Motivation

• Vectors can be multiplied in two ways. The first gives a number (a scalar), the second gives a vector.
• Let $\theta$ be the angle between $\vec{A}$ and $\vec{B}$. The component of $\vec{A}$ along $\vec{B}$ is $\mathrm{\parallel }\vec{A}\mathrm{\parallel }\cos \theta$.
• Multiplying that component by $\mathrm{\parallel }\vec{B}\mathrm{\parallel }$ gives the quantity used to define the scalar product.

Definition

• $$\vec{A}\cdot \vec{B}=\mathrm{\parallel }\vec{A}\mathrm{\parallel }\mathrm{\parallel }\vec{B}\mathrm{\parallel }\cos \theta ,0^{\circ }\le \theta \le 180^{\circ }$$

• The result is a real number, read "$A$ dot $B$".
• Its absolute value equals the area of the rectangle whose sides are $\mathrm{\parallel }\vec{B}\mathrm{\parallel }$ and the component of $\vec{A}$ along $\vec{B}$.

• To find $\theta$, both vectors must start from (or both end at) the same point. If they do not, translate one of them.
• If $\theta$ is acute the product is positive; if $\theta$ is obtuse it is negative; if $\theta =90^{\circ }$ it is zero.

✍️ Worked examples

• Example 1 The angle between $\vec{A}$ and $\vec{B}$ is $60^{\circ }$, and $\mathrm{\parallel }\vec{A}\mathrm{\parallel }=2\mathrm{\parallel }\vec{B}\mathrm{\parallel }=8$. Find $\vec{A}\cdot \vec{B}$.
• Solution:
• $\mathrm{\parallel }\vec{A}\mathrm{\parallel }=8$ and $\mathrm{\parallel }\vec{B}\mathrm{\parallel }=4$.
• $\vec{A}\cdot \vec{B}=8\times 4\times \cos 60^{\circ }=16$.
• Example 2 If $\mathrm{\parallel }\vec{A}\mathrm{\parallel }=6$, $\mathrm{\parallel }\vec{B}\mathrm{\parallel }=10$ and $\theta =135^{\circ }$, find $\vec{A}\cdot \vec{B}$.
• Solution:
• $\vec{A}\cdot \vec{B}=6\times 10\times \cos 135^{\circ }=60(-{\displaystyle \frac{\sqrt{2}}{2}})=-30\sqrt{2}\approx -42.4$.

Rule

• The component (scalar projection) of $\vec{A}$ in the direction of $\vec{B}$ is
  $$A_B=\frac{\vec{A}\cdot \vec{B}}{\mathrm{\parallel }\vec{B}\mathrm{\parallel }}=\mathrm{\parallel }\vec{A}\mathrm{\parallel }\cos \theta$$
• It is positive if $\theta$ is acute, negative if obtuse, and zero when the vectors are perpendicular.
• The projection vector of $\vec{A}$ on $\vec{B}$ is
  $${\text{proj}}_{\vec{B}}\vec{A}=\frac{\vec{A}\cdot \vec{B}}{\mathrm{\parallel }\vec{B}{\mathrm{\parallel }}^2}\vec{B}=A_B{\widehat{u}}_B$$
• Remember the component of $\vec{A}$ along $\vec{B}$ is not the same as the component of $\vec{B}$ along $\vec{A}$ unless $\mathrm{\parallel }\vec{A}\mathrm{\parallel }=\mathrm{\parallel }\vec{B}\mathrm{\parallel }$.

✍️ Worked examples

• Example 14 Find the component of the force $\vec{F}=2\widehat{i}-3\widehat{j}+5\widehat{k}$ in the direction of $\overrightarrow{AB}$, where $A(1,4,0)$ and $B(-1,2,3)$.
• Solution:
• $\overrightarrow{AB}=(-2,-2,3)$ and $\mathrm{\parallel }\overrightarrow{AB}\mathrm{\parallel }=\sqrt{17}$.
• $\vec{F}\cdot \overrightarrow{AB}=-4+6+15=17$.
• Component $={\displaystyle \frac{17}{\sqrt{17}}}=\sqrt{17}\approx 4.12$ force units.
• Example 15 Find the projection vector of $\vec{A}=(3,4,0)$ on $\vec{B}=(1,1,1)$.
• Solution:
• $\vec{A}\cdot \vec{B}=7$ and $\mathrm{\parallel }\vec{B}{\mathrm{\parallel }}^2=3$.
• Projection vector $={\displaystyle \frac{7}{3}}(1,1,1)=({\displaystyle \frac{7}{3}},{\displaystyle \frac{7}{3}},{\displaystyle \frac{7}{3}})$.
• Example 16:
• A cube of side 2 has one vertex at $O$ with edges along the axes.
• Find the component of the diagonal $\overrightarrow{OG}$ on the edge vector $\vec{e}=(2,0,0)$, where $G=(2,2,2)$.
• Solution:
• $\overrightarrow{OG}\cdot \vec{e}=4$ and $\mathrm{\parallel }\vec{e}\mathrm{\parallel }=2$, so the component is $2$.
• It equals the $x$-component of the diagonal, as expected for projection on an axis.