1) Definition

The dot product (or scalar product) of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as

\[
\boxed{\;\vec{A}\cdot\vec{B} \;=\; |\vec{A}|\,|\vec{B}|\,\cos\theta\;}
\]

where \(\theta\) is the angle \(0^\circ \le \theta \le 180^\circ\) between \(\vec{A}\) and \(\vec{B}\). The result is a scalar.

2) Geometrical Meaning

\(\vec{A}\cdot\vec{B}\) equals the magnitude of one vector times the projection of the other on it:

\(\vec{A}\cdot\vec{B} = |\vec{A}|\;(\text{Proj of } \vec{B} \text{ on }\vec{A}) = |\vec{B}|\;(\text{Proj of } \vec{A} \text{ on }\vec{B})\).

3) Component/Algebraic Form

If \(\vec{A} = A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\) and \(\vec{B} = B_x\hat{i}+B_y\hat{j}+B_z\hat{k}\), then

\[
\boxed{\;\vec{A}\cdot\vec{B} \;=\; A_xB_x + A_yB_y + A_zB_z\;}
\]

Also, \(|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}\) and \(|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2}\).

4) Properties & Identities

• Commutative: \(\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}\).
• Distributive: \(\vec{A}\cdot(\vec{B}+\vec{C}) = \vec{A}\cdot\vec{B}+\vec{A}\cdot\vec{C}\).
• Scalar multiplication: \((\lambda\vec{A})\cdot\vec{B} = \lambda(\vec{A}\cdot\vec{B})\).
• Self dot product: \(\vec{A}\cdot\vec{A} = |\vec{A}|^2\).
• Orthogonality: If \(\vec{A}\perp\vec{B}\), then \(\vec{A}\cdot\vec{B}=0\).
• Parallel/anti-parallel: If \(\theta=0^\circ\Rightarrow \vec{A}\cdot\vec{B}=|\vec{A}||\vec{B}|\); if \(\theta=180^\circ\Rightarrow \vec{A}\cdot\vec{B}=-|\vec{A}||\vec{B}|\).
• Cauchy–Schwarz (useful inequality): \(|\vec{A}\cdot\vec{B}|\le|\vec{A}||\vec{B}|\) (equality iff \(\vec{A}\) and \(\vec{B}\) are parallel).
• Cosine formula: \(\cos\theta=\dfrac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}\).
• Perpendicular test: \(\vec{A}\cdot\vec{B}=0 \iff \theta=90^\circ\) (non-zero vectors).

5) Angle Between Two Vectors

For non-zero vectors \(\vec{A},\vec{B}\):

\[
\cos\theta=\frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|},\quad
\theta=\cos^{-1}\!\left(\frac{\vec{A}\cdot\vec{B}}{|\vec{A}||\vec{B}|}\right).
\]

6) Projections

Scalar projection of \(\vec{A}\) on \(\vec{B}\):

\[
\text{comp}_{\vec{B}}\vec{A} \;=\; \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|}
\]

Vector projection of \(\vec{A}\) on \(\vec{B}\):

\[
\text{proj}_{\vec{B}}\vec{A} \;=\; \left(\frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\right)\vec{B}
\]

7) Work Done by a Constant Force

If a constant force \(\vec{F}\) moves a body by displacement \(\vec{d}\), the work done is

\[
\boxed{\;W=\vec{F}\cdot\vec{d}=Fd\cos\theta\;}
\]

Positive if \(0^\circ\le\theta<90^\circ\) (assistive), zero at \(90^\circ\) (no work), negative if \(90^\circ<\theta\le180^\circ\) (opposing).

8) Connection to Direction Cosines (nice-to-know)

If a unit vector \(\hat{u}\) makes angles \(\alpha,\beta,\gamma\) with the \(x,y,z\)-axes respectively, then
\(\hat{u}=\langle \cos\alpha,\cos\beta,\cos\gamma\rangle\) and
\(\hat{u}\cdot\hat{u}=1 \Rightarrow \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\).

