from the origin" which makes sense. But the magnitude of those three points or the coefficients of the plane do not equal to zero. I can't figure out the algorithm to find "d", does anyone know?

Thanks

2. Apr 27, 2014

### vanhees71

You find a normal vector $\vec{n}$ to the plane. Given the three points that's the cross product you mention in your posting. Then obviously the vector $\vec{x}$ is on the plane, if and only if
$$\vec{n} \cdot (\vec{x}-\vec{x}_0)=0$$
for any fixed vector $\vec{x}_0$ on the plane. You should carefully think about why this is the case. This finally answers the question, how to get the value $d$.

3. Apr 27, 2014

### ehild

You get the distance of the plane from the origin if you draw a line from the origin perpendicularly to the plane: The distance is the length of that line. See the attachment.
The projection of any vector of the plane on that normal is equal to the distane of the plane from the origin.

The projection of a vector a to a direction n is equal to the scalar product of a with the unit vector no in the direction of n. $a \cos(\alpha) =\vec a \cdot \vec{n_0}$

So the distance of the plane from the origin is $d=\vec {OP_0}\cdot\vec {n_0}$

ehild

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4. Apr 27, 2014

### LCKurtz

There isn't a value for $d$. It can be anything. Multiply the equation by a constant and you have a different number on the right side.

If you have a normal vector <a,b,c> then you know the equation of the plane is $ax+by+cz=D$ for some $D$. Various values of $D$ give parallel planes. If you plug in a point you know is on the plane, that will give you $D$ for that plane and that normal vector <a,b,c>.

5. Apr 27, 2014

### Ray Vickson

There are two fundamentally different cases: (i) d = 0; and (ii) d ≠ 0.

In case (i) the plane passes through the origin (0,0,0); in case (ii) the plane misses the origin, and in that case, d can be any nonzero number. The reason for this is that you can multiply both sides of the equation by a nonzero number and still have an equivalent equation (that is, the same plane). In other words, if you start with $ax + by + cz = d$ you can multiply by any nonzero number $k$ to get the equivalent equation $(ka)x + (kb)y + (kc)z = kd$. You could always take d = 1 on the right and then find a, b and c from the resulting three equations. If you don't like the appearance of the final solution, you can get a nicer answer by multiplying through by a convenient constant r and use $a_{new} = ra, b_{new} = rb, c_{new} = rc, d_{new} = r$.

6. Apr 28, 2014

### LCKurtz

Don't you all just love it when someone posts a question, gets several helpful replies, and never returns?