MCV4U The Dot Product: Notes and Practice

Pull a wagon $20$ m along a level path with a force of $50$ N. If the handle is level, all of your effort moves the wagon forward. Raise the handle to $60°$ and only the horizontal part of the pull, half of it, moves the wagon forward; the rest pulls upward and does no work along the path. Physics needed a way to multiply a force by a displacement that counts only the part of the force lined up with the motion. That multiplication is the DOT PRODUCT, and it turns out to answer purely geometric questions too: the angle between two vectors, whether they are perpendicular, and the shadow one casts on another.

Two definitions that agree

a \cdot b = ∣ a ∣∣ b ∣ cos θ a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

Geometric form (θ is the angle between the vectors drawn tail to tail, 0 to 180 degrees) and algebraic form (in 2-space, drop the third term).

The answer is a SCALAR, not a vector, which is why it is also called the scalar product. Why do the two forms give the same number? Draw $a$ and $b$ tail to tail; the third side of the triangle is $a - b$ . The cosine law says $∣ a - b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2∣ a ∣∣ b ∣ cos θ .$ Expanding the components instead gives $∣ a - b ∣^{2} = \sum (a_{i} - b_{i})^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2 (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}) .$ Comparing the two lines, $a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = ∣ a ∣∣ b ∣ cos θ .$

Fig 1Vectors a and b at angle θ, with the scalar projection |a|cos θ of a onto b marked as a shadow along b, and a · b = |a||b|cos θ. The inset compares a · b greater than 0, equal to 0 (perpendicular) and less than 0.

$a \cdot b > 0$ : the angle between them is acute or zero ( $0° \leq θ < 90°$ ).
$a \cdot b = 0$ (with neither vector zero): the vectors are perpendicular, also called ORTHOGONAL.
$a \cdot b < 0$ : the angle is obtuse or straight ( $90° < θ \leq 180°$ ).
Rearranging the geometric form gives the angle between any two non-zero vectors: $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ .

VideoA short introduction to the dot product: how to compute it and what it says about the angle between two vectors.Watch on YouTube

Properties

Property	Statement
Commutative	$a \cdot b = b \cdot a$
Distributive	$a \cdot (b + c) = a \cdot b + a \cdot c$
Scalars factor out	$(k a) \cdot (m b) = km (a \cdot b)$
Dot with itself	$a \cdot a = ∣ a ∣^{2}$
Unit vectors	$i \cdot i = j \cdot j = k \cdot k = 1$ , $i \cdot j = j \cdot k = i \cdot k = 0$
Not associative	$(a \cdot b) \cdot c$ is undefined: $a \cdot b$ is a scalar, and you cannot dot a scalar with a vector

Each property can be checked from the component form.

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In the Dot product and projection tab, drag a and b and watch the dot product (from components and from |a||b| cos θ), the angle between them, and the scalar and vector projections change sign as the angle passes through 90 degrees.

Projections

Shine a light straight down onto the line of $b$ and $a$ casts a shadow along it. The signed length of that shadow is the SCALAR PROJECTION of $a$ onto $b$ , equal to $∣ a ∣ cos θ$ . It is negative when the angle is obtuse (the shadow falls behind the tail). Turning the shadow into a vector along $b$ gives the VECTOR PROJECTION. Because $∣ a ∣ cos θ = \frac{a \cdot b}{∣ b ∣}$ , both are easy to compute from components.

scalar proj_{b} a = \frac{a \cdot b}{∣ b ∣} proj_{b} a = (\frac{a \cdot b}{∣ b ∣ ^{2}}) b = (\frac{a \cdot b}{b \cdot b}) b

Projection of a ONTO b: divide by the magnitude of b, the vector being projected onto.

VideoMr. Cioffi finds scalar and vector projections of one vector onto another, following the MCV4U course.Watch on YouTube

Work

A constant force $F$ (in newtons) acting through a displacement $d$ (in metres) does work $W = F \cdot d = ∣ F ∣∣ d ∣ cos θ$ joules. Back to the wagon: with $∣ F ∣ = 50$ N and $∣ d ∣ = 20$ m, a level handle gives $W = 1000$ J, a handle at $30°$ gives $1000 cos 30° \approx 866$ J, and a handle at $60°$ gives $1000 cos 60° = 500$ J. The lower the handle, the more of the same pull goes into moving the wagon forward. A force perpendicular to the motion does no work at all.

The Dot Product

Key things to remember

Two definitions that agree

Properties

Projections

Work

Worked examples

Where marks get lost

Flashcards 8

Practice