Sab Matsumoto: Hello, Kelly. Hello, students, your test number one is coming. It's going to cover the first five sections of the book. And so I will just briefly go over these review problems. Okay, so let me share my screen and we'll get started.
Sab Matsumoto: You will be able to access this checklist. Alright, so there's a checklist for test number. Oh, I
Sab Matsumoto: Basically called practice test number one. So if you look for the item called practice test number one under your module. And that's the chapter two module, you'll find this blank.
Sab Matsumoto: Sheet of paper. It's not blank sheet of paper. And what I mean by that. It's a sheet of paper that do does not have the answers.
Sab Matsumoto: Okay, so I have 13 questions here. I have several vectors and points and planes and all the questions are pretty much based on these and then later we have in addition to these know this and know that. Okay, so let's
Sab Matsumoto: I'm not going to go over everything in detail because I do have these answers written down under what practice just number one answer key, right. No, not so I'm not a very surprising title.
Sab Matsumoto: And so let's go over the questions one by one. All right. So number one, find the exact expression for the angle between PQ NPR PQ is right here and PR
Sab Matsumoto: You have to find out what keeps you and PR. Our first. All right, and then you may want to just ask questions like this to yourself. All right.
Sab Matsumoto: I'm supposed to find the angle between two vectors, what am I supposed to use. Okay. And the answer I hope you know is the dot product. Okay, so here's your p, p, q, and Dr. Alright, so you need to calculate these and then you know that the
Sab Matsumoto: What do you remember about the product. The product of two vectors you envy is equal to the
Sab Matsumoto: Normal you times the normal v times the cosine of the angle between them. So if you solve that for the cosine of the angle theta between them. You get you don't be divided by the normal view times the normal v. And so you
Sab Matsumoto: Apply that formula. And then you take the DOD product of these that's you know there's times there's three
Sab Matsumoto: Plus zero plus six. That's going to give you three plus six, which is nine over the square root of this and you know this is an okay answer another words.
Sab Matsumoto: If you want cancer to be the theta, which is the inverse cosine of this, I will give you full credit
Sab Matsumoto: Although you know when you think you can simplify, you probably should because you may get some additional information about the angle feta. Alright, so if you simplify this and I told you, you do not always have to
Sab Matsumoto: rationalize the denominator, but you can. And in this case, you get a very nice, very familiar number of square root of three over to as the cosine of the angle feta. And of course, from here.
Sab Matsumoto: You can conclude, of course, theta is between zero and pie or zero degrees and hundred 80 degrees so that angle theta has to be pi over six or 30 degrees.
Sab Matsumoto: And so now you can never get that information just by looking at these, right. So this is an advantage of using the dot product right okay number to find the area of the parallelogram, let's stop here. How do you find the area of parallelogram.
Sab Matsumoto: Where the two adjacent sides are to give them vectors. Okay, so you have already calculated PQ NPR and then you tell yourself the area of the parallel ground is
Sab Matsumoto: Yeah, it has to do with the cross product, right. And in fact, the cross product is a vector. So, that itself is not the wall.
Sab Matsumoto: Of the area but you take the known of the cross product and
Sab Matsumoto: From here you can take the cross product, right. So it's this times this minus this time that negative three minus zero, which is this
Sab Matsumoto: And the next one is this times this negative three minus this times this negative three minus negative six. And that's positive three. But remember this one.
Sab Matsumoto: Will take on an extra negative sign. So this will be negative three. The last one will be zero minus this times this. That's zero minus negative one times negative three. That's negative
Sab Matsumoto: Tree. Sorry. So that would be natively. So that's sort of an interesting vector here. You take the norm of that and that is
Sab Matsumoto: Square root of 27 or three square root of seven. Okay. And this will always be a positive number. And this you know these end up
Sab Matsumoto: Giving you the zero cross product, incidentally, if you recognize this as you know this. I mean, this is fine too. If you want to take the norm of this you can take that absolute value of that scale or out and figure out
Sab Matsumoto: The answer also is three. That's going to be the same answer. So in order to calculate the magnitude
Sab Matsumoto: If your vector is a scale or product of a vector, you can take the scale out. And then you can also find the answer that way.
