Unit Vectors and i & j Notation: Study with Video Lessons, Practice Problems & Examples

Vectors can be represented using I and j notation, where I hat and j hat are unit vectors in the x and y directions, respectively. To convert a vector into this notation, identify its components and express it as a combination of I and j. For any vector, the unit vector can be calculated by dividing the vector by its magnitude, ensuring it has a magnitude of 1. This method is essential for understanding vector operations and simplifies calculations in various applications.

i & j Notation

Welcome back, everyone. So in recent videos, we've been talking about vectors and various things associated with vectors, like their magnitude and direction. Now, what we're going to be learning about in this video is a new way that we can represent vectors using something called I and j notation. Now if this sounds scary, or like it's some kind of crazy code word here, don't sweat it, because it turns out, I and j notation is actually very straightforward. This is definitely a skill you're going to need to have in this course. So without further ado, let's get right into things.

Now let's say we have some kind of vector, vector v. And we're asked to represent this vector using I and j notation. Well, to do this, you first need to understand what a unit vector is. A unit vector is a vector that has a magnitude of 1, and points in any direction. So you can kind of think of it like a unit vector is the opposite of a scalar, where a scalar really only tells us a magnitude, a unit vector really only tells us a direction. Now, it turns out that these unit vectors that point specifically in the x and y direction, we give them special names. The unit vector that points in the x direction, we call this vector I hat, and the unit vector that points in the y direction, we call j hat. Now depending on what textbook you use, it's also possible that you'll see these vectors written as x hat and y hat. Now if you ever see this, just know that these mean the exact same thing. But for this video, we're going to use I and j, but it's also possible you see it as x hat and y hat.

Now going back to this example that we had where we had this vector down here. If we want to represent this using I and j notation, well, we can just use this idea of scaling vectors and using the tip to tail method that we've already learned about. So if I'm looking at my I vector, which I know points in the x direction, it has a magnitude of 1, I can see that I'm going to need 1, 2, 3, 4 I vectors to get to our vector v. And I can see that I'm also going to need 1, 2, 3 j vectors. So I need 4 I hat vectors and 3 j hat vectors to get this resultant vector, vector v. And that right there is how you can use this I j notation. This right here is the solution. So, really, I and j notation is just asking us what combination of I and j vectors do we need to get whatever vector we're looking for.

Now it turns out, you might also be given this vector in component form, and component form is something that we've been using throughout these videos. If you're given these vectors in component form, there's a very straightforward way of finding the I j notation. All you need to do is take your x component and multiply it by I hat, and then take your y component and multiply it by j hat. So if we have this vector here, which is 4, 3, you could just write this as 4I + 3J. And as you can see, that's exactly what we did here. So this is the idea of I j notation and how to convert from component form to I j notation. And it turns out, there actually are situations where this I j notation is going to be more useful than component form. But to understand this, as well as this just general concept of I and j notation, let's get some more practice using some examples.

So in these examples that we have down here, we're given these 2 vectors, vector u, which is 2, 4, and vector v, which is 1, 0. Now we're asked to rewrite these vectors using I and j notation, and we're going to start with example a. Example a asks us to find vector u, and I can see that vector u is 2, 4. So using I j notation, we're going to have 2i + 4j. And that right there is the solution. That's all there is to it. That's example a, right there. Now let's go ahead and try example b. Example b asks us to find vector v. I can see vector v is 1, 0, so using I j notation, we're gonna have 1 I + 0 J. Now I could say that this is my final answer, but it turns out I can simplify things further. 1 times I is just I, and 0 times j is 0. I + 0 is just I, so vector v is I, and that's the solution. And this is a situation where the I and j notation can actually be more useful than component form. Because notice how we were able to take a vector in component form, and we were able to compress it down and write it in a much more simple way using this new notation.

Now let's go ahead and try example c. Example c asks us to find vector u + vector v. Now vector u, I can see we already figured out. The I j notation is 2i + 4j, and vector v is just, well, I. Now how exactly could we add these vectors together? Well, it turns out using I j notation for any operation is the same idea as using component form. All you want to do is add, subtract, or multiply the individual components. So going back up to this example, well, what I can see here is that we have 2i and we have I. So adding these together, we'll get 3i. And I can see that we have 4j, and then we don't have any j's on this side, so we're just gonna be left with 4j. So 3i + 4j would be the solution to example c.

