Dot Product, Angles & Skew Lines | Class #10 | A2 Mathematics 9709

Generated automatically from the lecture captions, so spoken maths may be transcribed imperfectly.

Okay. So the last thing we we found the equation of a 3D line. I told you about the direction vector. The general form of the equation of a 3D line. It is this. R is equals to position vector on the line plus the parameter multiplied by the direction vector of the line. This is the general equation of a 3D line. And uh we'll do another this example 0 0 1 and you have 2 1 4 and we if we how do we find this the equation joining these two 3D points. So position vector is any one of the point right where relevant sorry what was it relative relative to the origin. So the coordinates as well I told you the

coordinates of a point are the same as the position vectors because both of them are relative to the origin. So this means you can directly you can use any one of these points as the position vector plus lambda and this is the direction vector. And how do we find the direction vector? Direction vector is simply let's suppose this is point A this is point B. So if you find the vector A B which will be equal to O B minus O A. So, O B it is 24 minus O A 01 and simplify. So, 2 1 3 this is your direction vector. That's it. This is it equation of a 3D line. You need any two points. So, you'll find direction

vector. You have already the position vector and this is this is it. Now the new thing finding the dot product dot this is also a very simple thing very easy dot product is a dot b vector a dot vector b. So for example if your A is 2 3 4 and vector vector B is -7 -2 and 1. How do we find A dot B A vector dot B vector. So you will write it like this as well. 2 3 4 dot in between and then -7 -2 and 1. And it is easier than the previous thing. You just multiply the x with the x. So 2 * - 7 plus multiply the y with the y 3 * -2 plus multiply the z with the z 4 * and then you add them all up. So -14

- 6 - 4. So -14 - 6 it is -20 - + 4 is -6. This is your dot product. So you multiply the x with the x, the y with the y and the zed with the zed and you add them all up. This is it. So if you have if I write it down the generalized format x1 y1 z1 dot x2 y2 z2 it will be equal to x1 x2 plus y1 y2 + z1 z2. This is it. Simple dot product nothing more. And the the thing that you would further use this dot product is find finding angle between two vectors. So if your vector is something like this and there another vector like this, they may be touching or may not be touching. Okay, it doesn't matter. It's

a 3D thing. If you visualize it, your vectors may cross each other or may not cross each other, but they would have an angle between them. This is your theta, the angle between them. And how do you find this? Using this dot product. So a dot b is equals to magnitude of a * magnitude of b into cos of theta and theta is the angle between them. So if I use the same question to find the angle between these two vectors. So you have already found the dot product. So the dot product of a dot b it is -6. So -6 it is equals to magnitude of a which will be square root of x² + y square + z square. So 2 square 3 squar 4

square multiplied by square root of the other one magnitude - 7² - 2² and 1 square into cos theta and then you will use your calculator use it. So -6 upon square<unk> of 4 + 9 + 16 * square roo<unk> of 49 + 4 + 1. So 0.441 something like 404 3 something. And this is equals to cos theta. And so now you will move the cos to the other side. So cos inverse of this answer is 113.8. This is the angle between the two vectors. This is the acute sorry this is the obuse angle. If they ask you for the abuse sorry acute angle you would just multiply sorry subtract it from 180 180 minus obuse. So if you subtract it from 180

180 minus so this is 66 two this is the answer. Sometimes they would ask you to find the acute angle between the two vectors. So if your answer is coming out as obtuse you will say just subtract it from 180. This is how you find the angle between two vectors. Copy this down. If you have any questions ask you have to learn this formula. Okay. Okay. Okay. Incantination. This is a very simple formula. In continuation to this thing, they would ask you to find angle between two lines, two 3D lines. So they would give you the two equations of the lines 3D lines and you would use the direction vectors.

So what would you use? The direction vector of line one dot with the direction vector of line two is equals to direction vector of line one magnitude direction vector of line two magnitude into cos theta. Let's find a question. Look at this. Question number 10, part B. They have given you the equations of two line. We know that this is the direction vector and this is sorry this is the uh this is this is your what was the name position vector on the line and this is the direction vector of the first line and this is the direction vector of the second line. So you would take now these two things.

