Vector addition and the dot product: from |a||b|cos θ to a₁b₁ + a₂b₂

Vector addition & dot product

The dot product is defined with lengths and an angle. Why does it come out as a₁b₁ + a₂b₂ in coordinates, where no angle appears?

Drag the tips of a and b. The board measures the angle between them on its own and computes |a||b|cos θ, then compares it with a₁b₁ + a₂b₂ from the coordinates: the two always agree. The parallelogram shows a + b, the red segment shows the projection of b onto a.

Drag the ringed points on the board.

High school
Parallelogram rule
Dot product
Projection
Perpendicular when a·b = 0

x_1

4.0

y_1

1.0

x_2

1.5

y_2

3.0

|\vec a||\vec b|\cos\theta = x_1x_2 + y_1y_2

$|\vec a|$: 4.123
$|\vec b|$: 3.354
$\theta$: 49.4
$|\vec a|\,|\vec b|\cos\theta$: 9.000
$x_1x_2 + y_1y_2$: 9.000
$|\vec b|\cos\theta$: 2.183
$|\vec a + \vec b|$: 6.801

angle between a and bacute: a·b > 0

Guided tour

1Two ways to the same number
The board measures $|\vec a|=4.123$ , $|\vec b|=3.354$ and the angle between them, $\theta=49.4^\circ$ , and multiplies: $|\vec a||\vec b|\cos\theta=9$ . Separately it takes the coordinates: $4\times1.5+1\times3=9$ . Drag either tip and the two keep agreeing.
2Head to tail
Walking along $\vec a$ and then along $\vec b$ lands at the same corner as $\vec b$ then $\vec a$ : the far corner of the parallelogram, $\vec a+\vec b=(5.5,\,4)$ . Coordinates simply add. Its length is $6.801$ , less than $|\vec a|+|\vec b|=7.477$ .
3Perpendicular: zero
Move $\vec b$ to $(-1,\,4)$ . The angle is $90^\circ$ , the projection shrinks to nothing and $\vec a\cdot\vec b=-4+4=0$ . The parallelogram is now a rectangle and $|\vec a+\vec b|^2=17+17=34$ : Pythagoras.
4Obtuse: negative
At $\vec b=(-3,\,1)$ the angle is $147.5^\circ$ . The projection of $\vec b$ falls behind $O$ , $|\vec b|\cos\theta=-2.668$ , and the dot product is $-12+1=-11$ . The sign of $\vec a\cdot\vec b$ tells you which side of the right angle you are on.

Explanation

1
Adding vectors
A vector is a displacement: how far and which way, wherever it starts. Doing displacement $\vec a$ and then $\vec b$ is one displacement, $\vec a+\vec b$ . Placing them head to tail in either order gives the two halves of a parallelogram, so $\vec a+\vec b=\vec b+\vec a$ .
In coordinates a displacement is (across, up), and acrosses add to acrosses, ups to ups:
$(x_1,y_1)+(x_2,y_2)=(x_1+x_2,\;y_1+y_2)$
2
The dot product and projection
The dot product is defined by lengths and the angle between the vectors:
$\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta$
Read it as length times projection: $|\vec b|\cos\theta$ is how far $\vec b$ reaches along the direction of $\vec a$ (the red segment), and that is multiplied by $|\vec a|$ . Work done by a force is exactly this: force times the distance moved along it.
3
Why the angle disappears in coordinates
Look at the triangle with sides $\vec a$ , $\vec b$ and $\vec a-\vec b$ and work out $|\vec a-\vec b|^2$ twice.
- With the angle. Drop the perpendicular from the tip of $\vec b$ to the line of $\vec a$ . Its foot is $|\vec b|\cos\theta$ from $O$ and the perpendicular has height $|\vec b|\sin\theta$ . Pythagoras on the right triangle with the tip of $\vec a$ :
$|\vec a-\vec b|^2=(|\vec a|-|\vec b|\cos\theta)^2+(|\vec b|\sin\theta)^2=|\vec a|^2+|\vec b|^2-2|\vec a||\vec b|\cos\theta$
- With coordinates. Pythagoras on the horizontal and vertical gaps:
$|\vec a-\vec b|^2=(x_1-x_2)^2+(y_1-y_2)^2=|\vec a|^2+|\vec b|^2-2(x_1x_2+y_1y_2)$
The two right-hand sides are equal, so
$|\vec a|\,|\vec b|\cos\theta=x_1x_2+y_1y_2$
The angle has not vanished: it is hidden in the coordinates. The first line is the law of cosines, and this argument is its proof.
4
What it is used for
- Perpendicular test. $\vec a\perp\vec b$ exactly when $x_1x_2+y_1y_2=0$ ; no angle needs measuring.
- Angle between vectors. $\cos\theta=\dfrac{x_1x_2+y_1y_2}{|\vec a|\,|\vec b|}$ .
- Length of a sum. Expanding $(\vec a+\vec b)\cdot(\vec a+\vec b)$ gives $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\,\vec a\cdot\vec b$ , so $|\vec a+\vec b|\le|\vec a|+|\vec b|$ , with equality only when the two point the same way.

Check yourself

Q01

For $\vec a=(2,\,3)$ and $\vec b=(3,\,-2)$ , what is $\vec a\cdot\vec b$ ?

Q02

If $\vec a\cdot\vec b<0$ , the angle between them is:

Q03

In the derivation of $x_1x_2+y_1y_2$ , which fact is used twice?

Keep exploring

Turn this lab into a videoUse this lab as the script for a narrated explainer you can share. The prompt below goes to the video creator, where you can edit it first.

Make the video

View · display