# Vectors



My notes of the excellent lectures of “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Description will use a plane $$\mathbb{R}^2$$, or space $$\mathbb{R}^3$$, but the same principles apply to higher dimensions.

Vectors are commonly displayed on the $$xyz$$-axis, with unit vectors $$\hat\imath\, \hat\jmath, \hat k$$.

Vectors do not have a start point, but do have a magnitude (length) and direction. They are described in terms of the unit vectors $$\hat\imath, \hat\jmath, \hat k$$, or using angle brackets notation. $$\vec{A} = \hat\imath\;a_1 + \hat\jmath\;a_2 + \hat\;k a_3 = \left\langle \;a_1,\;a_2,\;a_3\; \right\rangle$$

You can find the length of a vector $$|\vec{A}|$$, by applying the Pythagorean theorem twice. $$\shaded{ |\vec{A}| = \sqrt{(a_1)^2 + (a_2)^2 + (a_3)^2} } \nonumber$$

## Rotation

Let $$\vec{A}=\left\langle a_1, a_2\right\rangle$$, and let $$\vec{A}’$$ be $$\vec{A}$$ rotated over $$\frac{\pi}{2}$$. Then $$\shaded{ \vec{A}’=\left\langle -a_2, a_1\right\rangle } \label{eq:rotation}$$

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$. Then $$\vec{A}$$ plus $$\vec{B}$$ is defined as $$\shaded{ \vec{A}+\vec{B} = \left\langle a_1+b_1, a_2+b_2, a_3+b_3 \right\rangle } \nonumber$$

Geometric, the sum is the vector to the corner of the parallelogram.

## Scalar product

Let $$s$$ be a scalar, and $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$. Then the scalar product of $$s$$ and $$\vec{A}$$ is defined as $$\shaded{ s\;\vec{A} = \left\langle s\;a_1, s\;a_2, s\;a_3\right\rangle } \nonumber$$

Geometrically, it makes the vector longer or shorter.

## Dot-product

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$. The dot-product of $$\vec{A}$$ and $$\vec{B}$$ is defined as the scalar $$\shaded{ \vec{A} \cdot \vec{B} = \sum_i a_i\,b_i = a_1 b_1 + a_2 b_2 + a_3 b_3 } \nonumber$$

For a geometric interpretation, start with the dot-product of $$\vec{A}$$ with itself $$\vec{A}\cdot\vec{A} = |\vec{A}|^2 \cos 0 = |\vec{A}|^2 \label{eq:vecsquare}$$

Let $$\vec{C}=\vec{A}-\vec{B}$$, and expand $$|\vec{C}|^2$$ by applying $$\eqref{eq:vecsquare}$$ \begin{align} |\vec{C}|^2 &= \vec{C} \vec{C} = \left(\vec{A} – \vec{B} \right) \cdot \left(\vec{A} – \vec{B} \right) \nonumber \\ &= \vec{A}\cdot\vec{A} – \vec{A}\cdot\vec{B} – \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B} \nonumber \\ &= |\vec{A}|^2 + |\vec{B}|^2 – 2 \vec{A}\cdot\vec{B} \label{eq:expanded} \end{align}

Recall, the law of cosines from geometry.

$$c^2 = a^2 b^2 – 2 a b\cos\theta \nonumber$$

Apply the law of cosines to $$|\vec{A}|$$, $$|\vec{B}|$$ and $$|\vec{C}|$$ $$|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 – 2 |\vec{A}| |\vec{B}|\cos\theta \label{eq:lawofcos}$$

Combining equations $$\eqref{eq:expanded}$$ and $$\eqref{eq:lawofcos}$$ gives the geometric equation $$\shaded{ \vec{A}\cdot\vec{B} = |\vec{A}|\,|\vec{B}|\, \cos\theta } \nonumber$$

The dot-product can be used to compute length and angles in $$\mathbb{R}^3$$, or find components of $$\vec{A}$$ along unit vector $$\hat u$$ $$\shaded{ \vec{A}\cdot \hat u } \nonumber$$

## Determinant

### In 2 dimensions

Let $$\vec{A}=\left\langle a_1, a_2\right\rangle$$ and $$\vec{B}=\left\langle b_1, b_2\right\rangle$$. The $$\mathbb{R}^2$$-determinant is defined as \shaded{ \begin{align*} \mathrm{det}(\vec{A}, \vec{B}) &= \left|\begin{matrix} a_1 & a_2 \\ b_1 & b_2 \\ \end{matrix}\right| \\ &= a_1b_2-a_2b_1 \end{align*} } \nonumber

