3.13 Scalar and vector productsIB Maths: Applications and Interpretation HL: Revision notes

The scalar product (dot product) of $\mathbf{v}=(v_1,v_2,v_3)$ and $\mathbf{w}=(w_1,w_2,w_3)$ is a number: $\mathbf{v}\cdot\mathbf{w}=v_1w_1+v_2w_2+v_3w_3=|\mathbf{v}||\mathbf{w}|\cos\theta,$ where $\theta$ is the angle between the vectors when they are placed tail to tail. In two dimensions use two components. Example: $\mathbf{a}=(3,1,2)$ and $\mathbf{b}=(1,-2,4)$ give $\mathbf{a}\cdot\mathbf{b}=3-2+8=9$. Two non-zero vectors are perpendicular exactly when $\mathbf{v}\cdot\mathbf{w}=0$. Example: $(3,1,2)\cdot(-1,-1,2)=-3-1+4=0$.

Rearranging the definition gives $\cos\theta=\frac{\mathbf{v}\cdot\mathbf{w}}{|\mathbf{v}||\mathbf{w}|}.$ If $\cos\theta$ is positive the angle is acute; if it is negative the angle is obtuse. For the angle between $\mathbf{a}$ and $\mathbf{b}$ above, $\cos\theta=\frac{9}{\sqrt{14}\sqrt{21}}$, so $\theta=58.3^\circ$ (use $\cos^{-1}$ on your GDC). For two lines, use their direction vectors. The acute angle between the lines uses the absolute value, $\cos\theta=\frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$. Example: $\mathbf{d}_1=(2,1,-2)$ and $\mathbf{d}_2=(1,-2,2)$ give $\mathbf{d}_1\cdot\mathbf{d}_2=-4$, so $\cos\theta=\frac49$ and the acute angle is $63.6^\circ$.

The component of $\mathbf{a}$ acting in the direction of $\mathbf{b}$ is $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}=|\mathbf{a}|\cos\theta.$ The component of $\mathbf{a}$ acting perpendicular to $\mathbf{b}$, in the plane formed by the two vectors, is $\frac{|\mathbf{a}\times\mathbf{b}|}{|\mathbf{b}|}=|\mathbf{a}|\sin\theta$. Together they obey $\left(\text{parallel}\right)^2+\left(\text{perpendicular}\right)^2=|\mathbf{a}|^2$. Example: $\mathbf{F}=(4,-2,3)$ N and a pipe with direction $\mathbf{d}=(1,-2,2)$. The component along the pipe is $\frac{14}{3}=4.67$ N and the component perpendicular to it is $\frac{\sqrt{65}}{3}=2.69$ N. Check: $4.67^2+2.69^2=29=|\mathbf{F}|^2$.

The vector product (cross product) of $\mathbf{v}$ and $\mathbf{w}$ is a vector: $\mathbf{v}\times\mathbf{w}=|\mathbf{v}||\mathbf{w}|\sin\theta\,\mathbf{n},$ where $\mathbf{n}$ is the unit vector perpendicular to both, in the direction given by the right-hand screw rule (turn from $\mathbf{v}$ to $\mathbf{w}$). In components: $\mathbf{v}\times\mathbf{w}=\begin{pmatrix} v_2w_3-v_3w_2 \\ v_3w_1-v_1w_3 \\ v_1w_2-v_2w_1 \end{pmatrix}.$ Example: $(2,1,-1)\times(1,3,2)=(5,-5,5)$. Reversing the order reverses the direction: $\mathbf{w}\times\mathbf{v}=-(\mathbf{v}\times\mathbf{w})$. Dividing by the magnitude gives a unit vector perpendicular to both: $\frac{1}{\sqrt3}(1,-1,1)$. Your GDC can find vector products; always check the order of the vectors.

$|\mathbf{v}\times\mathbf{w}|$ is the area of the parallelogram with sides $\mathbf{v}$ and $\mathbf{w}$, and $\frac12|\mathbf{v}\times\mathbf{w}|$ is the area of the triangle. Example: a roof panel $ABC$ has $\overrightarrow{AB}=(3,2,-1)$ and $\overrightarrow{AC}=(1,3,3)$. Then $\overrightarrow{AB}\times\overrightarrow{AC}=(9,-10,7)$ and the area is $\frac12\sqrt{230}=7.58$ m$^2$. The shortest distance from $C$ to the line $AB$ is the height of the triangle: $\frac{|\overrightarrow{AB}\times\overrightarrow{AC}|}{|\overrightarrow{AB}|}=\frac{\sqrt{230}}{\sqrt{14}}=4.05$ m. This is the perpendicular component of $\overrightarrow{AC}$ relative to $\overrightarrow{AB}$. Not required: proofs of the general properties of the scalar and vector products.