P4.7 - Vectors

A vector records both magnitude and direction. In two dimensions it has two components; in three dimensions it has three. The components measure the signed change parallel to each coordinate axis.

Two vectors are equal when their corresponding components are equal, even if they start at different points. Thus $\begin{pmatrix}3\\-2\\5\end{pmatrix}$ and $3\mathbf i-2\mathbf j+5\mathbf k$ represent the same vector.

The magnitude $|\mathbf a|$ is the length of a vector. It follows from Pythagoras in two or three dimensions. Dividing a non-zero vector by its magnitude keeps its direction but changes its length to $1$.

For $\mathbf a=\begin{pmatrix}4\\-4\\2\end{pmatrix}$, $$|\mathbf a|=\sqrt{4^2+(-4)^2+2^2}=6.$$ A unit vector in the same direction is $$\widehat{\mathbf a}=\frac16\begin{pmatrix}4\\-4\\2\end{pmatrix}=\begin{pmatrix}2/3\\-2/3\\1/3\end{pmatrix}.$$ Its magnitude is $1$, which checks the calculation.

Vector addition combines successive displacements, while scalar multiplication changes a vector's length and possibly its direction. Both operations are performed component by component.

If $\mathbf a=\begin{pmatrix}2\\-1\\3\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}-4\\5\\1\end{pmatrix}$, then $$\mathbf a+\mathbf b=\begin{pmatrix}-2\\4\\4\end{pmatrix},\qquad 2\mathbf a-\mathbf b=\begin{pmatrix}8\\-7\\5\end{pmatrix}.$$ Geometrically, addition places vectors head-to-tail; the resultant joins the starting point to the finishing point.

The position vector of a point runs from the fixed origin to that point. If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$, the displacement from $A$ to $B$ is found by subtracting start from finish.

If $A=(2,-3,5)$ and $B=(8,1,-1)$, then $$\overrightarrow{AB}=\begin{pmatrix}8\\1\\-1\end{pmatrix}-\begin{pmatrix}2\\-3\\5\end{pmatrix}=\begin{pmatrix}6\\4\\-6\end{pmatrix}.$$ Reversing the journey gives $\overrightarrow{BA}=-\overrightarrow{AB}$.

Position vectors also encode division of a segment. If $B$ lies one third of the way from $A$ to $C$, then $$\mathbf b=\mathbf a+\frac13(\mathbf c-\mathbf a),$$ so $\mathbf c=3\mathbf b-2\mathbf a$. The coefficients reflect the stated direction and ratio.

The distance between two points is the magnitude of the displacement vector joining them. Subtract corresponding coordinates, square the changes, add them, and take the non-negative square root.

For $P=(1,-2,4)$ and $Q=(5,1,-8)$, $$\overrightarrow{PQ}=\begin{pmatrix}4\\3\\-12\end{pmatrix},$$ so $$PQ=\sqrt{4^2+3^2+(-12)^2}=\sqrt{169}=13.$$ The same distance results from $\overrightarrow{QP}$ because every component is squared.

A vector line is determined by one point and one non-zero direction vector. The parameter moves through every real value, generating every point on the line.

Through $C=(1,2,0)$ and $D=(3,-1,4)$, a direction is $\overrightarrow{CD}=(2,-3,4)$, so $$\mathbf r=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}2\\-3\\4\end{pmatrix}.$$ At $t=1$ this gives $D$, providing a quick check.

The scalar product converts two vectors into a number that measures directional alignment. Its component form calculates the value; its geometric form connects that value to the angle between the vectors.

For non-zero $\mathbf a=\begin{pmatrix}1\\2\\2\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}2\\0\\1\end{pmatrix}$, $$\mathbf a\cdot\mathbf b=4,\quad |\mathbf a|=3,\quad|\mathbf b|=\sqrt5,$$ so $$\theta=\cos^{-1}\!\left(\frac{4}{3\sqrt5}\right)\approx53.4^\circ.$$ For the acute angle between two lines, use the absolute value of the cosine because either direction vector may be reversed.

If non-zero vectors satisfy $\mathbf a\cdot\mathbf b=0$, then $\cos\theta=0$ and they are perpendicular. For example, $(1,2,2)\cdot(2,1,-2)=2+2-4=0$. This condition can locate the foot of a perpendicular by making a displacement vector dot a line's direction equal to zero.