MIT OCW

18.02 Multivariable Calculus

Equations of Lines

We have known that a line can be treated as the intersection of two planes, so it's okay to describe a line by two equations, but not that natural.
Here's another way, a line can be treated as the trajectory of a moving point.

e.g. Line through $Q_0(-1,2,2),Q_1(1,3,-1)$
Solution Select a point $Q(x,y,z)$ on the plane, and we should have $\mathbf{Q_0Q}=t\mathbf{Q_0Q_1}=(2,1,-3)$, for some $t$. Let's consider $t$ as a parameter and expand the formula that is $$\begin{bmatrix}x+1\\y-2\\z-2\end{bmatrix}=t\begin{bmatrix}2\\1\\-3\end{bmatrix}.$$
We can write $x,y,z$ as functions with respect to $t$ as the following $$\left\{\begin{aligned}x(t)&=&-1&+2t\\y(t)&=&2&+t\\z(t)&=&2&-3t\end{aligned}\right..$$

Intersection with a Plane

e.g. What's $Q_0(-1,2,2)$ and $Q_1(1,3,-1)$ like relative to $x+2y+4z=7$?
Same side / Opposite sides / One is on the plane.
Solution The plane $x+2y+4z=7$ is actually dividing the whole space into two regions where one fulfills $x+2y+4z>7$ and the other $x+2y+4z<7$. We just plugin these two points into the plane equation and check the equality.
For $Q_0$, $-1+2\times 2+4\times 2=11>7$; and for $Q_1$, $1+2\times 3+4\times(-1)=3<7$, so neither of them is on the plane and they are on the opposite sides.
To get the intersection, we just plug the parametric equation of the line into the plane equation that is $$x(t)+2y(t)+4z(t)=7,$$
after solving, we can get $t=\frac12$, and indicating the intersection is $(0,\frac25,\frac12)$.

Parametric Equations of Curves

Cycloid

The whole process is just a trick in geometry, the parametric equation for a cycloid with radius $a$ is
$$ \left\{ \begin{aligned} x(\theta)&=a(\theta-\sin(\theta))\\ y(\theta)&=a(1-\cos(\theta)) \end{aligned} \right.. $$
Let's observe the edge like the place near $\theta=0,2\pi,\cdots$. Without loss of generality, take $\theta=0$ as an example.
According to Taylor's theorem, $\sin(\theta)\sim\theta-\frac{\theta^3}{3!}$ and $\cos\theta\sim1-\frac{\theta^2}{2!}$, consider the slope near $\theta=0$ that is $\frac{y(t)}{x(t)}=\frac{1-\cos(\theta)}{\theta-\sin(\theta)}~\frac{\frac{\theta^2}{2!}}{\frac{\theta^3}{3!}}=\frac{3}{\theta}\to\infty$, indicating that the behavior near $\theta=0$ is to toggle some motion, which is obvious to be understood when we imagine a wheel rolling on the ground.

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.

MIT OCW

18.02 Multivariable Calculus

Matrices

Linear Transformation

If we want apply some operation on a vector $\mathbf{x}=(x_1,x_2)$ to change it into $\mathbf{x'}=(-x_2,x_1)$, the process can be described as a matrix.
Given a vector $\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}$ and a matrix $\mathbf{R}=\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$, a new vector will be generated after we apply the matrix $\mathbf{R}$ to $\mathbf{x}$ that is $$\mathbf{R}\mathbf{x}=\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}-x_2\\x_1\end{bmatrix}.$$

A more concrete example is that if we know how the target vector $\mathbf{u}=\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}$ is derived from $\mathbf{x}=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$, for instance $$\left\{\begin{aligned}u_1&=2x_1+3x_2+3x_3\\u_2&=2x_1+4x_2+5x_3\\u_3&=x_1+x_2+2x_3\end{aligned}\right.,$$ then we can intermediately write such operation as a matrix that is $$\mathbf{A}=\begin{bmatrix}2&3&3\\2&4&5\\1&1&2\end{bmatrix}.$$
We now have the following equation $$\mathbf{A}\mathbf{x}=\mathbf{u}.$$

