Dot Product (Scalar Product)

The dot product turns two vectors into a scalar. It measures how much of one vector lies along another — directly tied to work, projection, and angle.

Concept

Definition

For $\vec{A}$ and $\vec{B}$ with angle $\theta$ between them: $\vec{A}\cdot\vec{B} = |\vec{A}|\,|\vec{B}|\cos\theta = AB\cos\theta.$

Component Form

$\vec{A}\cdot\vec{B} = A_x B_x + A_y B_y + A_z B_z.$

Useful Identities

Geometric Meaning

$\vec{A}\cdot\vec{B}$ = (magnitude of $\vec{A}$) × (component of $\vec{B}$ along $\vec{A}$).

So if you project $\vec{B}$ onto $\vec{A}$, the result is $B\cos\theta$, and multiplying by $A$ gives the dot product.

Angle Between Vectors

$\cos\theta = \frac{\vec{A}\cdot\vec{B}}{AB}.$

Perpendicularity Test

$\vec{A}\cdot\vec{B} = 0$ iff $\vec{A} \perp \vec{B}$ (assuming non-zero vectors).

Applications

• Work: $W = \vec{F}\cdot\vec{d}$. Only the component of force along displacement does work.
• Power: $P = \vec{F}\cdot\vec{v}$.
• Magnetic flux: $\Phi = \vec{B}\cdot\vec{A}$.

Worked Example

Q1: $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\vec{B} = \hat{i} - 2\hat{j} + 2\hat{k}$. Find $\vec{A}\cdot\vec{B}$ and the angle between them.

Solution: $\vec{A}\cdot\vec{B} = (2)(1) + (3)(-2) + (-1)(2) = 2 - 6 - 2 = -6.$

$A = \sqrt{4+9+1} = \sqrt{14}$, $B = \sqrt{1+4+4} = 3$.

$\cos\theta = -6 / (3\sqrt{14}) = -2/\sqrt{14} \approx -0.535 \Rightarrow \theta \approx 122.3°$.

Q2: A constant force $\vec{F} = 4\hat{i} + 3\hat{j}$ N moves a particle through displacement $\vec{d} = 5\hat{i}$ m. Find the work done.

$W = \vec{F}\cdot\vec{d} = (4)(5) + (3)(0) = 20$ J.

Common Confusions

• "$\vec{A}\cdot\vec{B} = AB$." — Only if $\theta = 0$ (parallel vectors).
• "Dot product is a vector." — No, it returns a scalar.
• "$\vec{A}\cdot\vec{B} = 0$ implies $\vec{A} = 0$ or $\vec{B} = 0$." — No, it can also mean they are perpendicular.

Key Takeaways

• $\vec{A}\cdot\vec{B} = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z$.
• Returns a scalar (positive, negative, or zero).
• Zero iff perpendicular (for non-zero vectors).
• Work = $\vec{F}\cdot\vec{d}$.
• Commutative and distributive over addition.