The dot product of two vectors | Grade 10

You pull a sled along the snow with a rope. Only the part of your pull that points along the ground moves the sled; the part that lifts it up is wasted. Physicists capture this with a single number: force times distance times the cosine of the angle between them. Mathematicians turned the same idea into a product of two vectors whose result is a number: the dot product.

The angle between two vectors

Place two nonzero vectors $\vec{a}$ and $\vec{b}$ so that they start at the same point $O$ . The angle between them, written $(\vec{a}, \vec{b})$ , is the angle $\theta$ of the figure, with $0^\circ\le\theta\le 180^\circ$ .

The angle

\theta

between

\vec{a}

and

\vec{b}

If $\theta=0^\circ$ the vectors point the same way; if $\theta=180^\circ$ they point opposite ways; if $\theta=90^\circ$ they are perpendicular, written $\vec{a}\perp\vec{b}$ .

Definition of the dot product

The result is a number, not a vector. It is positive for an acute angle, negative for an obtuse angle and $0$ for a right angle. In particular $\vec{a}\cdot\vec{a}=|\vec{a}|^2$ (the angle is $0^\circ$ and $\cos 0^\circ=1$ ).

Properties

From these we get the identities $|\vec{a}+\vec{b}|^2=|\vec{a}|^2+2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2,\qquad |\vec{a}-\vec{b}|^2=|\vec{a}|^2-2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2.$

The dot product in coordinates

Two consequences follow at once:

$\vec{a}\perp\vec{b}\iff x_1x_2+y_1y_2=0$ (for nonzero vectors).
$\cos(\vec{a}, \vec{b})=\dfrac{x_1x_2+y_1y_2}{\sqrt{x_1^2+y_1^2}\,\sqrt{x_2^2+y_2^2}}$ .

Angles in a triangle

To find the angle at a vertex, use the two vectors that leave that vertex.

Triangle

ABC

Example Angles of a triangle

Find the angles of the triangle with $A(1, 2)$ , $B(3, 1)$ and $C(2, 4)$ .

Solution. $\overrightarrow{AB}=(2, -1)$ and $\overrightarrow{AC}=(1, 2)$ .

At $A$ : $\overrightarrow{AB}\cdot\overrightarrow{AC}=2-2=0$ , so $\angle A=90^\circ$ .
At $B$ : $\overrightarrow{BA}=(-2, 1)$ and $\overrightarrow{BC}=(-1, 3)$ . The dot product is $2+3=5$ , and $|\overrightarrow{BA}|=\sqrt5$ , $|\overrightarrow{BC}|=\sqrt{10}$ . So $\cos B=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt2}$ and $\angle B=45^\circ$ .
Angles add to $180^\circ$ , so $\angle C=45^\circ$ .

Key terms

English	Tiếng Việt
angle between two vectors	góc giữa hai vectơ
dot product	tích vô hướng
perpendicular	vuông góc
acute angle	góc nhọn
obtuse angle	góc tù
work (physics)	công
commutative	giao hoán
distributive	phân phối

Summary

$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos(\vec{a}, \vec{b})$ is a number; $\vec{a}\cdot\vec{a}=|\vec{a}|^2$ .
It is commutative and distributive, and $|\vec{a}\pm\vec{b}|^2=|\vec{a}|^2\pm2\vec{a}\cdot\vec{b}+|\vec{b}|^2$ .
In coordinates, $\vec{a}\cdot\vec{b}=x_1x_2+y_1y_2$ .
$\vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0$ , and the formula for $\cos$ gives angles.

Practice

Find $\vec{a}\cdot\vec{b}$ if $|\vec{a}|=5$ , $|\vec{b}|=2$ and the angle between them is $60^\circ$ .
Find $\vec{a}\cdot\vec{b}$ if $|\vec{a}|=4$ , $|\vec{b}|=3$ and the angle is $150^\circ$ .
Compute $\vec{a}\cdot\vec{b}$ for $\vec{a}=(1, -3)$ and $\vec{b}=(2, 5)$ .
Find $m$ so that $\vec{a}=(m, 4)$ and $\vec{b}=(2, -1)$ are perpendicular.
Find the angle between $\vec{a}=(1, 2)$ and $\vec{b}=(3, 1)$ .
Find $\angle A$ in the triangle with $A(-1, 0)$ , $B(2, 0)$ , $C(2, 3)$ .
Let $|\vec{a}|=2$ , $|\vec{b}|=3$ and $\vec{a}\cdot\vec{b}=-3$ . Find $|\vec{a}+\vec{b}|$ and $|\vec{a}-\vec{b}|$ .
A force of $20$ N moves an object $3$ m along a line, the force making an angle of $30^\circ$ with the motion. Find the work (exact value, then to $0.01$ J).
Challenge. Find a vector $\vec{b}$ perpendicular to $\vec{a}=(3, 4)$ with $|\vec{b}|=10$ .
Challenge. Show that if $|\vec{a}|=|\vec{b}|$ then $(\vec{a}+\vec{b})\perp(\vec{a}-\vec{b})$ . Check it for $\vec{a}=(3, 4)$ and $\vec{b}=(5, 0)$ .

Show solutions

$5\cdot 2\cdot\cos 60^\circ=10\cdot\frac12=5$ .
$4\cdot 3\cdot\cos 150^\circ=12\cdot\left(-\frac{\sqrt3}{2}\right)=-6\sqrt3$ .
$1\cdot 2+(-3)\cdot 5=2-15=-13$ .
$2m-4=0$ , so $m=2$ .
$\vec{a}\cdot\vec{b}=3+2=5$ , $|\vec{a}|=\sqrt5$ , $|\vec{b}|=\sqrt{10}$ . $\cos=\frac{5}{\sqrt{50}}=\frac1{\sqrt2}$ , so the angle is $45^\circ$ .
$\overrightarrow{AB}=(3, 0)$ and $\overrightarrow{AC}=(3, 3)$ . Dot product $9$ ; lengths $3$ and $3\sqrt2$ . $\cos A=\frac{9}{9\sqrt2}=\frac1{\sqrt2}$ , so $\angle A=45^\circ$ .
$|\vec{a}+\vec{b}|^2=4-6+9=7$ , so $\sqrt7$ . $|\vec{a}-\vec{b}|^2=4+6+9=19$ , so $\sqrt{19}$ .
$W=20\cdot 3\cdot\cos 30^\circ=60\cdot\frac{\sqrt3}{2}=30\sqrt3\approx 51.96$ J.
Let $\vec{b}=(x, y)$ . Then $3x+4y=0$ , so $\vec{b}=t(4, -3)$ . $|\vec{b}|=5|t|=10$ gives $t=\pm2$ . So $\vec{b}=(8, -6)$ or $\vec{b}=(-8, 6)$ .
$(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2=0$ . Check: $\vec{a}+\vec{b}=(8, 4)$ , $\vec{a}-\vec{b}=(-2, 4)$ and $-16+16=0$ . ✓ (This says the diagonals of a rhombus are perpendicular.)