Minor and Cofactor

For a matrix $\mathbf{A}$, delete the row and the column where a specific element $a_{ij}$ exactly lies, and we get a smaller matrix. Treat it as a determinant, and that's called the minor of $a_{ij}$.
The cofactor is almost the same as the minor, but have something to do with signs, which means that for the element $a_{ij}$, we get the cofactor of it by multiplying $(-1)^{i+j}$ before the minor.
Write the cofactor of some element $a_{ij}$ as $A_{ij}$, and compose a new matrix $$\mathbf{C}=\begin{bmatrix} A_{11} & A_{12} & A_{13} & \cdots \\ A_{21} & A_{22} & A_{23} & \cdots \\ A_{31} & A_{32} & A_{33} & \cdots \\ \vdots & \cdots & \cdots & \cdots \end{bmatrix},$$ which is called the cofactor matrix, namely comatrix.

Transpose

For a matrix $\mathbf{A}$, the transpose matrix of it $\mathbf{A}^{\mathsf{T}}$ is derived from $\mathbf{A}$ where one's rows are the other's columns in the same order, vice versa.
Concretely, take $\mathbf{A}=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ as an example, the tranpose matrix $\mathbf{A}^{\mathsf{T}}=\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix}$.

Adjugate

For a matrix $\mathbf{A}$, the adjugate matrix of it $\operatorname{adj}(\mathbf{a})=\mathbf{C}^{\mathsf{T}}$, where $\mathbf{C}$ is the comatrix.

Inverse

The inverse of a matrix $\mathbf{A}$ with the notation $\mathbf{A}^{-1}$ fulfills $$\mathbf{A}\mathbf{A}^{-1}=\mathbf{I},\mathbf{A}^{-1}\mathbf{A}=\mathbf{I},$$ where $\mathbf{I}$ is an identity matrix.
For an equation $\mathbf{A}\mathbf{X}=\mathbf{B}$, if we want to solve $\mathbf{X}$, we just apply $\mathbf{A}^{-1}$ to the both sides thus get $$\mathbf{A}^{-1}(\mathbf{A}\mathbf{X})=\mathbf{A}^{-1}\mathbf{B},$$ namely $\mathbf{X}=\mathbf{A}^{-1}\mathbf{B}$.
To calculate the inverse of a matrix $A$, we need to know the adjugate matrix $\operatorname{adj}(\mathbf{A})$ and the determinant of $\mathbf{A}$ to in the following formula $$\mathbf{A}^{-1}=\frac{\operatorname{adj}(\mathbf{A})}{\det(\mathbf{A})}.$$
e.g. Calculate the inverse of $\mathbf{A}=\begin{bmatrix}2&3&3\\2&4&5\\1&1&2\end{bmatrix}$.
Solution
First, we calculate the comatrix
$$\mathbf{C}=\begin{bmatrix} +\begin{vmatrix}4&5\\1&2\end{vmatrix}& -\begin{vmatrix}2&5\\1&2\end{vmatrix}& +\begin{vmatrix}2&4\\1&1\end{vmatrix}\\ -\begin{vmatrix}3&3\\1&2\end{vmatrix}& +\begin{vmatrix}2&3\\1&2\end{vmatrix}& -\begin{vmatrix}2&3\\1&1\end{vmatrix}\\ +\begin{vmatrix}3&3\\4&5\end{vmatrix}& -\begin{vmatrix}2&3\\2&5\end{vmatrix}& +\begin{vmatrix}2&3\\2&4\end{vmatrix} \end{bmatrix}=\begin{bmatrix} +3 & -(-1) & +(-2)\\ -3 & +1 & -(-1)\\ +3 & -4 & +(-2) \end{bmatrix}=\begin{bmatrix} 3 & 1 & -2\\ -3 & 1 & 1\\ 3 & -4 & 2 \end{bmatrix},$$
and tranpose it
$$ \mathbf{C}^{\mathsf{T}}=\begin{bmatrix} 3 & -3 & 3\\ 1 & 1 & -4\\ -2 & 1 & 2 \end{bmatrix}. $$
Easy to get $\det(\mathbf{A})=3$, therefore the inverse can be calculated
$$ \mathbf{A}^{-1}=\frac13\begin{bmatrix} 3 & -3 & 3\\ 1 & 1 & -4\\ -2 & 1 & 2 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 1\\ \frac13 & \frac13 & -\frac43\\ -\frac23 & \frac13 & \frac23 \end{bmatrix}. $$