So as you can see it's very straightforward. It's just like what we learned for component form. But now let's go ahead and try example d. Example d asks us to find the u - 2v. Now I'm first going to figure out what vector u is, and I can see that that's what we already calculated. It's 2i + 4j. Now to find vector 2v, well, 2v is just gonna be 2 times vector v. And you can see vector v is just I, so it's gonna be 2 times I, which could also just be written as 2I. So now that we have vector u and vector 2 v, to find u - 2v, well, we can just subtract these vectors we figured out. So we're going to have vector u, which is 2i + 4j, and this is all going to be minus vector 2v, which is 2i. Now what I can do is I can subtract the like components. So we're going to have 2I - 2I, which is 0. And then we're going to have 4j - nothing, which is just 4j. So we have 0 + 4j, which is just 4j. So this right here is the solution to example d. And as you can see, we were able to get a more compact simple vector by using I and j notation. So this is really what I and j notation is all about, and how you can do some basic operations. So hope you found this video helpful. Thanks for watching.

If vector $v⃗=12î-2ĵ$ and vector $u⃗=5î+20ĵ$ calculate $2v ⃗-2u ⃗$ using $î$ and $ĵ$ notation.

• A. $14î-44ĵ$
• B. $7î-18ĵ$
• C. $14î+44ĵ$
• D. $14î-36ĵ$

If vector $v ⃗=$ $11ĵ$ and vector $u⃗=$ $10î-25ĵ$ calculate $v⃗+\frac15u⃗$ using $î$ & $ĵ$ notation.

• A. $13î+14ĵ$
• B. $2î+14ĵ$
• C. $13î-5ĵ$
• D. $2î+6ĵ$

If vector $a⃗=20î$ and vector $b⃗=50ĵ$ calculate $a ⃗-b ⃗$ using $î$ and $ĵ$ notation.

• A. $-30ĵ$
• B. $20î-50ĵ$
• C. $30î̂$
• D. $20î+50ĵ$

Unit Vector in the Direction of a Given Vector

Welcome back, everyone. So, in the previous video, we talked about unit vectors. We got introduced to this general topic, and we specifically discussed vectors that point in the x and y directions, which were vectors I hat and j hat, respectively. Now, what we're going to be talking about in this video is how we can actually calculate a unit vector that points in any direction. Because it's possible that you're going to have a vector that doesn't just point to the right or up. You might have a vector that points in some other direction and we're going to talk about how to find unit vectors that point anywhere. So without further ado, let's just jump right into an example.

So let's say we have this example where we're given vector \( \mathbf{v} \) and vector \( \mathbf{v} \) is equal to \( 4i + 3j \). We're asked to find a unit vector \( \mathbf{\hat{v}} \) that points in the same direction as \( \mathbf{v} \). Now, since this vector is a combination of i's and j's we can't say that our unit vector is just going to be I or J. So to figure out what this unit vector is, it turns out there's actually a straightforward way of doing this. All you need to do is take whatever vector you're given and divide it by its magnitude, and that's going to give you the unit vector. And I think this makes logical sense, because if you take a number and divide it by itself, you would typically just get 1. And we know unit vectors have a magnitude of 1, so it makes sense that we would just need to divide this vector by its own magnitude. So this would be the equation for finding the unit vector.

And I noticed that we do have the vector given to us, but we don't have the magnitude. So that's what I'm going to calculate. Now to find the magnitude of a vector, you just need to take the square root of the x component squared plus the y component squared. I see the x component is 4 and that the y component is 3. Now, it turns out the square root of \( 4^2 + 3^2 \) this is all just going to come out to equal 5. So 5 is the magnitude of our vector \( \mathbf{v} \).

So now that we have our magnitude we can calculate \( \mathbf{\hat{v}} \). All we need to do is take our vector which is \( 4i + 3j \) and divide it by its own magnitude which we calculated to be 5. So I'm going to take each of these components 4 and 3 and I'm going to divide them by 5. So we'll have \( \frac{4}{5}i + \frac{3}{5}j \), and this right here is the unit vector \( \mathbf{\hat{v}} \) and the solution to this example.