Sab Matsumoto: All right. Number three, right, and equation of the sphere center and at p. So here's my p. So, you know, you will have an expression like x minus three and y minus zero and the z minus one and then
Sab Matsumoto: With radio six. So the other opposite side will be 36 that's a pretty easy question, right, this will be acceptable. I will not that it just disappears, but I can undo this.
Sab Matsumoto: So this is completely fine. I'll give full credit for this.
Sab Matsumoto: If you multiply this out, you will actually get this and that is also an acceptable answer, I will take either one. One is not better than the others. All right. Number four, find the unit vector in the direction of the sum of A plus B. How do you find the unit vector of
Sab Matsumoto: Vector. Okay, and you should be thinking of one word, what is that word.
Sab Matsumoto: Normalization okay that process of getting a unit vector in the direction of a given vector is called a normalization. Very simple. All you have to do.
Sab Matsumoto: Is to find a norm of this and divide that vector by it. Right. So first you add these two and then you divide it by the norm of the sun.
Sab Matsumoto: When you add these two, you get one native to one. And so you take, of course, what you will do is take the square root of one square plus negative two square plus one square and that's of course native I mean square root of six.
Sab Matsumoto: And you divide each component by that number, you can also write this this way, no problem with that. Okay, so this will also give you full credit
Sab Matsumoto: Number five. What is the projection of a on to be what is the magnitude of this projection for these, you probably should memorize learn the formula for the projection
Sab Matsumoto: Remember, it has a dot product on the top and bottom. And I just want to point out again, if you're trying to find the projection of a on to be almost everything you write is be right. So you have a.be on the top. Everything else is a BB BB BB right so a.me over beat up be times b and
Sab Matsumoto: Then you just calculate the dot product happens to be negative five and then the denominator is
Sab Matsumoto: It is a square of the magnitude of be and that turns out to be one, that's negative five over three and then you multiply that. So in other words, this whole thing is a scale, right. The only vector
Sab Matsumoto: Portion of this is this. That's your be and so you multiply that scale and negative five thirds would be and that is your answer. This is also acceptable, of course.
Sab Matsumoto: It's a small plane yourself now the projection, can the projection is of course the length of this and you know this can be taken that you can take out the negative five thirds
Sab Matsumoto: And find the absolute value. Everything else is one, so it's a square root of three.
Sab Matsumoto: Multiplied by 530 that's that's a fine answer, but there is a there's a formula for the scale or or the length of the project is sometimes it's called a scale or projection scale a projection simply means the length of the scale or length of the projection and
Sab Matsumoto: We had a I tell you what I think this is kind of important part. And I will just tell you what what we mean by this. Right. So if this is your A
Sab Matsumoto: And this is a be the projection and we're talking about this length right and that length is of course the normal way. That's the hypothesis.
Sab Matsumoto: Times the cosine of theta, right, I mean this is straight from day one of your trigonometry class.
Sab Matsumoto: And of course, that's okay except you don't know what cosine and feta is until you do the dog product and all that. So you can just go ahead and add this
Sab Matsumoto: As long as you do the same thing in the denominator. Well, see, that itself is done and now this is closely related to this formula. Of course I but this is the length of the time to use to have an absolute value here because it's a length, but this you can calculate right
Sab Matsumoto: So, this turns out to be in, again, you can see the top is being the absolute value of the norm of eight times a normal be times the cosine of feta that a B.
Sab Matsumoto: And divided by the length of p. So, if you use this formula, which is this, then of course, you get the same answer five thirds over the square root of three.
Sab Matsumoto: Okay, number six. Explain the direction and the magnitude of the cross product of any two three dimensional vectors. This is the kind of problem. I really like to ask you on your test.
Sab Matsumoto: And probably like on the final exam as well. Okay. Because, I mean, this is a concept question and it doesn't
Sab Matsumoto: Just, you know, telling me that you are you know how to calculate these things. It also tells me you understand what the cross product is ok so the cross product of two, three dimensional vectors is of course a three dimensional vector
Sab Matsumoto: And what I want you to tell me or what I want you to answer is that the the magnitude is the area of the parallelogram.