MIT OCW

18.02 Multivariable Calculus

Dot Product

Definition

For two $n$-dimensional vectors $\mathbf{a,b}$, their dot product is defined as $$\mathbf{a}\cdot\mathbf{b}=\sum a_ib_i$$ and is equivalent to the following definition:
$$\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$$ where $\theta$ is the angle between $\mathbf{a},\mathbf{b}$. (Definition 2 is more realistic in a plane or space.)

Application

  • Angles
    Derived from the definition 2, it's easy to get the cosine of the angle like $$\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$$ Thus given two concrete vectors in space, the angle between them can be easily found after their dot product is calculated by definition 1 and measured by Pythagoras.
  • Orthogonality Detection
    If $\mathbf{a}\cdot\mathbf{b}=0$, then $\cos\theta=0$, which implies that $\theta=\frac\pi2$, namely, $\mathbf{a}\perp\mathbf{b}$.
    Actually the sign of $\mathbf{a}\cdot\mathbf{b}$ is the same as $\cos\theta$ since $|a||b|$ has to be positive.
    e.g. What does the solution set of $x+2y+3z=0$ in space look like?
    Select a point $P(x,y,z)$ and the equation means that $\mathbf{OP}\cdot(1,2,3)=0$, which also means that $\mathbf{OP}\perp(1,2,3)$ and thus it's much more visual that the content is a plane.
  • Extract Components
    For a vector $\mathbf{a}$ and a unit vector $\hat\mathbf{e}$, the component on $\hat\mathbf{e}$'s direction is just $\mathbf{a}\cdot\hat\mathbf{e}$.

  • Area
    For a triangle composed of vectors $\mathbf{a},\mathbf{b}$, its area should be $S=\frac12|\mathbf{a}||\mathbf{b}|\sin\theta$. Anyhow $\sin\theta$ is actually not that easy to get. So we try to convert $\sin\theta$ to $\cos\theta'$ where $\sin\theta=\cos\theta'$, it's easy to construct a $\theta'=\theta-\frac\pi2$.
    Thus $S=\frac12|\mathbf{a}||\mathbf{b}|\cos\theta'$, meanwhile, we need to find some $\mathbf{a'}$ whose length is just $|\mathbf{a}|$ and the angle between $\mathbf{a'}$ and $\mathbf{b}$ should be $\theta'$.
    In Cartesian plane, a vector $\mathbf{x}=(x_1,x_2)$ becomes $\mathbf{x'}=(-x_2,x_1)$ after rotated counterclockwise.
    Therefore, if $\mathbf{a}=(a_1,a_2)$ and $\mathbf{b}=(b_1,b_2)$, we can find the substitute vector $\mathbf{a'}=(-a_2,a_1)$ and the area of the triangle is $S=\frac12|\mathbf{a'}\cdot\mathbf{b}|=\frac12|a_1b_2-a_2b_1|$. (The absolute value ensures that the area won't become negative.)
    For some polygon, we can just divide the polygon into some triangles and repeatedly use the method to get the whole area.

