The Scalar Product and Projections

In short: The scalar (dot) product multiplies two vectors to give a number: $a \cdot b = a_{1} b_{1} + a_{2} b_{2} = ∣ a ∣∣ b ∣ cos θ$ . It finds the angle between two vectors, and a dot product of zero means they are perpendicular.

Starting point: what you already know

Vectors in component form, magnitude $∣ a ∣ = x^{2} + y^{2}$ , and unit vectors (previous chapter).
The cosine rule $c^{2} = a^{2} + b^{2} - 2 ab cos C$ .
$cos 9 0^{\circ} = 0$ , and $cos θ < 0$ for obtuse angles.
Perpendicular lines have gradients that multiply to $- 1$ .

Two ways to calculate it

The answer is a number (a scalar), not a vector. The two formulas agree; the cosine rule shows why.

Perpendicular vectors

Useful properties

$a \cdot b = b \cdot a$
$a \cdot a = ∣ a ∣^{2}$
$a \cdot (b + c) = a \cdot b + a \cdot c$
$(k a) \cdot b = k (a \cdot b)$
$i \cdot i = j \cdot j = 1$ and $i \cdot j = 0$

Projections

The projection of $a$ onto $b$ is the "shadow" of $a$ on the line of $b$ , when light shines at right angles to $b$ .

%%FIG:projection%%

Scalar product

Formulas

a₁b₁ + a₂b₂
|a||b| cos θ

Results

Number, not a vector
Negative means obtuse
Zero means perpendicular

Properties

a · a = |a|²
Order does not matter

Projection

(a · b / b · b) b
Length |a| cos θ

Proofs

Diagonals of a rhombus
Angle in a semicircle

Year 12Proving geometry with vectors

The diagonals of a rhombus are perpendicular. Let the sides from one corner be $a$ and $b$ with $∣ a ∣ = ∣ b ∣$ . The diagonals are $a + b$ and $a - b$ . $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2} = 0$ So they are perpendicular.

The angle in a semicircle is a right angle. Let the centre be $O$ , the diameter run from $- a$ to $a$ , and $P$ have position vector $p$ with $∣ p ∣ = ∣ a ∣$ . Then $(p - a) \cdot (p + a) = ∣ p ∣^{2} - ∣ a ∣^{2} = 0$ .

Length of a sum. Expand a dot product: $∣ u + v ∣^{2} = (u + v) \cdot (u + v) = ∣ u ∣^{2} + 2 u \cdot v + ∣ v ∣^{2}$ . With $∣ u ∣ = 3$ , $∣ v ∣ = 5$ and $6 0^{\circ}$ between them, $u \cdot v = 7.5$ , so $∣ u + v ∣ = 9 + 15 + 25 = 7$ . Adding $∣ a + b ∣^{2}$ and $∣ a - b ∣^{2}$ gives $2∣ a ∣^{2} + 2∣ b ∣^{2}$ (the parallelogram law). For unit vectors at angle $2 θ$ , $∣ a + b ∣^{2} = 2 + 2 cos 2 θ = 4 cos^{2} θ$ .

Work done by a constant force $F$ over a displacement $s$ is $F \cdot s$ . A force $2 i + 5 j$ N moving an object by $3 i + j$ m does $6 + 5 = 11$ J of work.

A perpendicular vector. Swap the components and change one sign: $5 i - 12 j$ is perpendicular to $12 i + 5 j$ . Scale it to the length you need.

Triangle area. Find the angle between $O A$ and $O B$ from the dot product, then use $\frac{1}{2} ∣ a ∣∣ b ∣ sin θ$ .

Reflection in a line. The reflection of $b$ in the line through $O$ with direction $a$ is $2 proj_{a} b - b$ .

Distance from a point to a line. If the line through $O$ has direction $d$ , the scalar projection of $q$ onto it is $p = \frac{q \cdot d}{∣ d ∣}$ , and the distance is $∣ q ∣^{2} - p^{2}$ . The closest point on the line is $proj_{d} q$ .

Quick summary

$a \cdot b = a_{1} b_{1} + a_{2} b_{2} = ∣ a ∣∣ b ∣ cos θ$ , a number.
Use it to find the angle between vectors.
$a \cdot b = 0$ means perpendicular. Negative means obtuse.
$a \cdot a = ∣ a ∣^{2}$ .
Projection of $a$ onto $b$ : $(\frac{a \cdot b}{b \cdot b}) b$ .

Quiz

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Find $(2 i + 3 j) \cdot (4 i - j)$ .
1. A 11
2. B 5
3. C $8 i - 3 j$
4. D 14
Show answer
B: 5
$8 - 3 = 5$ .
The scalar product of two vectors is:
1. A a number
2. B a vector
3. C always positive
4. D always zero
Show answer
A: a number
That is why it is called the scalar product.
If $a \cdot b = 0$ for non-zero vectors, then they are:
1. A parallel
2. B perpendicular
3. C equal
4. D opposite
Show answer
B: perpendicular
$cos θ = 0$ , so $θ = 9 0^{\circ}$ .