# Exercise review: perpendicular-to-plane line

1. Jun 28, 2012

### bznm

Please, can you check the following execution is correct?
Problem text
I have a plane in affine space in R^4 described by two following equations:
$$\begin{Bmatrix}3x+y-z-q +1=0\\ -x-y+z+2q=0\end{Bmatrix}$$

I have the coords of a point P: (0,1,1,0)
Describe the locus of points Q such that line PQ is perpendicular to the plane.

My solution

Now I look for the line (which Q points belong to) perpendicular to the plane and passing by point P.
In order for a plane described by equation $$ax+by+cz+dq+e=0$$ and a line whose coefficients are $$l,m,n,t$$ to be perpendicular, this must be true:

$$\frac{a}{l}=\frac{b}{m}=\frac{c}{n}=\frac{d}{t}$$

So, the equations of line passing by generic point $$P(x_0,y_0,z_0,q_0)$$ is:

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}=\frac{q-q_0}{d}$$

Then, in my specific problem I get:

$$\begin{Bmatrix}x=y-1 \\ y=2-z \\ z=\frac{q}{2}+1 \end{Bmatrix}$$

Thank you very very much

2. Jun 28, 2012

### vela

Staff Emeritus
That equation doesn't describe a plane in ℝ4. A single equation eliminates one degree of freedom, so you still have three left. In other words, that's the equation describing a three-space. Note that the plane in the problem was specified using two equations.

It might help you to consider the analogous case in ℝ3. You're given a line L and a point P and asked to find all points Q such that PQ is perpendicular to L.

3. Jun 29, 2012

### bznm

Thanks for the helpful answer, you're right.
But the analoguos case in R^3 (which I know how to solve) is quite different from my exercise.
Which equation would I choose to get the normal vector?
Thanks again

4. Jun 29, 2012

### micromass

Staff Emeritus
Well, how would you solve it?

5. Jun 29, 2012

### bznm

6. Jun 29, 2012

### micromass

Staff Emeritus
That are the points perpendicular to a plane. That was not what vela asked. He asked about the points perpendicular to a line in $\mathbb{R}^3$. Do you know how to do that?? Your question is very analogous.

7. Jun 29, 2012

### bznm

Oops, I didn't realize... Coming back to vela answer, I'd find a point on the given line, I'd get the segment that connects it with given point, and working with projections I'd get the perpendicular segment between the two points, finally get the line. But can you help me seeing why the two questions are so similar?

8. Jun 29, 2012

### HallsofIvy

Staff Emeritus
The plane is defined by the two lines 3x+ y- z- q- 1= 0 and -x- y+ z+ 2q= 0. If we add the two equations, y and z both cancel leaving 2x+ q+ 1= 0 so, on this plane, x= -(1/2)q- 1/2. Putting that into the second equation, (1/2)q+ 1/2- y+ z+ 2q= -y+ z+ (5/2)q+ 1/2= 0. The coefficients of x, y, z, and q in that are 0, -1, 1, and 5/2, respectively so a normal to the plane is <0, -1, 1, 5/2>. Find parametric equations of the line through (0, 1, 1, 0) having direction vector <0, -1, 1, 5/2> and determine where that line crosses the plane.

9. Jun 29, 2012

### bznm

Perfect, thank you so much!
so is this correct?
$$L=(0,1,1,0)+ t(0,-1,1,5/2)$$

10. Jun 29, 2012

### vela

Staff Emeritus
I'm going to have to disagree with HallsofIvy here. Neither of those equations define lines in ℝ4. They define 3-spaces, and it's their intersection which is the plane. Also, in ℝ4, there is not one direction that is normal to a plane, unlike in ℝ3. For example, consider the xy-plane in ℝ4. Both the z-axis and the q-axis are perpendicular to the plane, but they clearly aren't multiples of each other.

11. Jun 30, 2012

### HallsofIvy

Staff Emeritus
Thanks, vela. You are right. For some reason it just never dawned on me that we were really talking about R4.

12. Jul 2, 2012

### bznm

Hi vela, hi HallsofIvy.
So, should I find the orthogonal complement of the intersection of two 3-dimensional subspaces?

Thanks a lot again