Magnitude and Unit Vectors

Previously, we learned about visualizing vectors. This lesson will cover how to find the unit vector.

What are Unit Vectors?

If you think back to pre-algebra, you will remember the concept of unit rates. Unit rates are when we compare the single ($1$) unit of one quantity to some rate of a second quantity. The most common real-world example of unit rates is when we look at unit pricing for produce in the grocery store. If we buy $5$ bananas for $\$0.95$, we can assume the unit price, or cost per banana, is $\$0.19$.

We can extend this concept into vectors! Typically, vectors can be of any magnitude or direction. Unit vectors can also be of any direction, but they have a very specific magnitude, $1$. We can examine this idea on a coordinate plane.

On the coordinate plane, the arrow is drawn from the origin to the point $(0,1)$. The distance, or magnitude, of this arrow is $1$. Because this arrow has both a magnitude, of $1$, and a direction, up, we can call this arrow a vector. All vectors that have a magnitude of $1$ are called unit vectors, which we denote as,

$$\hat{a}$$

, which we verbally say as "a hat". The "hat", $\hat{ \\\\}$ , denotes a unit vector. There is no unit vector formula because we use this designation.

(Video) What is a Unit Vector?

by Brian McLogan

This lesson by Brian McLogan talks about how to find a unit vector.

Summary

A vector is going to have an initial point, a direction, and a magnitude. Say we have our initial point at $(0,0)$, at the origin. If we were to draw a vector into the first quadrant, we can find the magnitude by looking at the final point of the vector.

To find the distance,

$$||v|| = \sqrt{(v_1)^2 + (v_2)^2} = \sqrt{6^2 + 6^2} = \sqrt{72}$$

If we go back to the unit circle, you'll remember that there were one reason why we were able to easily find angles: the distance between the origin and any point on the unit circle is $1$. So, what we're doing is creating right angles with a hypotenuse of $1$.

Let's draw a unit circle on our existing vector.

So, the unit vector has the same direction as the original vector, it just has a magnitude of $1$, which we can see from the unit circle on the graph.

(Video) Magnitude of a Vector

by Mario's Math Tutoring

This video by Mario's Math Tutoring talks about how to find the magnitude of a vector.

Summary

The magnitude of a vector is how long it is, which represents how much force the vector has. The larger the vector, the more force it has.

To find the magnitude, ||v||, we use the following formula for a vector, $<x,y>$,

$$\sqrt{x^2 + y^2}$$

This formula is derived from the Pythagorean Theorem!

Magnitude of a Vector

A vector has two main components: a magnitude and a direction. The magnitude is the distance of a vector, or how long the vector is. We've mentioned the distance of lines before using the distance formula. We're going to use a similar idea when looking at the magnitude of a vector, except this formula was derived from the Pythagorean Theorem. We can designate magnitude of a vector, $\vec{v}$, with the double bars: $||\vec{v}||$. You may also see magnitude designated with the single bar, $|\vec{v}|$.

To find the magnitude, we find the square root of the sum of the components of the vector. For vector $\vec{v}$, $<a,b>$, the magnitude would be:

$$||\vec{v}|| = \sqrt{a^2 + b^2}$$

Example: Say we had the vector $\vec{a}$, $<3,4>$. What is the magnitude of $\vec{a}$?

Solution: We can plug in the components of vector $\vec{a}$ in the formula to solve for magnitude!

$$\begin{align} ||\vec{a}|| &= \sqrt{(3)^2 + (4)^2} \\ &= \sqrt{9 + 16} \\ &= \sqrt{25} \\ &= 5 \\ \end{align}$$

So, the magnitude of this vector is $5$!

Magnitude of a Vector

We know that a vector has both magnitude and direction. We can calculate the magnitude, or length, or a vector by using a formula.

A vector is a vector - so it will always give you a magnitude and a direction.

Example

Find the magnitude of the vector, $\vec{d}$, where $\vec{d} = <3,4>$.

To solve this problem, we will plug in our values into the formula we learned above.

$$\begin{align} ||\vec{d}|| &= \sqrt{3^2 + 4^2} \\ ||\vec{d}|| &= \sqrt{9 + 16} \\ ||\vec{d}|| &= \sqrt{25} \\ ||\vec{d}|| &= 5 \\ \end{align}$$

So, the magnitude, or length, of vector $\vec{d}$ is $5$!

Other Topics with Vectors

Get ready to be a master of vectors! You can learn about the dot products, vector operations, the cross product, and vector projections.