# MIT OCW

## 18.02 Multivariable Calculus

### Equations of Planes

#### Expression

The equation of a plane should be in the form of `$ax+by+cz=d$`

.

**e.g.** Plane through origin with normal vector `$\mathbf{N}=(1,5,10)$`

.

**Solution** Select a point `$P(x,y,z)$`

on the plane and we have `$\mathbf{OP}\cdot\mathbf{N}=0$`

, that is `$x+5y+10z=0$`

.

**e.g.** Plane through `$P_0(2,1,-1)$`

with normal vector `$\mathbf{N}=(1,5,10)$`

.

It's slightly different but what we will do is almost the same. Select a point `$P(x,y,z)$`

on the plane and we have `$\mathbf{P_0P}\cdot\mathbf{N}=0$`

, that is `$(x-2)+5(y-1)+10(z+1)=0$`

, simplified as `$x+5y+10z=-3$`

.

Observe these two examples, their coefficients before `$x,y,z$`

are exactly the components of the normal vector `$\mathbf{N}$`

, the only difference is just the constant. Here the constant can be understood as some offset to origin, since there is no offset when the constant is `$0$`

, namely through the origin.

After we know the fact, we can solve the second example in an easier way. Because we've known the form in the second example is `$x+5y+10z=d$`

, we just plug `$P_0(2,1-1)$`

into it and solve the value of `$d$`

, finally we get the whole equation.

#### Position

If we know a vector and the equation of a plane, we can judge their relationship by the above trick.

**e.g.** Given a vector `$\mathbf{v}=(1,2,-1)$`

and a plane `$x+y+3z=5$`

, what's their relationship? Parallel / Perpendicular / Neither.

**Solution** The normal vector of the plane is `$\mathbf{N}=(1,1,3)$`

, and we find that `$\mathbf{N}\cdot\mathbf{v}=0$`

. Caution! It DOES NOT mean `$\mathbf{v}$`

is perpendicular to the plane, instead, what we check is `$\mathbf{v}$`

and the *normal* vector `$\mathbf{N}$`

of the plane. Therefore, it oppositely suggests that the vector `$\mathbf{v}$`

is parallel to the plane.

### Linear Systems

Think about a `$3\times 3$`

linear system, for example, `$$\left\{\begin{alignedat}{4} x & {}+{} & & {}{} & z & {}={} & 1 \\ x & {}+{} & y & {} {} & & {}={} & 2 \\ x & {}+{} & 2y & {}+{} & 3z & {}={} & 3\end{alignedat}\right..$$`

These three equations are all planes, and what's the meaning of a solution?

In geometry, we can only find only one solution when the first two planes intersect in a line, and the line intersects with the third plane with a point. To solve this, a good way is to use matrix, a.k.a. `$\mathbf{A}\mathbf{X}=\mathbf{B}\Leftrightarrow\mathbf{X}=\mathbf{A}^{-1}\mathbf{B}$`

.

Anyhow, there exists some weird situtations and the method doesn't work. Generally, the solution of the system have four possibilities:

- No solutions
- One point (unique solution)
- A line (infinite solutions)
- A plane (infinite solutions)

If two planes are parallel, then there exists no solutions at all.

If three planes are all the same, then there exists infinite solutions, namely a plane.

If two planes intersect in a line, and the thrid plane is parallel to the line (not contained), then there exists no solutions.

If two planes intersect in a line, and the thrid plane is contain the line, then there exists infinite solutions, namely a line.

#### Invertible Matrices

What's above method going wrong in algebra? The formula `$\mathbf{A}^{-1}=\frac{\operatorname{adj}(\mathbf{A})}{\det(\mathbf{A})}$`

is definitely correct, but what will happen if `$\det(\mathbf{A})=0$`

? We say that those matrices whose determinant `$\det(\mathbf{A})\neq 0$`

are invertible and they have inverse, the others are not invertible and don't have inverse.

#### Homogeneous Systems

A homogeneous system has the form `$\mathbf{A}\mathbf{X}=0$`

，we can find that no matter what `$\mathbf{A}$`

is, obviously `$\mathbf{X}=0$`

is always a solution, which is called a trivial solution.

If `$\det(\mathbf{A})\neq 0$`

, `$\mathbf{X}=0$`

is the unique solution.

If `$\det(\mathbf{A})=0$`

, then `$\det(\mathbf{N_1},\mathbf{N_2},\mathbf{N_3})=0$`

, which means their normal vectors are coplanar. Then there exists nontrivial solutions like `$\mathbf{X}=\mathbf{N_1}\times\mathbf{N_2}$`

.

#### General Cases

For a system `$\mathbf{A}\mathbf{X}=\mathbf{B}$`

- if
`$\det(\mathbf{A})\neq 0$`

, then it has unique solution - if
`$\det(\mathbf{A})=0$`

, then it has either no solutions or infinite solutions.

The more details will be found when trying to solve the system by elimination and substitution, but if we know `$\det(\mathbf{A})=0$`

and find a solution, then it has to have infinite solutions; and when we get something contradictory, then it turns out that the system has no solutions.