Unit Vectors and i & j Notation - Video Tutorials & Practice Problems

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 gonna be learning about in this video is a new way that we can represent vectors using something called INJ 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 ing 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 when 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 one and points in any direction. So you can kind of think of it like a unit vector is the opposite of a SCR where a SCR 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 gonna 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 one, I can see that I'm going to need 12, 34 I vectors to get to our vector V. And I can see that I'm also going to need 123 J vectors. So I need four I hat vectors and three JH vectors to get this result in vector vector V. And that right there is how you can use this IJ 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 giving these vectors in component form, there's a very straightforward way of finding the IJ 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 four comma three, you could just write this as four I plus three J. And as you can see, that's exactly what we did here. So this is the idea of IJ notation and how to convert from component form to IJ notation. And it turns out there actually are situations where this IJ 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 two vectors vector U which is two comma four and vector V which is one comma zero. 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 two comma four. So using IJ notation, we're going to have two I plus four J. 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 one comma zero. So using IJ notation, we're gonna have one I plus zero J. Now I could say that this is my final answer. But it turns out I can't simplify things further. One times I is just I and zero times J is zero, I plus zero 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 plus vector V. Now vector U I can see we already figured out the IJ notation is two I plus four J and vector V is just, well, I now how exactly could we add these vectors together? Well, it turns out using IJ notation for any operation is the same idea as using component form. All you want to do is add subtractor multiply the individual components. So going back up to this example, well, what I can see here is that we have two I and we have I, so adding these together will get three I and I can see that we have four J and then we don't have any J's on this side. So we're just gonna be left with four J. So three I plus four J 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 asked us to find a U minus two V. 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 two I plus four J. Now to find vector two V, well, two V is just gonna be two times vector V. And you can see the vector V is just I, so it's gonna be two times I, which could also just be written as two I. So now that we have vector U and vector two V defined U minus two V. Well, we can just subtract these vectors. We figured out it's going to have vector U which is two I plus four J and this is all going to be minus vector two V which is two I. Now, what I can do is I can subtract the light components. So we're going to have two I minus two I which is zero. And then we're going to have four J minus nothing, which is just four J. So we have zero plus four J which is just four J. 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 INJ notation. So this is really what INJ notation is all about and now 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, Ihat and J hat respectively. Now, what we're gonna 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 V and vector V is equal to four I plus three J, we're asked to find a unit vector V hack that points in the same direction as V. Now, since this vector is a combination of I's and J's, we can't say that our unit vector is just gonna 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 one and we know unit vectors have a magnitude of one. 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 four and that the Y component is three. Now it turns out the square root of four squared plus three squared. This is all just gonna come out to equal five. So five is the magnitude of our vector V. So now that we have our magnitude, we can calculate V hat. All we need to do is take our vector which is four I plus three J and divide it by its own magnitude, which we calculated to be five. So I'm gonna take each of these components four and three and I'm going to divide them by five so about 4/5 I hat plus 3/5 J hat. And this right here is the unit vector V hat. And the solution to this example. And so what happens is if we go over here to our graph, this vector V hat is going to look something like this unit vectors always have a magnitude of one, 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 we, this vector specifically points 4/5 in the I direction and it points 3/5 in the J direction. And that's what gives you the unit vector V hat. 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 one. So let's go ahead and try this example where it asks us to find if V hat 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 the hat. 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 can see that the X component is 4/5 and they need to add this to the Y component, which is three. 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 have the square root and then we have four squared, which is 16 divided by five squared, which is 25. And that's going to be plus three squared, which is nine divided by five squared, which is 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 plus nine, that's equal to 25. So we have the square root of 25 divided by 25. Now 25/25. Well, any number divided by itself is just gonna be equal to one and the square root of one is just one. So notice how we got that our magnitude for V hat is one 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. So 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, you have 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, you ha 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 negative I plus two J. And you can also write that as negative one I plus two J. So that way we can clearly see what the X and Y components are. And then it's all going to be divided by the magnitude of vector U. The magnitude of view 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 negative one. So we're gonna have negative one squared plus the Y component, which is two. Now the square root of negative one squared, well, or I should just say negative one squared comes up to positive one and then two squared comes out to four. So I have the square root of one plus four, which is the square root of five. So that means that the magnitude of U is the square root of five. 