# Linear Algebra Matrix Transformation to plane

1. Oct 26, 2014

### FlorenceC

Find the matrix for the transformation that projects each point in R3 (3-D) perpendicularly onto the plane 7x + y + 3z = 0 .

The attempt at a solution is attached for question 1 (actually instructor's solution)

I kind of understand it but ...
why is n <dot> v = equation of the plane?
Does v represent all of the possible points of R^3 (certainly does not seem so...) which is projected to the normal?
I understand the v-projection is there to get the projection of v onto the plane because we cannot directly project to the plane right? But why do we want v' what does v' represent and how is that the solution?

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Last edited: Oct 27, 2014
2. Oct 27, 2014

### Staff: Mentor

There is no solution attached. I moved this post from the Homework & Coursework section because it is less about how to do the problem than it is about general questions about vectors and projections onto vectors.

3. Oct 27, 2014

### RUber

$\vec n \cdot \vec v = \|\vec n\| \|\vec v\| \cos \theta$.
In short, if $\vec n$ is a unit vector, in your example $\frac {\vec n}{\| \vec n \|}$, you can draw the right triangle with v as the hypotenuse and n the adjacent leg with theta the angle between them.
This is why the dot product is used for projections, but the result is just a length (scalar). When you are projecting onto the normal vector, you have to multiply the length you found for that projection by the unit normal.
This is not on the plane, but the part of $\vec v$ that is normal to the plane.
If you subtract that piece off of the original vector, what is leftover is the part of the vector that lies in the plane. That is what you are calling v'.
Using this method, v is any vector pointing to any x,y,z in $\mathbb{R}^3$.

4. Oct 28, 2014

### HallsofIvy

Staff Emeritus
It isn't. But I think what you meant to say is that "the equation of the plane is n dot v= 0." where n is the normal to the plane and v is the "position vector" of a point in the plane. Also, this only applies to the case of a plane that contains the origin- the only kind of plane that is a "subspace". Since the plane contains the origin, the "position vector" of a point in the plane lies in the plane itself. Since "v" is normal to the plane, it is perpendicular to every vector in the plane. In particular, n is perpendicular to vector v so their dot product is 0.

No, v represents (is the position vector of) all points in the plane.

5. Oct 28, 2014

### RUber

Remember that any vector can be written as the sum of two (or more) orthogonal vectors. In this example, you are breaking $\vec v$ into two parts, one perpendicalar to the plane (i.e. parallel to the normal) and the projection $\vec v'$ that lies in the plane.
$\vec v = (\vec v \cdot \hat n) \hat n + \vec v'$ where $\hat n = \frac {\vec n }{\| \vec n \|}$.
Solving for the projection into the plane gives you the equation
$\vec v'=\vec v-(\vec v \cdot \hat n) \hat n$.
If you wanted to project directly only the plane, you would first need to find the basis vectors for the plane.
A simple example is if you have a vector $\vec M = a\hat x + b\hat y + c\hat z \in \mathbb{R}^3$ and you wanted to project onto the x,y plane. The standard basis vectors for the x,y plane are $\hat x, \hat y$.
You could project $\vec M$ onto the basis vectors one at a time and sum them, i.e.:
$(\vec M \cdot \hat x )\hat x = a \hat x$
$(\vec M \cdot \hat y )\hat y = b \hat y$
$\vec M ' = a \hat x + b \hat y$
Or you could notice that $\hat z$ is normal to the x,y plane. In fact, the x,y plane is defined in $\mathbb{R}^3$ by $z=0$.
Then using the method above, simply subtract off the part of $\vec M$ which is normal to the plane and the result will be the projection into the plane.
$(\vec M \cdot \hat z )\hat z = c \hat z$
$\vec M ' = \vec M -(\vec M \cdot \hat z )\hat z = a \hat x + b \hat y$.
Both methods get you to the same answer, but when you are given the normal vector, it should be clear which approach is more efficient.