Vectors

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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 $$ \shaded{ \vec{A}’=\left\langle -a_2, a_1\right\rangle } \label{eq:rotation} $$

Addition

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.