A ** Unit Vector ** has a magnitude of 1 and is denoted like \(\hat{u}\). You can make a unit vector in the same
direction of a vector by dividing by the magnitude:

\( \hat{u} = \frac{\vec{u}}{|\vec{u}|} \)

By multiplying a vector by a scalar equal to the magnitude, we are able to stretch the vector until it has a magnitude of 1
but keep the direction. You can say that the vecetors are **Collinear** or **Parallel Vectors **.

Write a unit vector in the direction of \( \vec{u}= 10 \; \; \; 70^\circ \; S \; of \; W\).

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Write a unit vector in the direction of \( \vec{u} = \left<4, -2\right>\).

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The **Standard Basis** unit vectors are in the direction of the the x, y and z axes:

\( \hat{\imath} = \left<1,0,0\right> \)

\( \hat{\jmath} = \left<0,1,0\right> \)

\( \hat{k} = \left<0,0,1\right> \)

Later, we will see how we can use these unit vectors to create linear combinations of other vectors.

We can also use ** Unit Vectors ** to convert a **Geometric Vector** to an **Algebraic Vector** when
there is information about how the vector interacts. For example, consider the force along a rope holding up a beach
volleyball net. The force acts in the direction of the rope. If we know the magnitude of the force, we can convert it
into an **Algebraic Vector**.

First, we measure the x and y-components of the rope. We can make a position vector \(\vec{p}\) which represents the a vector along the physical rope:

\(\vec{p} = \left<-10, -10 \right> [cm]\)

Next, let's find the ** Unit Vector ** \(\hat{p}\):

\(\hat{p} = \frac{\vec{p}}{|\vec{p}|} = \frac{1}{\sqrt{(-10)^2+(-10)^2}} \vec{p} \)

\(\hat{p} = \frac{1}{10\sqrt{2}} \left<-10, -10 \right> \)

Now, the force in the rope will also run in the direction of the rope. That means:

\(\hat{F} = \hat{p} = \frac{1}{10\sqrt{2}} \left<-10, -10 \right> \)

We know that the force vector is:

\( \vec{F} = \hat{F} |\vec{F}|\)

That is to say, the force vector has a magnitude \(|\vec{F}|\) and runs in the direction of \(\hat{F}\).

If we know the magnitude is \( 50 [N] \) then the vector is:

\( \vec{F} = \hat{F} |\vec{F}|\)

\( \vec{F} = \frac{1}{10\sqrt{2}} \left<-10, -10 \right> (50)\)

\( \vec{F} = \left<\frac{-500}{10\sqrt{2}}, \frac{-500}{10\sqrt{2}} \right>\)

\( \vec{F} = \left<\frac{-50}{\sqrt{2}}, \frac{-50}{\sqrt{2}} \right> \; [N]\)

Remeber, the \(\frac{1}{10\sqrt{2}}\) and \(50\) are scalar multiples.

We can double check the magnitude of \(F\):

\( |\vec{F}| = \sqrt{\left(\frac{-50}{\sqrt{2}}\right)^2 + \left(\frac{-50}{\sqrt{2}}\right)^2} \)

\( |\vec{F}| = \sqrt{\frac{50^2}{2} + \frac{50^2}{2}} \)

\( |\vec{F}| = \sqrt{50^2} \)

\( |\vec{F}| = 50 \; [N]\)