Sab Matsumoto: With those two vectors, being a Jason science and the direction is the white perpendicular. This is very important. Right. And we have used this fact, over and over again already in first part of this chapter in order to get something perpendicular to to given non
Sab Matsumoto: parallel vectors, you take the cross product. So the direction of the cross product is perpendicular, orthogonal
Sab Matsumoto: Synonymous to eat each vector and in accordance with the right hand rule. And so again, you should remember what the right hand rule is if this is you. And if this is V
Sab Matsumoto: Remember it your, your right hand is sort of like this. It's a terrible right hand. All right. But anyway, you're wrapping your fingers around this way, like that. And so your thumb.
Sab Matsumoto: Is pointing upward and that this is the right angle and this is also the right angle, and that is the answer to this question. Make sure you know that the
Sab Matsumoto: Cross product is primarily used to find something or something or to both and given vectors fairly important okay and Number seven, find the volume of the apparent little piped stop here, the volume of the parallel typed. As soon as you use, you see that you should
Sab Matsumoto: Remind yourself. Okay. It is about that scale or triple product. And so you take the scale a product which is a triple product. And basically, I mean, this is it. Right. It's a.be cross. See, but it also turns out to be b dot c cross A.
Sab Matsumoto: It's also see DOD a cross me. I know it's kind of weird. You can find because, see, first, and here's your because, see, and then you take the DOD product but
Sab Matsumoto: I hope you didn't. I hope you've got this in the lecture videos you can do this in two steps. But remember, when you take the product of you know this and us. It's basically you can replace this first row with your a
Sab Matsumoto: And you take the determinant of a three by three matrix. And that's when give you the same thing. The answer is 25 now this
Sab Matsumoto: This particular situation that triple product turns out to be negative. Okay. And so you just take the absolute value. And the reason
Sab Matsumoto: This is turns out to be negative is because, well, you know, be be cross. See if you did see cross be then it would not have been negative somehow we did this orientation in in the opposite way. And so what this is showing is and
Sab Matsumoto: How do I say this, you know, you have this parallel type like this. Right. And there's an order in which you have to label these and unless you follow the order. The answer turns out to be negative.
Sab Matsumoto: It's basically sort of calculating the backwards, the opposite, you know, but you don't have to worry about why this is negative. It just happened to be that this particular order. We picked be cross see as opposed to see cross be and give you a negative brownie.
Sab Matsumoto: All right. Number eight fine parametric equations. Yeah, okay, this is starting 2.8 sorry 2.5 right find parametric equations of the nine through q and r
Sab Matsumoto: Okay, so I see queue and they see our. And so let's find how you would do that QR is this thing, right. So you remember what you have to do here is I mean this is a picture. You have to. And this is our. It doesn't matter if you find QR or our Q
Sab Matsumoto: This is the victor. We use which is parallel to the line of course right so you need a point and you need a vector parallel to the line in order to write down the equations.
Sab Matsumoto: While you found QR and you found actually two points, but you can just use one point if you pick this point. This is the set of parametric equations, you get
Sab Matsumoto: If you pick this point, you get this parametric equations. This of parametric equations. Either one is completely acceptable. And as you know, as he goes from negative infinity to positive infinity. These will give you the exact same sets same set of points. And so the only differences.
Sab Matsumoto: If you plug in t is equal to y here, you actually get
Sab Matsumoto: This right when t is equal to zero is equal to one note in there. So those will give you the same point. And so that is the only difference. You don't want you just fine.
Sab Matsumoto: Okay, number nine, find the equation of the plane right so i think let's go back to the equation. The problem eight if you have to come up the equations of law align a parametric equations online, you should know it looks like this.
Sab Matsumoto: Right, and then you should know one point that will give you these three coordinates and then you need a vector that will give you the coefficients for the t. And so you need one vector
Sab Matsumoto: And the point if that's how you can find particular line. If you're looking for plane, though, you need what in order to describe a plane or to come up with an equation of that plane that you need one normal vector and point right so keep these things in mind.