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 negative one I plus two J that's going to be divided by the square root of five and dividing each of these components by the square root of five. We're going to have negative one over the square root of five I plus two over the square root of five J. So this is what the vector ends up being for a 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 five. Now, what this is going to do is get the square roots to cancel in each of the denominators. So for vector U hat I'm going to end up with one times the square root of five, which is just the square root of five divided by five I. And then this is negative and that's going to be plus two times the square root of five divided by five J. 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. So in this example, we're told if vectors A equal negative 13 and vector B equals 47 and C is equal to four times vector A plus two times vector B find the unit vector in the direction of C. Now there's a lot going on in this problem. But the way to make this as straightforward as possible is going to be to break this up into steps. Now, the first step that I'm going to do is I'm first going to see if I can find vector C. The second step is going to be to find the magnitude of C because I know that finding the magnitude is something we're going to need to do for finding the unit vector. And then my third step is going to be to find the unit vector C have. And this right here is the ultimate goal of this problem where that's in the direction of C. So let's start by figuring out what vector C is now to do this, I can see that C right here is four A plus two B. So what we really need to do is just the operations that we see. Now what I can do is plug vector A in which is negative 13 and I can plug vector B in which is 47. So these are the vectors plugged in. And what I need to do is scale these vectors by the numbers that we have out in front. So we're going to have four times negative one and then four times three, multiplying these by each component and four times negative one is negative four and four times three is 12. Then this is gonna be added to two times four and then two times seven. So we're going to have two times four which is eight and two times seven, which is 14. Now, what I need to do from here is add the corresponding components. So we're going to add the negative four and the eight and that's going to give us positive four. Then we're going to add the 12 and the 14, well, 10 plus 14 is 24 plus two. That's gonna be 26. So this is vector C and now we've gone ahead and completed this first step where we found vector C. Now, our next step is going to be to find the magnitude of C to find C's magnitude. What I'm going to do is use the equation for the magnitude of a vector, which is the square root of the X component squared plus the Y component squared. Or basically just another version of the Pythagorean theorem with vectors. Now, using this, I'm going to plug in the X and Y components of C which we actually found with this vector. You can see the X component is four. So we're going to have four squared plus the Y component squared, which is 26 squared. Now, four squared that's equal to 16 and 26 squared turns out to be 676 16 plus 676 is 692. And it turns out that this actually does simplify. If you recall how to simplify square roots, we need to find numbers in here that can square. And it turns out that this is equal to the square root of two times two times 100 and 73 sorry two times two times 173 and two times two is the same thing as two squared. So what we can do is split this up to have the square root of two squared times the square root of 173 the square on the square root will cancel. Meaning all we're going to end up with is two times the square root of 100 73 as our simplified radical. So now we've gone ahead and found the magnitude of vector C. Our last step is going to be to find the unit vector, which is ultimately what we're trying to do in this problem. Well, I can see that the way to do this is to just use everything that we have and combine it into the equation for the unit vector. The unit vector is going to be vector C divided by the magnitude of C. We can see that we have vector C which is what we calculated over here. It's 426. And then we can see that we also have the magnitude of C which is two times the square root of 100 and 73. So what I can do is rewrite this by dividing this into each of the components. So I can write this as four divided by two times the square root of 100 and 73. And then that's going to be 26 divided by two times the square root of 100 73. Now, I just wanted to point out right here, you could do this using IJ notation if you find this to be easier. I'm just using the brackets because that's what we have in the problem. Either way it should actually work and give you the right answer. Now, what I can do is simplify because four is the same thing as two times two and 26 is the same thing as two times 13. So this two is gonna cancel with that one and that two is gonna cancel with this one. So we're going to get end up getting the vector two over the square root of 100 and 73. And then that's gonna be 13 over the square root of 100 73. Now, at this point, all we need to do is rationalize both the denominators and to rationalize the denominators just take the top and bottom of the fractions and multiply them by the square root of 100 and 73. This will just clear the fractions in the denominators and give us a more complete answer. So the final answer in our unit vector C half is going to be two times the square root of 100 and 73/173. And then it's going to be 13 times the square root of 100 73 over 100 and 73. We can do that in the brackets there. And this right here is our unit vector and the solution to the problem. So this right here is the answer and that's how you can solve this. So these are all the steps that you need to do when you're trying to find a unit vector where there are some complex operations. And it's not a very simple magnitude. But as you can see if you just follow the same pattern for all of these types of problems, you can get to a final answer. So I hope you found this video helpful. Thanks for watching.