9) Solved Examples

10) Common Mistakes & Tips

• Forgetting that the dot product result is a scalar, not a vector.
• Using degrees vs radians inconsistently in calculators when computing \(\cos\theta\).
• Mixing up projection formulas: scalar vs vector projection.
• Assuming \(\vec{A}\cdot\vec{B}=0\) implies either \(\vec{A}=\vec{0}\) or \(\vec{B}=\vec{0}\) — not true; it indicates perpendicular (unless one is zero).
• For work done, ensure \(\theta\) is the angle between force and displacement, not between other directions.

11) Practice Questions

1. Compute \(\vec{A}\cdot\vec{B}\) for \(\vec{A}=\langle 4,-3,1\rangle\), \(\vec{B}=\langle -2,5,2\rangle\). Find the angle between them (to the nearest degree).
2. If \(\vec{u}=\langle a,1,2\rangle\) and \(\vec{v}=\langle 3,-1,1\rangle\) are perpendicular, find \(a\).
3. Find the scalar and vector projection of \(\vec{P}=\langle 7, -1, 2\rangle\) on \(\vec{Q}=\langle 1,2,2\rangle\).
4. A 12 N force acts at \(60^\circ\) to the displacement of 5 m. Find the work done.
5. Show that \(|\vec{A}\cdot\vec{B}|\le|\vec{A}||\vec{B}|\) using \((\vec{A}-\lambda\vec{B})\cdot(\vec{A}-\lambda\vec{B})\ge 0\) for all real \(\lambda\). (Hint: quadratic in \(\lambda\) has \(\Delta\le 0\).)
6. Vectors \(\vec{a}=\langle 2,\,p,\,1\rangle\) and \(\vec{b}=\langle 1,\,2,\,p\rangle\) make an acute angle. Find the range of \(p\). (Use \(\vec{a}\cdot\vec{b}>0\).)
7. If \(\vec{A}=\langle 0,3,-4\rangle\) and \(\vec{B}\) is a unit vector such that \(\vec{A}\cdot\vec{B}=2\), find the scalar projection of \(\vec{A}\) on \(\vec{B}\) and one possible \(\vec{B}\).
8. Let \(\vec{r}=\langle x,y,z\rangle\) and \(\hat{n}\) be a unit vector. Show that \(\text{proj}_{\hat{n}}\vec{r}=(\vec{r}\cdot\hat{n})\hat{n}\).
9. If \(\vec{A}\cdot\vec{B}=5\), \(|\vec{A}|=4\), \(|\vec{B}|=3\), find \(\theta\) and \(\text{proj}_{\vec{B}}\vec{A}\).
10. Prove that for any vectors \(\vec{A},\vec{B}\): \(|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A}\cdot\vec{B}\).

12) Answer Key (Practice)

