Can direction stay fixed while length changes?

A vector can point in a useful direction but be too long for the job. Can we shorten it without turning it?

Start with $\mathbf v=(3,4)$. Its horizontal and vertical components form the two short sides of a right triangle. By the Pythagorean theorem, the arrow has length $\sqrt{3^2+4^2}=5$.

Keep a record of your investigation

Use $k=1$ and $\mathbf v=(3,4)$ as the reference. Before each trial, write the predicted components, length, and direction relative to the original arrow. Change only $k$, keeping $\mathbf v$ fixed. Record those three observations and compare them with the reference; separate a change in length from a reversal of direction.

Find a one-unit arrow

Predict the length when $\mathbf v$ is multiplied by 2, by $1/2$, and by $1/5$. Which choice should produce length 1?

Try the first two values using Scale k, then select k = 0.2. The dashed arrow keeps the original vector visible.

Now predict what k = −1 will do. Will the length be −5? What direction remains at k = 0?

A number that measures vector length

The Euclidean norm, written $\|\mathbf v\|$, is the length of the vector. In two dimensions,

For more coordinates, add all the squared components before taking the square root. For example, $\|(2,-1,2)\|=\sqrt{4+1+4}=3$.

The norm is a nonnegative number, while the vector is a list of components. A negative component does not make a negative norm: its square contributes positively.

Scaling multiplies length by the absolute value of the scale:

This explains both stretching and reversal in one rule.

Normalize: keep direction, make length one

For a nonzero vector, divide every component by its norm:

This operation is called normalization. The divisor is positive, so direction stays the same. The resulting length is $\|\mathbf v\|/\|\mathbf v\|=1$.

For $\mathbf v=(3,4)$, normalization gives $(3/5,4/5)$. Check its squared length: $9/25+16/25=1$.

Not every unit vector is a standard coordinate unit vector. $(3/5,4/5)$ has length 1 but points between the coordinate axes. The standard vectors point along those axes.

The zero vector cannot be normalized this way: its norm is zero, and division by zero is undefined. Choosing an arbitrary direction would introduce information the zero vector does not contain.

Distance starts with a difference

To measure the distance between points $P$ and $Q$, first find the displacement $Q-P$, then take its norm. For $P=(1,2)$ and $Q=(4,6)$,

The norm of $Q$ alone would measure distance from the origin, not from $P$.

Transfer to a new direction

Normalize $\mathbf w=(-5,12)$. Then find a vector in the same direction with length 2.

Design your own comparison

Choose a positive scale within the explorer’s range that we have not used. Predict the result, test it, and record the components and length. Then change only the sign of that scale. What stays the same, and what reverses?

Choose a new nonzero vector on paper and repeat your comparison. Use its components to explain whether your rule depends on the particular vector $(3,4)$.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.