Sab Matsumoto: For number nine. Oh, yeah. So one. Another thing about number nine is you're supposed to find the equation containing three points. Okay, so you have p, q, and r
Sab Matsumoto: And so how do you find this normal n well remember which problem was that it is this problem. That gives us a clue to this one, right.
Sab Matsumoto: Because how do you find any that is perpendicular to
Sab Matsumoto: This plane well find PQ and find PR and then you take the cross product. And that's exactly what I do PQ NPR, take the cross product of these, which we already did, by the way, in an earlier problem.
Sab Matsumoto: And so you can plug this in to the formula. But here's a little trick. And I think you should be familiar with this trick because this will make questions a lot easier sometimes these numbers are really big.
Sab Matsumoto: Now all you need is one normal vector that happens to what I mean by a normal vector is it must be perpendicular to the plane that I would like to describe
Sab Matsumoto: And all that means is any scale or product and this is okay. Well, simply put,
Sab Matsumoto: Tried to make divide everything by us nice number and let's simplify this. You think you can do that. What would be a good constant to divide each coordinate by
Sab Matsumoto: I mean, I would think of like negative three. So let's go ahead and use this
Sab Matsumoto: And. Okay. And so using this and 111 and easy point 00 for you plug in 004 here 111 here and then make sure you have the right hand side, because that's what makes this into an equation.
Sab Matsumoto: And then this is your answer. Okay, now you can also use instead of our you can use P OR Q and this part will look very different. Right. But once you simplify
Sab Matsumoto: To this kind of a form where one side is equal to zero, whether you use P OR Q that will look exactly the same. In fact, you can use native native the native three
Sab Matsumoto: All right. And if you come up with this form, you will just have this whole thing, multiply by negative three. So at the very end you can divide the whole equation by negative three. Anyway, you will always end up with this simple form. And so that is
Sab Matsumoto: The probably the best solution here even though I will still accept you know this solution here as well.
Sab Matsumoto: Okay, so how far do we go. That was number nine. Let's go to number 10 find an expression for the angle between playing one plane to I've given you these two planes here.
Sab Matsumoto: So try to imagine a plane crossing like this. Okay, this is a nice way to think about this plane one plane to
Sab Matsumoto: What is what's the what's an expression for the angle between them. Now remember, you're trying to measure this angle here, but that's the angle between these two normal vectors as well.
Sab Matsumoto: So all you have to do is to come up with the an expression for the angle between in one and two and n one and then two are, it's right there. I mean, you don't even have to
Sab Matsumoto: Work hard, or think hard just read off these coefficients right and one will be too negative one, one and two is one, three, negative one. Why, why, because these are these corporations are the normal vectors.
Sab Matsumoto: Right, this is just fabulous. It's very nice. The way it works here. So, playing one. Here's a normal vector, he is playing to the normal vector. Take the well.
Sab Matsumoto: Trying to find the angle between them. And that's exactly what you are doing on problem number one.
Sab Matsumoto: So the cosine of the angle is the dog product divided by the norm of for the first times the normal. The second now. The only difference is. Now be careful with us. The only differences.
Sab Matsumoto: Here I did have the absolute value on top. Why not well because two vectors can possibly be apart by an obtuse angle right we always measure this less, less than 180 degrees.
Sab Matsumoto: All right. If you go to two planes meeting with each other. Where are my two planes with meeting with each other. You know what the differences
Sab Matsumoto: Yeah. And when two planes meet each other. Okay, you can measure this angle or this angle. Well, you always made you the small angle. That's how we talk, I
Sab Matsumoto: Mean when you when you see two angles two planes meeting at let's say 30 degrees here.
Sab Matsumoto: We normally don't say, well, the angle between them is hundred and 50 degrees you don't choose that part. So,
Sab Matsumoto: When two planes meet, we can always assume that dangle of intersection is that acute angle. I mean, it could be a right angle, but other than that.