1. \(\vec{A}\cdot\vec{B}=4(-2)+(-3)5+1(2)=-8-15+2=-21.\)
   \(|\vec{A}|=\sqrt{16+9+1}=\sqrt{26},\;|\vec{B}|=\sqrt{4+25+4}=\sqrt{33}.\)
   \(\cos\theta=\frac{-21}{\sqrt{26\cdot 33}}\Rightarrow \theta\approx 137^\circ.\)
2. \(\vec{u}\cdot\vec{v}=3a + (1)(-1) + 2(1)=0 \Rightarrow 3a-1+2=0 \Rightarrow a=-\tfrac{1}{3}.\)
3. \(\vec{P}\cdot\vec{Q}=7\cdot1+(-1)\cdot2+2\cdot2=7-2+4=9,\;|\vec{Q}|^2=1^2+2^2+2^2=9.\)
   Scalar projection \(= \dfrac{9}{|\vec{Q}|}=\dfrac{9}{3}=3.\)
   Vector projection \(= \dfrac{9}{9}\vec{Q}=\langle 1,2,2\rangle.\)
4. \(W=Fd\cos\theta = 12\times 5\times \cos 60^\circ = 60\times \tfrac12=30\text{ J}.\)
5. Discriminant \(\le 0 \Rightarrow\) yields \( (\vec{A}\cdot\vec{B})^2 \le |\vec{A}|^2|\vec{B}|^2\Rightarrow |\vec{A}\cdot\vec{B}|\le |\vec{A}||\vec{B}|.\)
6. \(\vec{a}\cdot\vec{b}=2\cdot1 + p\cdot2 + 1\cdot p = 2 + 3p.\) Acute \(\Rightarrow \vec{a}\cdot\vec{b}>0 \Rightarrow 2+3p>0 \Rightarrow p>-\tfrac{2}{3}.\)
7. Scalar projection is given as \(2\) (since \(\vec{B}\) is unit). One possible \(\vec{B}\) is any unit vector with \(\vec{A}\cdot\vec{B}=2\).
   For instance, take \(\vec{B}=\frac{1}{5}\langle 0,3,4\rangle=\langle 0,\tfrac{3}{5},\tfrac{4}{5}\rangle\), then \(\vec{A}\cdot\vec{B}=0\cdot 0 + 3\cdot \tfrac{3}{5} +(-4)\cdot\tfrac{4}{5}=\tfrac{9-16}{5}=-\tfrac{7}{5}\neq 2\).
   Adjust: choose \(\vec{B}\) along \(\vec{A}\): \(\hat{A}=\dfrac{1}{5}\langle 0,3,-4\rangle\Rightarrow \vec{B}=\hat{A}\Rightarrow \vec{A}\cdot\vec{B}=|\vec{A}|=5\) (too large).
   Instead, pick \(\vec{B}\) with components \(\langle b_1,b_2,b_3\rangle\) satisfying \(b_1^2+b_2^2+b_3^2=1\) and \(3b_2-4b_3=2\).
   One solution: set \(b_1=0\). Solve \(3b_2-4b_3=2\) and \(b_2^2+b_3^2=1\) ⇒ \(b_2=\tfrac{2}{5},\, b_3=-\tfrac{3}{5}\).
   Hence \(\vec{B}=\langle 0,\tfrac{2}{5},-\tfrac{3}{5}\rangle\) works and \(\vec{A}\cdot\vec{B}=0\cdot0+3\cdot\tfrac{2}{5}+(-4)\cdot(-\tfrac{3}{5})=\tfrac{6+12}{5}= \tfrac{18}{5}=3.6\) (still not 2).
   Try \(b_2=\tfrac{2}{\sqrt{13}}, b_3=\tfrac{1}{\sqrt{13}}\Rightarrow 3b_2-4b_3=\dfrac{6-4}{\sqrt{13}}=\dfrac{2}{\sqrt{13}}.\)
   To make \(3b_2-4b_3=2\), take \(b_2=\dfrac{2}{\sqrt{13}}\cdot \sqrt{13}/1 =2\) invalid (exceeds unit).
   Compact approach: Choose a unit vector with given dot: let \(\hat{a}=\dfrac{\vec{A}}{|\vec{A}|}=\dfrac{1}{5}\langle 0,3,-4\rangle\).
   Take \(\vec{B}= \cos\phi\,\hat{a} + \sin\phi\,\hat{a}_\perp\) (any unit \(\hat{a}_\perp\perp \hat{a}\)). Then \(\vec{A}\cdot\vec{B}=|\vec{A}|\cos\phi=5\cos\phi=2\Rightarrow \cos\phi=\tfrac{2}{5}\).
   Thus one possible \(\vec{B}\) exists for any \(\hat{a}_\perp\). (Students may construct a specific \(\vec{B}\) numerically.)
8. From definition of projection with unit \(\hat{n}\): \(\text{proj}_{\hat{n}}\vec{r}=(\vec{r}\cdot\hat{n})\hat{n}\). (Direct substitution.)
9. \(\cos\theta=\dfrac{5}{4\cdot 3}=\dfrac{5}{12}\Rightarrow \theta=\cos^{-1}\!\left(\tfrac{5}{12}\right).\)
   \(\text{proj}_{\vec{B}}\vec{A} = \left(\dfrac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\right)\vec{B}=\left(\dfrac{5}{9}\right)\vec{B}.\)
10. Expand \(|\vec{A}+\vec{B}|^2=(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})=|\vec{A}|^2+|\vec{B}|^2+2\vec{A}\cdot\vec{B}.\)