So 2 - 1 3 dot I'm doing the part B dot 1 - 4 is equals to magnitude so square<unk> 2² is 4 1 square is 1 3 square is 9 square root 1 + 1 + 16 into cos theta and look they're asking you to find the acute angle. So let's see if if it is coming out is acute or not. So for the dot product 2 * 1 + - 1 * - 1 is 1 + 12 upon everything on the right side the square roots they will go over there and divide. So this is 14 and square root 18. So use a calculator cos inverse 2 + 1 + 12 upon under root 14 times under root 18. So 19.1 this is coming out as acute. So leave it be. This is your answer. Three marks.

That's it. Easy. Three marks. Any questions? This is one thing finding the equ the angle between vectors. Angle between two lines or angle between two vectors. Next thing in vectors is simultaneously solving two equations of 3D line. We know whenever we solve simultaneously what do we get? The point of intersection whenever we solve two equations simultaneously we are working for the point of intersection. So we know how do we find the point of intersection the simultaneous solving in the linear 2D lines the y is equ= to mx plus c. the simple ones we do the simple substitution method elimination method

and we get the point of intersection. Now over here there are three cases that you need to understand that are possible when you are trying to solve two 3D lines together. So the case one that they intersect this is the basic one. So if you look at in 3D space you have the X Y Z all three and there is a line separately going over like this and there's another line separately going like this and they are intersecting at one point. You will find the point of intersection and there would only be one point of intersection. This is when they exactly intersect at one point. Second one they are parallel.

They are parallel and not overlapping. For example, you your room is a 3D space. Okay, your room it has length, breadth and height all three of them. So if you have a line which is going diagonal from one corner to another and you have another line which is this is your floor. This is your floor. If you understand this is your floor. I'm just trying to make it logical. So this is your one line which is diagonal to your floor. And there is another 3D line which is diagonal to the what do you say this English what do we call English roof >> roof roof >> roof h roof so look these two lines are parallel and

they are not overlapping so there is you can see no point of intersection ction. Third case they are overlapping this you won't see this much but this is another case that is possible so that's why I'm just uh giving you as an example but you won't see this much but you will never see this but this is one thing that is possible so that's giving you this thing. So let's suppose there is a line which is from one corner to another corner like this and there is another line also the same overlapping. So this means in this case you have infinite number of solutions. Every point on every line on both of the lines

is a point of intersection. So this is infinite points of intersection. You won't get a question like this but this is a possibility. So that's why I'm just I'm just giving you as an uh knowledge. You do you may use this knowledge later on in your university case. They are skew. Skew means that they are not intersecting but they are not parallel as well. For example, you have one line red one on the base and you have another line like this. So you can see that the or you have another line like this. So like this. So this is on the floor. This is this is the corner and this is the corner. Okay.

And this this is the corner. This is the corner. So you can see the two lines they will not intersect and they are not parallel as well. Both have different planes of movement. So they are not intersecting. no intersection and not parallel as well. So these kinds of two lines are called skew lines. Okay. Now one more thing I forgot to tell you in the case number two when the two lines are parallel and they are not overlapping and whenever they are parallel their direction vectors would be same. direction vector 1 would be k * direction vector 2. I told you whenever there are two uh parallel vectors they

are scalar multiples of each other. I told you this. So the direction vector of the line one let's suppose this is line one this is line two they both will have the relationship between them would be a scalar multiple. Same would be over here as well. direction vectors of the two lines will be scalar vectors of each other. Scalar multiples of each other. In the fourth case, it is a build separate thing. Any questions over here? >> Yes sir. >> Let me show you this thing over here as well. Sir, >> the point of the section. Let's suppose this is one of the from one corner to another corner of the

room. If it is if this is a cuboid something sorry a cube and from one corner to this corner there would be a point of intersection. If this is a cube this is actually a cube something like that in a cube. So you have the three dimensions X, Y and Z and you're just uh joining the corners together and you will have definitely a point of intersection. So this is one point of intersection. Okay, any questions? All good. >> Now this concept is very important. They would ask you to find the point of inter intersection or they will ask you to tell if the if the lines are parallel or not or if they ask you the to tell

that the lines are skew or not. This is what they would ask you the most. Look and the two lines have given show that the lines are skew or show that the there is a point of intersection between the two lines. So that is something we need to now find or do a question. We have time. Yes, we have time. Now this is 3D world. I cannot make a question on my own. I don't have a very a graphic card installed in me very big processor. I will have to use operations that they have given you. Okay. So they have given you look the lines L and M have vector equations. So L line let us rewrite it. So this will become 1