Sab Matsumoto: It should be the smaller on the two angles. And that's why we take the absolute value because the acute angles have the cosine and the value of cosine of an acute angle is always positive, and it is the opposite of the corresponding
Sab Matsumoto: Complimentary angle right so the CO sign on failure is this. So the answer for theta is the inverse cosine of that. Okay. And that's a very small number, by the way, which tells me that what does it tell you if mango is small.
Sab Matsumoto: This number is close to zero. And so the theta is going to be close to 90 degrees if it is zero, then, of course, theta is 90
Sab Matsumoto: Okay, number 11 is next, we're almost done with this find out right parametric equations for the line of intersection of playing one and playing too.
Sab Matsumoto: So how do you do this, find the parametric equation of the line of intersection. We're talking about what line. Are we talking about clearly this line I well
Sab Matsumoto: This line here so that what I mean, again, what do you have to know or what do you have to have in order to write equations, the parametric equations have a line. I need one vector
Sab Matsumoto: And the point, right, and we don't have either of them right now.
Sab Matsumoto: So first we have to find a point of intersection between this plan that plane and there of course infinitely many of them because every point on this nine is clearly an intersection point
Sab Matsumoto: And then the direction, the vector can be found by what what it has to be perpendicular
Sab Matsumoto: To end one since this plane and this nine nice in the plane one. This also has to be perpendicular to enter because it has to be on this plane.
Sab Matsumoto: Not the nine of intersections also a part of p two. Right. So you have to take the cross product of and one and two. In order to find this vector along the line or parallel to this night. And so that's what I do. First, let's take the easier part. Okay. Because what's easy in us.
Sab Matsumoto: Yes. Yeah, so you take the cross product of this right and then you get your vector that's perpendicular to both and one and two and that part is done. That's an easy part.
Sab Matsumoto: The second thing is to find any common point of intersection. And so all you have to do is to set up the system of linear equations where you have these two
Sab Matsumoto: Equations of lines right and then find just any point any point. Okay. And in this case, you see, you notice plus z and minus z. So you can just, you know, at the two equations. This will give you that one equation.
Sab Matsumoto: And that's pretty much it. What's the solution. Well, you know, there's no one solution, right, because the solution is really that infinite many points which constitutes the line. So at this point, this is the only requirement. So you just draw a picture point. Okay, and what point
Sab Matsumoto: You can XP zero or why is there. I think it's easier in this case to the XP equal to zero because then why is one
Sab Matsumoto: And you plug in these numbers and that will give you Z's equal to for now this is not the answer, of course, right, because this is just a way to find one point.
Sab Matsumoto: And once you decide what x and y are you have to plug this into one of these equations. I don't care. Which one are you plug
Sab Matsumoto: Zero and one for x and y, respectively. And then you find z is equal to five. So you found 1.015 to be on the line of intersection
Sab Matsumoto: And the line of intersection is parallel to this vector. Now you can write your equations. So they're technically, there's a zero here, of course, right, and then one five and then negative 237 appear as coefficients for our team.
Sab Matsumoto: Alright.
Sab Matsumoto: Then the last two questions have to do with Sunday's right. All right, so these are pretty important question as well.
Sab Matsumoto: Find that this distance between our and playing too. So you have this plane to and our which is 00 for how do you do this. Okay, so this is
Sab Matsumoto: Nice picture and then you i personally draw a picture like this every time.
Sab Matsumoto: I do a distance problem because then that makes it easier for me to remember the formula. And if I forget the formula that will start drawing picture like this, you know, and this is the this is the
Sab Matsumoto: Do I need this. How do I need, how do I find this well this is going to be. And here's angle that's also an angle right so that he has to be the the hypothesis, times the cosine of angle. And that's how we remember this formula.
Sab Matsumoto: Or you can just know your formula here, and that is the DOD so so pick
Sab Matsumoto: I forgot to tell you first pick some random P first on the plane. Right. That's a weird way to start a problem but pick any point and then find P are
Sab Matsumoto: In this case PR 002 because I pick this point here. And so now all you have to do is to take the cross product and PM, sorry, the dot product of PR and there's normal vector, which we know is one, three, negative one.
