- In the previous article, we covered the major aspects of vector representation
- In the given article, we will extend our knowledge to some other vector operations which are useful in geometry and will definitely be relevant if you take IB Math Analaysis & Approaches subject.
For $\mathbf{v}=\binom{x}{y}$, the magnitude is found using Pythagoras: $$|\mathbf{v}|=\sqrt{x^2+y^2}$$
If $\overrightarrow{AB}=\binom{-7}{-4}$ then $$|\overrightarrow{AB}|=\sqrt{(-7)^2+(-4)^2}=\sqrt{65}\approx 8.06 \text{ (3 s.f.)}$$
- If the question asks for an answer "correct to 3 s.f.", round only at the end.
- Keep the exact form (like $\sqrt{65}$) during working to avoid rounding error.
- So, $$\mathbf{u}\cdot\mathbf{v}=u_xv_x+u_yv_y$$
- A key geometric link is: $$\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta$$ where $\theta$ is the angle between the vectors.
- If vectors are perpendicular, then $\theta=90^\circ$ and $\cos 90^\circ=0$, so $$\mathbf{a}\cdot\mathbf{b}=0$$
- Conversely, if $\mathbf{a}\cdot\mathbf{b}=0$, then either one vector is the zero vector, or the vectors are perpendicular.
Do not conclude "perpendicular" from a zero dot product unless you also consider the possibility of a zero vector.
- Rearrange the dot product formula: $$\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$$
- Then $$\theta=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)$$
- Let $\mathbf{u}=\binom{3}{-2}$ and $\mathbf{v}=\binom{2}{4}$.
- Dot product: $$\mathbf{u}\cdot\mathbf{v}=3\cdot2+(-2)\cdot4=6-8=-2.$$
- Magnitudes: $$|\mathbf{u}|=\sqrt{3^2+(-2)^2}=\sqrt{13},\quad |\mathbf{v}|=\sqrt{2^2+4^2}=\sqrt{20}$$
- Angle: $$\cos\theta=\frac{-2}{\sqrt{13}\sqrt{20}}$$
- Then compute $\theta$ using $\cos^{-1}$ on a calculator.
Because equality of vectors means equality of components, vector equations often become simultaneous equations.
- If $$\binom{2x}{y}=\binom{7x-10}{3-x},$$
- then match components: $$2x=7x-10,\quad y=3-x.$$
- Solve the scalar equations to find $x$ and $y$.
- To show $\overrightarrow{AB}$ is parallel to $\mathbf{v}$, you can show $$\overrightarrow{AB}=k\mathbf{v}$$
- for some scalar $k$. In components, this means the ratios of corresponding components match (when the components are nonzero).
- For parallel vectors, use scalar multiplication rather than angles.
- It is usually faster to show one vector is a multiple of the other.
- If $\mathbf{u}=\binom{6}{-9}$, write a shorter parallel vector with integer components.
- Without calculating any angles, decide whether $\binom{1}{4}$ and $\binom{8}{-2}$ are perpendicular (use the dot product).