Unit Vectors

A vector of length one

A unit vector is a vector whose length is exactly 1. It carries a direction and nothing else, which is why it is used wherever only the direction matters: the way a camera points, the way a wind blows.

Every vector other than the zero vector has a unit vector pointing the same way. Take a = (3, 4). By Pythagoras its length is $|a|=\sqrt{3^2+4^2}=\sqrt{25}=5$|a| = √(3² + 4²) = √25 = 5.

Divide by the length

To shrink a to length 1, divide each component by the length 5: (3 ÷ 5, 4 ÷ 5) = (0.6, 0.8), which is $\left(\frac{3}{5},\frac{4}{5}\right)$(3/5, 4/5) as fractions. This is the unit vector in the direction of a. It is written â and read “a-hat”.

Dividing by 5 is the same as multiplying by the scalar $\frac{1}{5}$1/5. A positive scalar keeps the direction and multiplies the length, so â points exactly the way a points, and its length is $5\times \frac{1}{5}=1$5 × 1/5 = 1.

Check it with Pythagoras: $\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=\sqrt{1}=1$√(0.6² + 0.8²) = √(0.36 + 0.64) = √1 = 1. As fractions, $\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1$9/25 + 16/25 = 25/25 = 1.

a = (3, 4), of length 5, and â = (0.6, 0.8), of length 1. The short arrow lies along the long one and ends on the ring of radius 1.

Every direction has one

Draw every unit vector from the origin and their heads trace the circle of radius 1, because each one is 1 long. The unit vector at an angle $\theta$θ above the x-axis ends at the point $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ) on that circle.

The direction of (3, 4) is about 53.1° above the x-axis, and cos 53.1° and sin 53.1° are 0.6 and 0.8 to one decimal place: the same unit vector, found from the angle instead of from the length.

A radius of 1 at 53°, close to the direction of (3, 4). Its two legs, 0.60 across and 0.80 up to two decimal places, are the components of the unit vector. Drag the point round the circle: the components change with the direction, and the length stays 1.

Length times direction

Turned around, the division says that any vector is its length times its unit vector: a = |a| × â. For a, that is 5 × (0.6, 0.8) = (3, 4), the vector we started with.

This gives a vector of any length in a chosen direction. In the direction of a, a vector of length 10 is 10â = (6, 8), and one of length 2.5 is 2.5 × (0.6, 0.8) = (1.5, 2).

When the length is not a whole number

The vector (1, 2) has length $\sqrt{1^2+2^2}=\sqrt{5}$√(1² + 2²) = √5. Its unit vector is $\left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)$(1/√5, 2/√5), which is (0.45, 0.89) to 2 decimal places. With the denominators rationalized, it is $\left(\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\right)$(√5/5, 2√5/5).

Negative components keep their signs. (−5, 12) has length $\sqrt{25+144}=\sqrt{169}=13$√(25 + 144) = √169 = 13, so its unit vector is $\left(-\frac{5}{13},\frac{12}{13}\right)$(−5/13, 12/13), still pointing left and up.

The zero vector, (0, 0), has no unit vector. Its length is 0, and dividing by 0 has no meaning; it has no direction to keep.

The usual mistakes

Leaving the vector as it is. (3, 4) points the right way, but its length is 5, not 1.

Dividing only one component. (0.6, 4) is not a multiple of (3, 4), so it points in a different direction.

Dividing by the sum of the components instead of the length. $\left(\frac{3}{7},\frac{4}{7}\right)$(3/7, 4/7) has length $\frac{5}{7}$5/7, not 1, because the length of (3, 4) is 5, not 3 + 4 = 7.

Giving a point instead of a step. After finding a step along a unit vector, add it to the starting position to find where it ends.

Pointing a camera

In the application below, a camera at C must point at a basket at G. The vector from C to G is G minus C, destination minus start. Dividing it by its length gives the unit vector that sets the camera’s direction, and 5 times that unit vector is a step of exactly 5 m along the same line, added to the position of C.