And so what happens is if we go over here to our graph, this vector \( \mathbf{\hat{v}} \) is going to look something like this. Unit vectors always have a magnitude of 1, but it's going to point in the direction of the vector that we already have. And what this tells us what we just calculated is that this vector specifically points \( \frac{4}{5} \) in the I direction, and it points \( \frac{3}{5} \) in the j direction, and that's what gives you the unit vector \( \mathbf{\hat{v}} \).

Now something else that we might need to do is figure out or prove to ourselves that this is actually a unit vector. And how could we do that? Well, you could do that by figuring out if the magnitude of this vector is truly equal to 1. So let's go ahead and try this example where it asks us to find if \( \mathbf{\hat{v}} \) in this example above is actually a unit vector. So let's see what happens here. Well, what I'm going to do is calculate the magnitude of \( \mathbf{\hat{v}} \). And to find the magnitude of any vector, you just need to take the square root of the x component squared plus the y component squared.

Now I can see that the x component is \( \frac{4}{5} \), and I need to add this to the y component which is \( \frac{3}{5} \). Now what I can do is I can square both the numerator and denominator for each of these fractions and add them together. So we're going to have the square root, and then we have \( 16 \) divided by \( 25 \), that's going to be plus \( 9 \), divided by \( 25 \). Now notice we got the same denominator in both cases. And since we have the same denominator, we can just add the numerator straight across. Now \( 16+9 \), that's equal to \( 25 \). So we'll have the square root of \( 25 \) divided by \( 25 \). Now \( 25 \) over \( 25 \), well, any number divided by itself is just going to be equal to 1, and the square root of 1 is just 1.

Notice how we got that our magnitude for \( \mathbf{\hat{v}} \) is 1, and that means this is a unit vector. So the solution we calculated up here is correct. So this is the main idea of finding unit vectors in any direction. All you need to do is take whatever vector you have and divide it by its magnitude, and then you can use this strategy of finding the magnitude of your unit vector to check to make sure you did things right.

Hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.

Unit Vector in the Direction of a Given Vector Example 1

Hey, everyone. Let's give this problem a try. So here we're asked to find the vector û, the unit vector, in the direction of vector u. Now to solve this problem, we're really just trying to calculate this unit vector. And recall for the unit vector û, it's going to be vector u divided by the magnitude of u. This is always the equation that you use when trying to find the unit vector. Now I can see that we are given vector u that's -i + 2j. And you can also write that as -1i + 2j so that way we can clearly see what the x and y components are, and then that's all going to be divided by the magnitude of vector u. Now the magnitude of u is something we don't have yet, so that's what we're gonna have to calculate. Now to find the magnitude of any vector, what you want to do is take the square root of the x component squared plus the y component squared, and this is how you can calculate it.

Now what I can see from the situation is that our x component is what's in front of the i, -1. So we're gonna have -1 squared plus the y component, which is 2. Now the square root of -1 squared, well, I should just say -1 squared comes out to positive one, and then 2 squared comes out to 4. So I have the square root of 1 + 4 which is the square root of 5. So that means that the magnitude of u is the square root of 5. Now from here, now that we found the magnitude of u, what I need to do is divide that into the vector on top. So we're going to have -1i + 2j, that's going to be divided by the square root of 5. And dividing each of these components by the square root of 5 we're going to have -1 over the square root of 5i, plus 2 over the square root of 5j.

So this is what the vector ends up being for our unit vector, but notice that we end up with a square root in the denominator of each of these fractions. Typically, we don't want our final answer to have this radical on bottom. So what I'm going to do is rationalize the denominator by multiplying the top and bottom of both fractions by the square root of 5. Now what this is going to do is get these square roots to cancel in each of the denominators. So for vector û, I'm going to end up with 1 times the square root of 5, which is just the square root of 5, divided by 5i, and then this is negative. And that's going to be plus 2 times the square root of 5 divided by 5j. And this right here is the unit vector and the solution to this problem. So hope you found this video helpful. Thanks for watching.

Find the unit vector in the direction of $a⃗=6î+3ĵ$.

• A. $â=3\surd5ĵ$
• B. $â=\frac{2\surd5}{5}î-\frac{\surd5}{5}ĵ$
• C. $â=\frac{2\surd5}{5}î+\frac{\surd5}{5}ĵ$
• D. $â=\frac{2\surd5}{5}î+\frac{3\surd5}{5}ĵ$

Find the unit vector in the direction of $v⃗=12î-35ĵ$.