0 look there's nothing 1 0 and minus 2 plus s 213 and line m r is equals to now this is the thing we are learning how to find the point of intersection 6 - 5 4 + t 1 - 2 1 and we want to find the point of intersection. So definitely we would solve them simultaneously. But before simultaneously solving them, I will just combine the two equations. Oh, sorry. Simplify this thing. 1 + 2 s 0 + s - 2 + 3 s and this is r which is 6 + t - 5 - 2 t 4 + t. And now we solve them. Equate them together. 1 + 2 s - 2 + 3 s. Look at what I'm doing. And now you would form three equations. Your your equation from the first part

would be 2 s - t is = to 6 - 1 is 5. From here will be s + 2t is equals to minus 5. And from this one 3 s minus t it is equ= to 2 + 4 is 6. You form these three equations. Equation one, equation two, equation three. Now choose any two of the equations and solve them simultaneously. It's up to you. You can choose the first and third, second or third, first or second. It's up to you. So I am choosing right now. I'll use the calculator. Okay, I want to save time. So I am choosing equation number one and equation number two. and I'll use the calculator. So equation number one is 2 -1 and 5 and equation number two is 1 2 and - 5.

So you are getting your value of s is coming out as 1 and t is coming out as minus3. Now is the important part. You find you use any two equations and you solve you solve for the two variables. And now you you see if the check if the third also satisfies also satisfies these values. So the third equation is 3 * s is 1 - - 3 is equ= to 6. So it is coming out as 3 + 3 is 6. 6 is equ= to 6. So yes it is satisfying and when it is satisfying they are intersecting. there is a point of intersection you can find it. If if the third equation does not satisfy then this will tell you that the two lines are skewed. There is no point of

intersection. But if the third equation is also satisfied then you would have a point of intersection right now. Yes it they are intersect it is satisfying. So yes there is a point of section otherwise if not if not if not satisfying the third one then it would be skew. So in this case it is satisfying and then they're asking you to find the position vector as well for the point of intersection. So now you can use any one of them s1 or t minus 3 and substitute over there or here and find the value of r. So now this would become the point of the section would be now equal to 1 + 2 * s is 1 1 and - 2 + 3 * 1

which will come out as 3 1 1. This is your answer. This is the point of intersection. Okay, copy this down. Everything right now is important. Okay, so now you know this as well. There's only one more thing left now to discuss. Show that line. Okay, let's do this one as well. do not intersect. Do not intersect means that the lines are skew. So you will do the complete same thing and at this step we will see that the equation that the two values that we are getting will not satisfy the third equation and that will show that they are not intersecting. So the equation 1 is 2 + s -1 + s and 4 - s and line two

- 2 - 2 t 2 + t and 1 plus t. Okay, I have just taken the x from here and x from here. y from here and y from here. z from here and z from here. And now you will solve them simultaneously. You will equate each other together. So 2 + s - 1 + s 4 - s - 2 - 2 t 2 + t 1 + t and you would form the two the three equations. So this would become s + 2t is equ= to - 2 - 2 - 4 s - t 2 + 1 is 3 then - s minus t it is equals to 1 - - 4 is - 3. So use equation one and equation two. Let's suppose using equation one and equation two values of s and t are coming out as let's see 1 2 4 1 - 1 and 3 so s is 2 upon Okay.

