The scalar (dot) product of two vectors is a number, defined geometrically through the angle $\theta$ ($0\le\theta\le\pi$) between them and computed algebraically from components:
$$\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta=a_1b_1+a_2b_2+a_3b_3.$$
The geometric form explains why the dot product matters; the component form is how you actually evaluate it. Equating the two is the engine behind every angle, projection and perpendicularity result below.
Properties
- Commutative: $\vec a\cdot\vec b=\vec b\cdot\vec a$.
- Distributive over addition: $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$.
- Scalar factors pull out: $(\lambda\vec a)\cdot\vec b=\lambda(\vec a\cdot\vec b)=\vec a\cdot(\lambda\vec b)$.
- Self dot product: $\vec a\cdot\vec a=|\vec a|^2$ (since $\theta=0$).
- Standard basis: $\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$ and $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$.
- Sign of $\theta$: $\vec a\cdot\vec b>0$ means the angle is acute, $<0$ means obtuse, $=0$ means right-angled.
Angle between two vectors
Solving the definition for $\cos\theta$ gives the standard formula:
$$\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|}=\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\,\sqrt{b_1^2+b_2^2+b_3^2}}.$$
Perpendicularity (orthogonality)
Two non-zero vectors are perpendicular exactly when their dot product vanishes:
$$\vec a\perp\vec b\iff\vec a\cdot\vec b=0.$$
This is the single most-used test in the chapter — it turns "are these perpendicular?" and "find $\lambda$ so they are perpendicular" into a one-line algebra problem.
Projection of a vector
The scalar projection of $\vec a$ on $\vec b$ (the signed length of the shadow $\vec a$ casts along $\vec b$) is
$$\text{proj}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{|\vec b|}=\vec a\cdot\hat b.$$
It is positive when $\theta$ is acute and negative when obtuse. The corresponding vector projection is $\left(\dfrac{\vec a\cdot\vec b}{|\vec b|^2}\right)\vec b$. The projection of $\vec a$ on a coordinate axis is simply the matching component (e.g. on $\hat i$ it is $a_1$).
Useful identities (resolution / component form)
Expanding $|\vec a\pm\vec b|^2=(\vec a\pm\vec b)\cdot(\vec a\pm\vec b)$ gives two identities used constantly:
$$|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\,\vec a\cdot\vec b,\qquad |\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\,\vec a\cdot\vec b.$$
A neat corollary: $|\vec a+\vec b|=|\vec a-\vec b|$ if and only if $\vec a\cdot\vec b=0$, i.e. the diagonals of a parallelogram are equal exactly when it is a rectangle.
Deeper Insight — the dot product measures alignment. Strip away the algebra and $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ is a single statement: how much of one vector points along the other. That one idea unifies the whole topic — the angle formula is it rearranged, perpendicularity is the case "no alignment" ($\cos\theta=0$), and projection is literally the aligned length. In physics it is exactly the work done by a force ($W=\vec F\cdot\vec d$): only the component of force along the displacement counts. If you remember "dot product $=$ alignment", you rarely need to memorise the sub-formulas separately.
Find $\vec a\cdot\vec b$ for $\vec a=\hat i+2\hat j+3\hat k$, $\vec b=2\hat i-\hat j+\hat k$.
Solution$=1\cdot2+2\cdot(-1)+3\cdot1=2-2+3=3.$
Answer: $3$
Are $\vec a=\hat i+\hat j$ and $\vec b=\hat i-\hat j$ perpendicular?
Solution$\vec a\cdot\vec b=1\cdot1+1\cdot(-1)=0$. Since the dot product is $0$, yes, they are perpendicular.
Answer: Yes, $\vec a\perp\vec b$ (since $\vec a\cdot\vec b=0$).
Find the angle between $\vec a=\hat i$ and $\vec b=\hat i+\hat j$.
Solution$\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dfrac{1}{1\cdot\sqrt2}=\dfrac{1}{\sqrt2}$, so $\theta=45^\circ.$
Answer: $\theta=45^\circ$
Find the projection of $\vec a=2\hat i+3\hat j$ on $\vec b=\hat i$.
SolutionProjection $=\dfrac{\vec a\cdot\vec b}{|\vec b|}=\dfrac{2}{1}=2.$
Answer: $2$
Find the angle between $\vec a=\hat i+\hat j+\hat k$ and $\vec b=\hat i-\hat j+\hat k$.
Solution$\vec a\cdot\vec b=1-1+1=1$ and $|\vec a|=|\vec b|=\sqrt3$. So $\cos\theta=\dfrac{1}{\sqrt3\cdot\sqrt3}=\dfrac13$, giving $\theta=\cos^{-1}\!\left(\dfrac13\right).$
Answer: $\theta=\cos^{-1}\!\left(\dfrac13\right)$
Find $\lambda$ so that $\vec a=2\hat i+\lambda\hat j+\hat k$ is perpendicular to $\vec b=\hat i-2\hat j+3\hat k$.
SolutionSet $\vec a\cdot\vec b=0$: $2(1)+\lambda(-2)+1(3)=0\Rightarrow 5-2\lambda=0\Rightarrow\lambda=\dfrac52.$
Answer: $\lambda=\dfrac52$
The projection of $\vec a=\hat i+3\hat j+7\hat k$ on $\vec b=7\hat i-\hat j+8\hat k$.
Solution$\vec a\cdot\vec b=7-3+56=60$ and $|\vec b|=\sqrt{49+1+64}=\sqrt{114}$. Projection $=\dfrac{60}{\sqrt{114}}.$
Answer: $\dfrac{60}{\sqrt{114}}$
If $|\vec a|=3,\ |\vec b|=4$ and $\vec a\cdot\vec b=6$, find the angle between $\vec a$ and $\vec b$.
Solution$\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dfrac{6}{3\cdot4}=\dfrac12$, so $\theta=60^\circ.$
Answer: $\theta=60^\circ$
If $|\vec a|=5,\ |\vec b|=3$ and the angle between them is $60^\circ$, find $|\vec a+\vec b|$.
Solution$\vec a\cdot\vec b=5\cdot3\cdot\cos60^\circ=\dfrac{15}{2}$. Then $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=25+9+15=49$, so $|\vec a+\vec b|=7.$
Answer: $|\vec a+\vec b|=7$
Show that if $|\vec a+\vec b|=|\vec a-\vec b|$ then $\vec a\perp\vec b$.
SolutionSquaring both sides: $|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b$. This forces $4\vec a\cdot\vec b=0$, i.e. $\vec a\cdot\vec b=0$, so $\vec a\perp\vec b.$
Answer: $\vec a\perp\vec b$
Three unit vectors satisfy $\vec a+\vec b+\vec c=\vec 0$. Find $\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$.
SolutionTake the dot product of the relation with itself: $|\vec a+\vec b+\vec c|^2=0$, i.e. $|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$. Since each magnitude is $1$, $3+2S=0$, so $S=-\dfrac32.$
Answer: $\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=-\dfrac32$
Find a unit vector perpendicular to both $\vec a=\hat i+\hat j$ and the $x$-axis direction... instead, find the value of $\vec a\cdot\vec b$ where $\vec a=2\hat i-\hat j+2\hat k$ and $\vec b$ is the unit vector along $\hat i+2\hat j+2\hat k$.
Solution$\hat b=\dfrac13(\hat i+2\hat j+2\hat k)$. Then $\vec a\cdot\hat b=\dfrac13\big(2(1)+(-1)(2)+2(2)\big)=\dfrac13(2-2+4)=\dfrac43.$ This is also the projection of $\vec a$ on $\vec b$.
Answer: $\vec a\cdot\hat b=\dfrac43$