Q.Forces of magnitudes $5\sqrt2$ and $10\sqrt2$ units acting in the directions $3\hat i+4\hat j+5\hat k$ and $10\hat i+6\hat j-8\hat k$, respectively, act on a particle which is displaced from the point with position vector $4\hat i-3\hat j-2\hat k$ to the point with position vector $6\hat i+3\hat j-\hat k$. Find the work done by the forces.

The dot product $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ and cross product $\vec a\times\vec b$ (magnitude $|\vec a||\vec b|\sin\theta$, direction perpendicular to both) are not just formulas — placed carefully, they reprove classical geometry theorems and compute two real mechanical quantities.

Vector proofs in geometry and trigonometry. The standard technique: put a convenient point (a triangle's vertex, a circle's centre) at the origin, write every other point as a position vector, and express every needed segment as a difference of position vectors. Then:

• Perpendicularity is proved by showing a dot product is $0$. E.g. if $O$ is a circle's centre and $M$ is the midpoint of chord $AB$ (position vectors $\vec a,\vec b$ with $|\vec a|=|\vec b|=r$), then $\vec{OM}=\tfrac{\vec a+\vec b}2$ and $\vec{AB}=\vec b-\vec a$, and $\vec{OM}\cdot\vec{AB}=\tfrac12(|\vec b|^2-|\vec a|^2)=0$.
• Equal lengths / rectangles come from expanding $|\vec p\pm\vec q|^2=|\vec p|^2\pm2\vec p\cdot\vec q+|\vec q|^2$ and comparing.
• Areas come from $\text{area of a parallelogram}=|\vec p\times\vec q|$ for adjacent sides $\vec p,\vec q$; a triangle is half that, and a general quadrilateral with diagonals $\vec d_1,\vec d_2$ has area $\tfrac12|\vec d_1\times\vec d_2|$ (proved by splitting along one diagonal, since the two triangles on either side add — same-sense cross products — instead of subtracting).
• Compound-angle identities ($\cos(\alpha\mp\beta)$, $\sin(\alpha\pm\beta)$) drop out of dotting or crossing two unit vectors $\hat i\cos\alpha+\hat j\sin\alpha$ and $\hat i\cos\beta\pm\hat j\sin\beta$ placed at angles $\alpha,\pm\beta$ to the $x$-axis. …