### In 3 dimensions

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$ and $$\vec{C}=\left\langle c_1, c_2, c_3\right\rangle$$. The $$\mathbb{R}^3$$-determinant is defined as \shaded{ \begin{align*} \mathrm{det}(\vec{A}, \vec{B}, \vec{C}) &= \left|\begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix}\right| \\ &= a_1 \left|\begin{matrix} b_2 & b_3 \\ c_2 & c_3 \end{matrix}\right| – a_2 \left|\begin{matrix} b_1 & b_3 \\ c_1 & c_3 \end{matrix}\right| + a_3 \left|\begin{matrix} b_1 & b_2 \\ c_1 & c_2 \end{matrix}\right| \end{align*} } \nonumber

### Area of a parallelogram

Let $$\vec{A}=\left\langle a_1, a_2\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2\right\rangle$$.

The area of the parallelogram shown above is calculated as width $$\times$$ height. $$\mathrm{area}_\triangle = |\vec{A}| |\vec{B}| \sin\theta \label{eq:triangle}$$

Change from $$\sin\theta$$ to $$\cos\theta$$ so it fits the dot-product.

Obtain $$\vec{A}’$$ by rotating $$\vec{A}$$ over $$\frac{\pi}{2}$$, see equation $$\eqref{eq:rotation}$$. Apply $$sin\;\theta = \cos(\tfrac{\pi}{2}-\theta)$$ $$\left. \begin{array}{l} \theta ‘ = \tfrac{\pi}{2} – \theta \\ \cos(\tfrac{\pi}{2}-\theta) = \sin\theta \end{array} \right\} \Rightarrow \cos(\theta’) = sin(\theta) \label{eq:sincos}$$

Substitute $$\eqref{eq:sincos}$$ in $$\eqref{eq:triangle}$$ $$\mathrm{area} = |\vec{A}’| \cdot |\vec{B}| \cos\theta = \tfrac{1}{2}\vec{A}’\cdot \vec{B}$$

Expand the dot-product between $$\vec{A}’$$ and $$\vec{B}$$, and find the determinant \begin{align*} \mathrm{area} &= \left\langle -a_2, a_1 \right\rangle \cdot \left\langle b_1, b_2 \right\rangle \\ &= \left( a_1 b_2 – a_2 b_1 \right) \\ &= \left|\begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \end{array}\right| \end{align*}

The area of a parallelogram follows $$\shaded{ \mathrm{area} = \mathrm{det}\left(\vec{A},\vec{B}\right) } \label{eq:area}$$

## Cross-product

Let $$\vec{A}=\left\langle a_1, a_2, a_3\right\rangle$$, and $$\vec{B}=\left\langle b_1, b_2, b_3\right\rangle$$. The cross product of $$\vec{A}$$ and $$\vec{B}$$ in $$\mathbb{R}^3$$ is defined as the pseudo determinant vector \shaded{ \begin{align*} \vec{A}\times\vec{B} &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} a_2 & a_3 \\ b_2 & b_3 \end{array} \right| – \hat\jmath \left| \begin{array}{cc} a_1 & a_3 \\ b_1 & b_3 \end{array} \right| + \hat k \left| \begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \end{array} \right| \end{align*} } \nonumber

Theorem:

• the area of the parallelogram from the vectors $$\vec{A}$$ and $$\vec{B}$$ is $$|\vec{A}\times\vec{B}|$$
• the direction of $$\vec{A}\times\vec{B}$$ is perpendicular to the plane of the parallelogram.

The direction of the vector $$|\vec{A}\times\vec{B}|$$ is determined by the right-hand rule

For example: $$\hat\imath\times\hat\jmath=\hat k$$ \begin{align*} \hat\imath\times\hat\jmath &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right| – \hat\jmath \left| \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right| + \hat z \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| \\ &= \hat z \end{align*}

Some properties

The right-hand rule shows that $$\shaded{ \vec{A}\times\vec{B}=-\vec{B}\times\vec{A} }$$

The parallelogram of $$\vec{A}\times\vec{A}$$ has area zero. $$\vec{A}\times\vec{A}=\vec{0}$$

### Volume in space

Let $$\vec{A}, \vec{B}, \vec{C}$$ in space $$\mathbb{R}^3$$.

