3.13 The scalar productIB Maths: Analysis and Approaches HL: Revision notes

For $\mathbf{v}=(v_1,v_2,v_3)$ and $\mathbf{w}=(w_1,w_2,w_3)$ the scalar product (dot product) is $\mathbf{v}\cdot\mathbf{w}=v_1w_1+v_2w_2+v_3w_3=|\mathbf{v}||\mathbf{w}|\cos\theta,$ where $\theta$ is the angle between the vectors, placed tail to tail. The answer is a number, not a vector. Example: $(2,-1,3)\cdot(1,4,2)=2-4+6=4$.

• $\mathbf{v}\cdot\mathbf{w}=\mathbf{w}\cdot\mathbf{v}$ (commutative)
• $\mathbf{u}\cdot(\mathbf{v}+\mathbf{w})=\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{w}$ (distributive)
• $(k\mathbf{v})\cdot\mathbf{w}=k(\mathbf{v}\cdot\mathbf{w})$
• $\mathbf{v}\cdot\mathbf{v}=|\mathbf{v}|^2$

These let you expand brackets like algebra. For example $|\mathbf{a}+\mathbf{b}|^2=(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})=|\mathbf{a}|^2+2\,\mathbf{a}\cdot\mathbf{b}+|\mathbf{b}|^2.$ With $|\mathbf{a}|=3$, $|\mathbf{b}|=4$ and an angle of $60^\circ$: $9+12+16=37$, so $|\mathbf{a}+\mathbf{b}|=\sqrt{37}$.

Rearranging the definition: $\cos\theta=\frac{\mathbf{v}\cdot\mathbf{w}}{|\mathbf{v}||\mathbf{w}|},\qquad 0\le\theta\le\pi.$ For the angle of a triangle at a vertex $B$, use the two vectors starting at $B$: $\overrightarrow{BA}$ and $\overrightarrow{BC}$. A negative scalar product means the angle is obtuse.

For non-zero vectors:

• Perpendicular: $\mathbf{v}\cdot\mathbf{w}=0$, because $\cos90^\circ=0$.
• Parallel: $|\mathbf{v}\cdot\mathbf{w}|=|\mathbf{v}||\mathbf{w}|$, because $\cos\theta=\pm1$.

To find an unknown making two vectors perpendicular, set the scalar product equal to 0 and solve. Example: $(2,2,-1)\cdot(-1,t,2)=2t-4=0$ gives $t=2$.

Proof (angle in a semicircle). Let $AB$ be a diameter of a circle centred at $O$, with $\overrightarrow{OA}=\mathbf{a}$, so $\overrightarrow{OB}=-\mathbf{a}$. For any other point $P$ on the circle, $|\mathbf{p}|=|\mathbf{a}|$. Then $\overrightarrow{AP}\cdot\overrightarrow{BP}=(\mathbf{p}-\mathbf{a})\cdot(\mathbf{p}+\mathbf{a})=|\mathbf{p}|^2-|\mathbf{a}|^2=0$, so $A\hat{P}B=90^\circ$.

Work done. A constant force $\mathbf{F}$ moving an object through $\mathbf{d}$ does work $\mathbf{F}\cdot\mathbf{d}$. The part of $\mathbf{F}$ along $\mathbf{d}$ is $\lambda\mathbf{d}$ with $\lambda=\frac{\mathbf{F}\cdot\mathbf{d}}{\mathbf{d}\cdot\mathbf{d}}$; what is left, $\mathbf{F}-\lambda\mathbf{d}$, is perpendicular to $\mathbf{d}$ and does no work.