And t is - 7 upon 3. And if you put this this thing in the third equation, so minus 2 upon 3 minus - - 7 upon 3 is it equal to -3? I don't think so. minus - - 2 upon 3 + 7 upon 3. So 5 upon 3. So look 5 upon 3 is not equal to minus 3. So therefore it is skew there is no point of intersection. If the third one is also satisfied then they are intersecting fully. Now this is a little bit interesting. They will ask you to find the perpendicular distance from a point to a line. Sorry, I said two in my mind but I did not write it down. No. Okay. So this is your 3D line. Let's suppose line one and they have given you a

position vector external position vector O X is the position vector and what are they asking you? They are asking you to get the perpendicular distance between the line and the and the this what do you say >> and the point between the line and so this is the interesting thing. Now one more thing I forgot to tell you that when two vectors are perpendicular when two vectors are perpendicular so their dot product comes out as zero always. Why? Because your formula is what? Your formula is a do b is equals to a magnitude b magnitude and cos theta. And if your theta is 90, perpendicular means that the angle between the

vector if it is 90. So if the vectors are perpendicular, so cos 90 is zero. Look, we know cos 90 is zero. So cos 90 is zero. So this means that this complete thing becomes zero. So you are left off with the dot product being equal to zero. Similar now we will use this this thing this logic to find the perpendicular distance between a line and an external point. So what do we do? I think we'll have to do it very quickly. Only 10 minutes left. So this is the arbitrary point on this line which we don't know this will be let's you can you can uh you can call it let's change the name you can call it o x

so this point o x you can write it down in terms of your parameter in terms of parameter you can parameter your s or lambda whatever it is you can write it down and then you can find the vector ax ax will be this vector and ax will be equal to o x minus o a so you have the oa they have already given you the point external point oa and you will find the o x the point o x in terms of the parameter it is a general term of o x And then this ax you can see this vector ax will be perpendicular to direction vector of the line. So this means that ax dot product with the direction vector of the line will be equal to zero

because when two vectors are perpendicular their dot product is zero. And then there you we would find using this find the parameter find the value of the parameter when this happens and then you can use this value of parameter to get the point of x or ax and then you find the Perpendicular distance would be will be this ax magnitude. So if you have the value of the parameter that when this is happening you will substitute it over there and you will find the vector ax and then you will find the magnitude of the vector ax that would give you the perpendicular distance between the between the external point and a line.

This is the background. These are the steps that you will follow. Sorry. M. >> So this is your first step. This is your second step, third step, fourth step and then fifth step. Okay. Let's do one question. So look they have given you a point external point -411 and they have given you the equation of a line and they're asking you to get the perpendicular distance between from P the external point to the line L. So this is your straight line and this is your external point P O P is given to you. So first you will find the this point O X in terms of the parameter. So r it is equals to 1 + 2 lambda

3 + lambda and -4 + 3 lambda. And therefore in terms of this lambda I can find the point of intersection between the perpendicular dropped from the P to the line. It will be the same 1 + 2 lambda 3 + lambda and - 4 + 3 lambda. generalized coordinate for this point. Now we can find px. px will be o x minus o p. So o x we have found 1 + 2 lambda 3 + lambda - 4 + 3 lambda minus op. They gave us op -1 4 11. And if you simplify you will get -1 - - 1 would be 2 2 + 2 lambda uh 3 - 4 -1 + lambda and -4 - 11 - 15 + 3 lambda. Now this vector px px with the dot of of the direction vector of line. So this is your px

>> and dot with the direction vector of line. The direction of vector direction vector of line is this the part with the parameter 213. And because these two are perpendicular the line the px and the line this would equal to zero and you would form you would solve it 2 * 2 + 2 lambda. So 4 + 4 lambda plus -1 + lambda plus - 45 + 9 lambda is equals to 0. And you can solve this find the value of lambda. So -4 + 1 + 9 what become 15 14 lambda and 4 - 1 - 45 - 42 is equals to 0. And therefore your lambda come out as comes out as 42 upon 14 three. So now you know the value of lambda. So if you use the value of lambda over here

you will get the equation of px. Sorry you will get the vector px. So 2 + 6 is 8 - 1 + 3 2 and -5 + 9 - 6. This is your px vector and the perpendicular distance would be this vector px magnitude which will be equal to under root 8² + 4 + 36. So 2 10.20 this is your answer. >> Okay. Okay. I'm stopping the recording. This is it for today's class. In the next class, we'll just do past papers of this weather.