Sab Matsumoto: Right. There it is. Take the cross dot product of this and this and that's going to give you a negative two, and then you divide it by the, the length of the normal vector
Sab Matsumoto: Right. And so that is your answer, of course, you have to have the absolute value in the front because theta i mean the top because say that could be negative.
Sab Matsumoto: Be obtuse. And in which case the cosine would be negative. But remember, if you forget this, though. I mean, you can always remember the is the high partners, which is the normal P, Q,
Sab Matsumoto: Times the cosine of fader right and remember this data is the same as this data here, because these are the ultimate interior angles.
Sab Matsumoto: Well, you know, that's good. But if you just put in on the top and bottom in the normal, then you can tell this thing here. And by the way, this is supposed to be absolute value. This thing here is the dot product. Okay.
Sab Matsumoto: And so you can easily recall or recreate or we construct the formula on the spot. Even if you forget. And that's why I drew a picture like this. And then finally, number 13 use your answer from problem eight above. That's your apartment to equations on the line.
Sab Matsumoto: QR and find the distance between P, Q, R. So how do you do that, here's the picture you have p
Sab Matsumoto: And then you have a Q R Q is somewhere like here, right. And so here's what you need to find out. I need to find the distance here. How do you find the distance what. So, by the way.
Sab Matsumoto: In the equations of lines, what's obvious is not normal vector, it's a vector parallel to it right so in order to find. I mean, you know, this vector. And if you pick a random point in this case are then you know PR
Sab Matsumoto: And then the angle between those two well to loan vectors is this part here.
Sab Matsumoto: And with respect to that part, the distance we want is actually the opposite, right, and that's why you use the one that's involved with cosine, because the distance distance d
Sab Matsumoto: Is the length PQ, or sorry PR times the sign of angle feta, and the sign of the angles theta is the one that is in with the cross product. In fact, across rather seen this case I have V and I have p sorry RP right so we cross our paths will cross RP, what is V cross RP
Sab Matsumoto: The norm of the vector. In fact, the magnitude of this cross product, turns out to be. Remember it is the normal v times the norm of the second
Sab Matsumoto: Vector times the sign of the angle between them. Right. And so you can tell that the look. D is again just looking at this, it's really the length of PQ
Sab Matsumoto: Times the sign of failure. Right. That's what you get from this triangle. And so you kind of wish you know
Sab Matsumoto: It would be nice if I can just put a be here because then I will have this expression on the top.
Sab Matsumoto: Yeah, I can just go ahead and do that as long as I do the same thing on the bottom. And this tells you, then the top is v cross PQ, the normal that divided by the normal v.
Sab Matsumoto: Again, if you forget this formula, you can just reconstruct it on the spot. If you remember how to do that. So, and of course for your test, you probably want to memorize this. And the thing is, you don't have to memorize this forever after this test, you probably can't forget that.
Sab Matsumoto: Because you can also recreate this and calculus three is not about finding distances. Alright, so you probably will not be using this very often for the rest of this semester.
Sab Matsumoto: But for this test. I want you to know how to do 12 and 13 okay so finally know the operations and what comes out. For instance, if you take the product, you know, the answer is not a vector, right. It's a number
Sab Matsumoto: You should also know how to sketch in describe our region in three dimensional space. So if I say z is equal to seven. What is that what that's going to be a horizontal plane right if x is equal to y.
Sab Matsumoto: Z is irrelevant. What do you get, it's going to be a vertical wall. Okay. And it's going to be a vertical wall that is on top of this line. So in the three dimensional case, it'll look like this and
Sab Matsumoto: But anyway, so. No, no, no. Some basic equations in three dimensions that
Sab Matsumoto: Through this path, but you have this checklist or this is the practice test one
Sab Matsumoto: Okay, as well as this solution without all the messy handwriting. So, and this is found in the module for chapter two. And so I hope you will study and be ready for your test. Okay. Have a wonderful day, and I will see you
Sab Matsumoto: Soon.
