AI Tutor
When game developers program a laser beam, they need it to travel at exactly the same speed no matter where you aim. How do you keep a vector's exact direction while forcing its length to be exactly 11?
The answer is normalization β€” the process of scaling any non-zero vector down to a length of 11 without changing its path.
So what does a vector of length 11 look like mathematically?

What Is a Unit Vector?

A unit vector is any vector whose length, or magnitude, is exactly 11. We write a unit vector in the direction of v\mathbf{v} with a small caret called a "hat": u^\mathbf{\hat{u}} or v^\mathbf{\hat{v}}.
To find u^\mathbf{\hat{u}}, we multiply the original vector v\mathbf{v} by the scalar reciprocal of its magnitude: u^=1βˆ₯vβˆ₯v=vβˆ₯vβˆ₯\mathbf{\hat{u}} = \frac{1}{\|\mathbf{v}\|} \mathbf{v} = \frac{\mathbf{v}}{\|\mathbf{v}\|}
Why does this guarantee a length of 11? Because pulling the positive scalar 1βˆ₯vβˆ₯\frac{1}{\|\mathbf{v}\|} outside the magnitude gives βˆ₯u^βˆ₯=∣1βˆ₯vβˆ₯∣βˆ₯vβˆ₯=βˆ₯vβˆ₯βˆ₯vβˆ₯=1\|\mathbf{\hat{u}}\| = \left|\frac{1}{\|\mathbf{v}\|}\right| \|\mathbf{v}\| = \frac{\|\mathbf{v}\|}{\|\mathbf{v}\|} = 1.
Let's put this formula to work on a concrete two-dimensional vector.

Normalizing in Two Dimensions

Suppose you have the vector v=⟨3,βˆ’4⟩\mathbf{v} = \langle 3, -4 \rangle. First, compute its magnitude using the distance formula: βˆ₯vβˆ₯=32+(βˆ’4)2=9+16=5\|\mathbf{v}\| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5.
Next, divide every component of v\mathbf{v} by 55: u^=⟨3,βˆ’4⟩5=⟨35,βˆ’45⟩\mathbf{\hat{u}} = \frac{\langle 3, -4 \rangle}{5} = \left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle