Vectors: Components, Dot Product and Geometry
Core concepts
Work with vectors in component form and as directed segments. Know addition, scalar multiplication, magnitude $|\mathbf{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$, unit vectors, and the dot product $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta=a_1b_1+a_2b_2+a_3b_3$. Use it to find angles, test perpendicularity ($\mathbf{a}\cdot\mathbf{b}=0$), and compute projections. Understand position vectors, displacement $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$, parallel vectors as scalar multiples, and geometric proofs using vector methods.
Traps
- Direction matters: $\overrightarrow{AB}=-\overrightarrow{BA}$.
- The dot product gives a scalar, not a vector.
- $\mathbf{a}\cdot\mathbf{b}=0$ means perpendicular only when neither vector is zero.
- Do not "cancel" vectors in equations; use components or dot with a suitable vector.
- For angle between vectors, use the dot product, not the cross product (not required here).
Rule of thumb
To prove points are collinear, show $\overrightarrow{AB}=k\overrightarrow{AC}$ for some scalar $k$ with a common point $A$. To prove perpendicular lines, show the dot product of direction vectors is zero.
Worked example
Let $\mathbf{a}=(1,2,-2)$ and $\mathbf{b}=(3,0,4)$. Then $\mathbf{a}\cdot\mathbf{b}=3+0-8=-5$, $|\mathbf{a}|=\sqrt{1+4+4}=3$, $|\mathbf{b}|=\sqrt{9+16}=5$. So $\cos\theta=\frac{-5}{15}=-\tfrac13$, giving $\theta\approx109.5^\circ$. The negative cosine signals an obtuse angle between the vectors.