The volume equals the area of the base times the height. The area base follows from equation $$\eqref{eq:area}$$. The height is the component of $$\vec{A}$$ that is perpendicular to the base. Call the direction perpendicular to the base unit vector $$\hat n$$. $$\mathrm{volume} = |\vec{B}\times\vec{C}|\;(\vec{A}\cdot\hat n) \label{eq:volume1}$$

The unit vector $$\hat n$$ can be derived from the cross-product of $$\vec{B}$$ and $$\vec{C}$$. To make it a unit vector, we divide by its length. $$\hat n = \frac{\vec{B}\times\vec{C}}{|\vec{B}\times\vec{C}|} \nonumber$$

Substitute this back in $$\eqref{eq:volume1}$$ \begin{align*} \mathrm{volume} &= \bcancel{|\vec{B}\times\vec{C}|}\;\left(\vec{A}\cdot \frac{\left(\vec{B}\times\vec{C}\right)}{\bcancel{|\vec{B}\times\vec{C}|}}\right) \\ &= \vec{A}\ \cdot\ \left(\vec{B}\times\vec{C}\right) \end{align*}

This equals the determinant of $$\vec{A}, \vec{B}, \vec{C}$$, the so called “triple product” rule $$\shaded{ \mathrm{det}\left(\vec{A},\vec{B},\vec{C}\right) =\vec{A}\ \cdot\ \left(\vec{B}\times\vec{C}\right) } \label{eq:tripleproduct}$$

Because $$a_1 \left| \begin{matrix} b_2 & b_3 \\ c_2 & c_3 \end{matrix} \right| – a_2 \left| \begin{matrix} b_1 & b_3 \\ c_1 & c_3 \end{matrix} \right| + a_3 \left| \begin{matrix} b_1 & b_2 \\ c_1 & c_2 \end{matrix} \right| \ = \left\langle a_1, a_2,a_3 \right\rangle \cdot \ \left( \hat\imath \left| \begin{array}{cc} b_2 & b_3 \\ c_2 & c_3 \end{array} \right| – \hat\jmath \left| \begin{array}{cc} b_1 & b_3 \\ c_1 & c_3 \end{array} \right| + \hat k \left| \begin{array}{cc} b_1 & b_2 \\ c_1 & c_2 \end{array} \right| \right) \nonumber$$

The volume in space described by $$\vec{A}$$, $$\vec{B}$$ and $$\vec{C}$$ follows as $$\shaded{ \mathrm{volume } = \mathrm{det}\left(\vec{A},\vec{B},\vec{C}\right) } \nonumber$$

### Equation of a plane from points

Find the plane that contains the points $$p$$, $$q$$ and $$r$$.

#### Solution

Consider $$\overrightarrow{qr}$$, $$\overrightarrow{qs}$$ and $$\overrightarrow{qp}$$ that form a parallelepiped. if these vectors are in the same plane, the parallelepiped will be flat. In other words, it will have no volume.

If $$p$$ is in the $$qrs$$-plane, the determinant should be $$0$$. $$\shaded{ \mathrm{det}\left( \overrightarrow{qp}, \overrightarrow{qr}, \overrightarrow{qs} \right) = 0 } \nonumber$$ with $$q$$, $$r$$ and $$s$$ known, and $$p$$ unknown, this equation will give the expression in $$x,y,z$$ for the plane.

#### A more intuitive solution

Let a “normal vector” $$\overrightarrow n$$ be a vector perpendicular to the plane. Then $$p$$ is the plane when $$\overrightarrow{qp} \perp \overrightarrow n$$. Therefore the dot-product $$\overrightarrow{qp}\cdot \overrightarrow{n} = 0 \label{eq:moreintuitive}$$

$$\overrightarrow{n}$$ equals $$\overrightarrow{pr} \times \overrightarrow{qs}$$. Substituting this in equation $$\eqref{eq:moreintuitive}$$ $$\overrightarrow{qp} \cdot \left( \overrightarrow{pr} \times \overrightarrow{qs} \right) = 0 \nonumber$$

Applying the triple product equation $$\eqref{eq:tripleproduct}$$ gives the condition $$\shaded{ \mathrm{det}\left( \overrightarrow{qp}, \overrightarrow{pr}, \overrightarrow{qs} \right) = 0 }$$ with $$q$$, $$r$$ and $$s$$ known, and $$p$$ unknown, this equation will give the expression in $$x,y,z$$ for the plane.