# Thread: orthogonal planes

1. ## orthogonal planes

I have to find equation to the plane that contains line (-1,1,2) +(3,2,4)t and is perpendicular to plane 2x+y-3x+4=0

2. ## Re: orthogonal planes

the plane will contain direction vector $\displaystyle <2,1,-3>$ correct? because that is the normal vector of the given

3. ## Re: orthogonal planes

Now, let's hope I don't toss you an incorrect one.

One way to describe a line is by giving a point on that line and one direction vector which describes the orientation of the line in space.
One way to describe a plane is by giving one point that is on the plane and two direction vectors that describe the orientation of the plane in space.

The general form of the equation for the line that is built in this fashion is:
$\displaystyle \vec{q}=\vec{p}+t\vec{d}$
where t is scalar that varies over the entire set of real numbers. The vector $\displaystyle \vec{p}$ is simply the point that is on the line. And the vector $\displaystyle \vec{d}$ gives the direction from that point. By varying t over the real numbers we can get in $\displaystyle \vec{q}$ the coordinates of any point that is on that line.

The situation with the plane is similar, but we need two direction vectors (cannot be colinear, or we are not describing a plane):
$\displaystyle \vec{q}=\vec{p}+r\vec{u}+s\vec{v}$
Here r and s vary over the real numbers.

We can describe a line in 2-dimensional space by alternative means, giving a point on the line and a perpendicular (to the line) vector.
And we can do the same in 3-dimensional space for the plane. We need a point and perpendicular (to the plane) vector.
We call these vectors normal. One line, respectively plane, can have many different normal vectors. All of them are though, colinear. For example if [1,1] is normal vector for some line, so are [2,2], [3,3],[4,4] etc.

The equation of the line, built this way, looks like this:
$\displaystyle (\vec{p} - \vec{q})\vec{n}=\vec{0}$
Where $\displaystyle \vec{p}$ is the known point on the plane, and $\displaystyle \vec{n}$ is a normal vector to that plane. Any point $\displaystyle \vec{q}$ that statisfies the equation is on the plane.

Both types of equation of the plane (or line) are convenient for slightly different purposes. The first one gives very quick way of generating arbitrary points, laying on the described plane (think 3D computer graphics applications). The second is way more useful in giving an answer if some specific point lies on the plane, described by the equation.

In addition, normal vectors have that wonderful property that if you have the "traditional" equation of the plane
$\displaystyle ax+by+cz=d$
the coefficients in front of the variables x, y, and z readily form normal vector to that plane. This is possible because any two planes
$\displaystyle ax+by+cz=d_1$ and $\displaystyle ax+by+cz=d_2$ (same coefficients, different constant d) are parallel (try to solve the system of these equations and check what do you get for their intersection). So normal vector to $\displaystyle ax+by+cz=d$ will be normal to $\displaystyle ax+by+cz=0$ too. And the normal vector to this plane is simply (a,b,c), because for any (x,y,z) on that plane [a,b,c] . [x, y, z] (sorry for writing the dot product so ugly) must be 0, otherwise the equation $\displaystyle ax+by+cz=0$ won't hold.

Now to the problem:

We are given a line, described by point and direction vector and that line is in our plane. We therefore have a point in our mystery plane [-1,1,2] and a direction vector [3,2,4]. We need another direction vector and we are almost ready. We are given perpendicular plane by it's "traditional" equation and we can immediately read off a normal vector [2,-1,3] this normal vector is the second direction vector we need. The equation is therefore

[-1,1,2]+r[3,2,4]+s[2,-1,3]

How we transfer from this equation to the "traditional" form? We need normal vector $\displaystyle \vec{n}$ such that
(1) $\displaystyle \vec{n}.[3,2,4]=0$
(2) $\displaystyle \vec{n}.[2,-1,3]=0$

which gives a nice system:

$\displaystyle 3n_1+2n_2+4n_3=0$
$\displaystyle 2n_1-n_2,+3n_3=0$

Tree variables with two equations that guarantees many valid solutions, all laying on one line.

$\displaystyle 3n_1+2n_2+4n_3=0$
+
$\displaystyle 4n_1-2n_2,+6n_3=0$

gets

$\displaystyle 7n_1+10n_3=0 \Leftrightarrow n_1=(-10/7)n_3$

gets

$\displaystyle 2(-10/7)n_3 - n_2 + 3n_3=0 \Leftrightarrow -n_2 + (-20/7)n_3 + (21/7)n_3=0 \Leftrightarrow$
$\displaystyle n_2=(1/7)n_3$

Choose for example $\displaystyle n_3=7$ which leads to [-10, 1, 7] and check

[-10,1,7].[3,2,4]=0
[-10,1,7].[2,-1,3]=0

We have our $\displaystyle \vec{n}=[-10,1,7]$ and we have our $\displaystyle \vec{p}=[-1,1,2]$

$\displaystyle (\vec{p} - \vec{q})\vec{n}=\vec{0}$ becomes
$\displaystyle [-1,1,2].[-10,1,7] - \vec{q}.[-10,1,7]=0 \Leftrightarrow \vec{q}[-10,1,7]=25$

So we have:

$\displaystyle -10q_1+q_2+7q_3=25$ change $\displaystyle q_1, q_2 and q_3$ with $\displaystyle x, y, z$

And then check how many mistakes have I made in the calculations

4. ## Re: orthogonal planes

For starters the normal vector of the perpendicular plane is not [2,-1,3], but [2,1,-3].
The idea is important