• A. $v̂=\frac{12}{37}î-\frac{35}{37}ĵ$
• B. $v̂=37î̂$
• C. $v̂=35î-12ĵ$
• D. $v̂=\frac{35}{37}î-\frac{12}{37}ĵ$

Unit Vector in the Direction of a Given Vector Example 2

Welcome back, everyone. Let's try solving this example. In this example, we're told if vector a equals negative 13, and vector b equals 47, and c is equal to 4 times vector a plus 2 times vector b, find the unit vector in the direction of c. There's a lot going on in this problem, but to make this as straightforward as possible, I'm going to break it up into steps. The first step is to find vector c. The second step is to find the magnitude of c, because finding the magnitude is necessary for finding the unit vector. The third step is to find the unit vector ĉ, which is the ultimate goal of this problem and is in the direction of c. Let's start by determining what vector c is. We see that c is 4a+2b. We need to perform the operations we see here. Let's plug vector a in, which is negative 13, and vector b, which is 47. We need to scale these vectors by the numbers in front of each. So, we have 4 times negative 13, resulting in negative 4, and 4 times 13, giving us 12. These results are added to 2 times 4 and 2 times 7. So, 2 times 4 is 8 and 2 times 7 is 14. Adding the corresponding components results in 4 for the x-component and 26 for the y-component. This is vector c, completing the first step.

The next step is to find the magnitude of c. We use the equation for the magnitude of a vector, which is the square root of the x-component squared plus the y-component squared, similar to the Pythagorean theorem with vectors. For vector c, the x-component is 4, so we have 4 squared, plus the y-component squared, which is 26 squared. This results in 16 and 676, respectively, for a total of 692. Simplifying, we find 2 squared times 173, leading us to 2 times the square root of 173 as the simplified magnitude of vector c.

The final step is to find the unit vector ĉ. The unit vector is vector c divided by its magnitude. We have vector c as 4, 26 and its magnitude as 2 times the square root of 173. Dividing each component by the magnitude, we can simplify the vector components. Rationalizing the denominators of the fractions yields the final unit vector ĉ as 2√173⁄173 for the x-component and 13√173⁄173 for the y-component. These are all the steps you need when trying to find a unit vector in complex operations. I hope you found this video helpful. Thanks for watching.

Here’s what students ask on this topic:

What is i and j notation in vectors?

i and j notation is a way to represent vectors using unit vectors. The unit vector $i_{^}$ points in the x-direction, and the unit vector $j_{^}$ points in the y-direction. Any vector can be expressed as a combination of these unit vectors. For example, a vector with components (4, 3) can be written as $4i_{^}+3j_{^}$. This notation simplifies vector operations and calculations.

How do you convert a vector to i and j notation?

To convert a vector to i and j notation, identify its x and y components. Multiply the x component by the unit vector $i_{^}$ and the y component by the unit vector $j_{^}$. For example, a vector with components (2, 4) can be written as $2i_{^}+4j_{^}$. This method helps in visualizing and performing vector operations more easily.

How do you find a unit vector in the same direction as a given vector?

To find a unit vector in the same direction as a given vector, divide the vector by its magnitude. The magnitude of a vector $(a,b)$ is calculated as $\sqrt{a^2+b^2}$. For example, for vector $(4,3)$, the magnitude is 5. The unit vector is $(\frac{4}{5},\frac{3}{5})$, or $\frac{4}{5}{i}_{^}+\frac{3}{5}{j}_{^}$.

What is the difference between component form and i and j notation?

Component form represents a vector as an ordered pair (x, y), showing its x and y components directly. i and j notation, on the other hand, expresses the vector as a combination of unit vectors $i_{^}$ and $j_{^}$. For example, a vector with components (3, 2) in component form is written as $(3,2)$, while in i and j notation, it is $3i_{^}+2j_{^}$. Both forms convey the same information but are used in different contexts for convenience.

How do you add vectors using i and j notation?

To add vectors using i and j notation, add their respective components. For example, if vector $\mathrm{u}=2i_{^}+4j_{^}$ and vector $\mathrm{v}=1i_{^}$, then $\mathrm{u}+\mathrm{v}=(2+1)i_{^}+(4+0)j_{^}$, which simplifies to $3i_{^}+4j_{^}$. This method is straightforward and aligns with component-wise addition.