Uses of the Scalar Product (Cambridge (CIE) A Level Maths: Pure 3): Revision Note

Uses of the scalar product

This revision note covers several applications of the scalar product for vectors – namely, how you can use the scalar product to:

• find the angle between vectors or lines

• test whether vectors or lines are perpendicular

• find the closest distance from a point to a line

How do I find the angle between two vectors?

• Recall that a formula for the scalar (or ‘dot’) between vectors $\mathbf{a}$ and $\mathbf{b}$ is

$\mathbf{a}\mathbf{·}\mathbf{b}=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\cos \theta$

• where $\theta$ is the angle between the vectors when they are placed ‘base to base’

  • that is, when the vectors are positioned so that they start at the same point

• We arrange this formula to make $\cos \theta$ the subject:
  

• To find the angle between two vectors

  • Calculate the scalar product between them

  • Calculate the magnitude of each vector

  • Use the formula to find $\cos \theta$

  • Use inverse trig to find $\theta$

How do I find the angle between two lines in 3D?

• To find the angle between two lines, find the angle between their direction vectors

•  For example, if the lines have equations $\mathbf{r}={\mathbf{a}}_1+s{\mathbf{d}}_1$ and $\mathbf{r}={\mathbf{a}}_2+t{\mathbf{d}}_2$, then the angle $\theta$ between the lines is given by

$\theta ={\cos }^{-1}\left(\frac{{\mathbf{d}}_1\mathbf{·}{\mathbf{d}}_2}{\left|{\mathbf{d}}_1\right|\left|{\mathbf{d}}_2\right|}\right)$

How do I know if vectors or lines are perpendicular?

• Two (non-zero) vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular if, and only if, $\mathbf{a}\mathbf{·}\mathbf{b}=0$

  • If the a and b are perpendicular then:

    • $\theta =90°\Rightarrow \cos \theta =0\Rightarrow \left|\mathbf{a}\right|\left|\mathbf{b}\right|\cos \theta =0\Rightarrow \mathbf{a}\mathbf{·}\mathbf{b}=0$

  • If  $\mathbf{a}\mathbf{·}\mathbf{b}=0$ then:

    • $\left|\mathbf{a}\right|\left|\mathbf{b}\right|\cos \theta =0\Rightarrow \cos \theta =0\Rightarrow \theta =90°\Rightarrow$ a and b are perpendicular

  • For example, the vectors $2i-3j+5k$ and $-4i-j+k$  are perpendicular since

$\left(2i-3j+5k \right)·\left(-4i-j+k\right)=2\times \left(-4\right)+\left(-3\right)\times \left(-1\right)+5\times 1=-8+3+5=0$

How do I find the shortest distance from a point to a line?

• Suppose that we have a line $l$ with equation $\mathbf{r}=\mathbf{a}+t\mathbf{d}$  and a point $P$ not on $l$

• Let $F$ be the point on $l$ which is closest to $\mathit{P}$ (sometimes called the foot of the perpendicular)

  • Then the line between $F$ and $P$ will be perpendicular to the line $l$

• To find the closest point $F$

  • Call $\mathbf{f}=\overrightarrow{\mathrm{OF}}$ and $\mathbf{p}=\overrightarrow{OP}$

  • Since $F$ lies on $l$, we have $\mathbf{f}=\mathbf{a}+t_0\mathbf{d}$, for a unique real number $t_0$

  • Find the vector $\overrightarrow{\mathit{F}\mathit{P}}$ using $\mathbf{p}-\mathbf{f}$

  • $\overrightarrow{\mathit{F}\mathit{P}}$ is perpendicular to $\mathit{d}$ so form an equation using $\left(\mathbf{p}-\mathbf{f}\right)·\mathbf{d}=0$

  • Solve this equation to find the value of $t_0$

  • Use the value of $t_0$ to find $\mathbf{f}$

• The shortest distance between the point  and the line  is the length

• Note that the shortest distance between the point and the line is sometimes referred to as the length of the perpendicular