Determinant (2D, 3D)

Definition

For vectors $\mathbf{a}=(a_1,a_2),\mathbf{b}=(b_1,b_2)$, the determinant $$\det(\mathbf{a},\mathbf{b})=\begin{vmatrix}a_1 & a_2\\b_1 & b_2\end{vmatrix}=a_1b_2-a_2b_1.$$
Having such symbolic notation, we can write the area expression above as $$S=\frac12\begin{Vmatrix}a_1 & a_2 \\ b_1 & b_2\end{Vmatrix}=\frac12|\det(\mathbf{a},\mathbf{b})|.$$
For a vector $\mathbf{a}=(a_1,a_2,a_3),\mathbf{b}=(b_1,b_2,b_3),\mathbf{c}=(c_1,c_2,c_3)$, the determinant $$\det(\mathbf{a},\mathbf{b},\mathbf{c})=\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}=a_1\begin{vmatrix}b_2 & b_3 \\ c_2 & c_3\end{vmatrix}-a_2\begin{vmatrix}b_1 & b_3 \\ c_1 & c_3\end{vmatrix}+a_3\begin{vmatrix}b_1 & b_2 \\ c_1 & c_2\end{vmatrix}.$$

Geometry

The determinant $\det(\mathbf{a},\mathbf{b},\mathbf{c})$ is just the volume of parallelepiped boxed by $\mathbf{a},\mathbf{b},\mathbf{c}$.

Application

  • Plane Equation
    Suppose there are three different points $A,B,C$ (determining a plane exactly), and we want to know the equation of that plane. Select a point $P(x,y,z)$ on the plane then the parallelepiped boxed by $\mathbf{AP},\mathbf{AB},\mathbf{AC}$ should have no volume, which generates the equation that is $\det(\mathbf{AP},\mathbf{AB},\mathbf{AC})=0$, you can simplify it as long as you'd love.

Cross Product

Definition

Given two vectors $\mathbf{a}=(a_1,a_2,a_3),\mathbf{b}=(b_1,b_2,b_3)$, their cross product $\mathbf{a}\times\mathbf{b}=\begin{vmatrix} \hat\mathbf{i} & \hat\mathbf{j} & \hat\mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$.

Geometry

If the area of the parallelogram composed of $\mathbf{a},\mathbf{b}$ is $S$, then $|\mathbf{a}\times\mathbf{b}|=S$. In terms of the direction, it holds that $\mathbf{a}\times\mathbf{b}\perp\mathbf{a}$ and $\mathbf{a}\times\mathbf{b}\perp\mathbf{b}$, namely, the cross product is perpendicular to the plane which $\mathbf{a},\mathbf{b}$ determine.
Concretely, the actual direction has two possible results, which can be determined uniquely through Right-hand Rule.
Memorization Given a special case that is the $\hat\mathbf{i}\times\hat\mathbf{j}=\hat\mathbf{k}$.

Application

  • Volume
    For a parallelepiped boxed by $\mathbf{a},\mathbf{b},\mathbf{c}$, its volume is exactly the base area times the height, more importantly, the base is actually a parallelogram. Without loss of generality, suppose the base is $\mathbf{b}$ and $\mathbf{c}$, thus the base area is $|\mathbf{b}\times\mathbf{c}|$.
    However, the height should be converted to a component on some "vertical" direction, namely which is perpendicular to the base.
    We can use the cross product again and change it into some unit vector to get the direction that is $$\hat\mathbf{n}=\frac{\mathbf{b}\times\mathbf{c}}{|\mathbf{b}\times\mathbf{c}|}.$$
    Therefore the height is $\mathbf{a}\cdot\hat\mathbf{n}$.
    Finally, the volume can be expressed as $$V=|\mathbf{b}\times\mathbf{c}|\left(\mathbf{a}\cdot\frac{\mathbf{b}\times\mathbf{c}}{|\mathbf{b}\times\mathbf{c}|}\right)=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}).$$ and it's equivalent to $\det(\mathbf{a},\mathbf{b},\mathbf{c})$

  • Plane Equation
    The same prerequisites as above, and $P$ is still on the plane. Given the vector $\mathbf{n}$ which is perpendicular to the plane, it holds that $\mathbf{AP}\cdot\mathbf{n}=0$ since $\mathbf{n}$ should be perpendicular to any vector on the plane, where$\mathbf{n}=\mathbf{AB}\times\mathbf{AC}$.