For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

## Discussion

## Practice Questions

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## Table of Contents

## Transcription

## Related Books

### More on Planes

- Recall that a plane can be represented by two points, →x and →p, on the plane and a vector →N normal to it by the equation (→x − →p) ×→N = 0. We can utilize our three points to find →N.
- Since (5,3, − 1) and (2, − 7,1) are points on the plane, we obtain [ (5,3, − 1) − (2, − 7,1) ] ×→N = 0 or (3,10, − 2) ×→N = 0. Similarly [ (4,9, − 6) − (2, − 7,1) ] ×→N = 0 or (2,16, − 7) ×→N = 0.
- We have (3,10, − 2) ×→N = 0 and (2,16, − 7) ×→N = 0, letting →N = (a,b,c) gives (3,10, − 2) ×(a,b,c) = 0 and (2,16, − 7) ×(a,b,c) = 0.
- Simplifying results in the system of equations

, letting c = 1 yields3a + 10b − 2c = 0 2a + 16b − 7c = 0

. Note that choosing c = 1 is arbitrary (except for c = 0).3a + 10b = 2 2a + 16b = 7 - Solving our system of equations results in a = − [19/14], b = [17/28] and c = 1. Our vector normal to the plane is →N = ( − [19/14],[17/28],1 ).
- The standard form of our plane is then [ ( x,y,z ) − (2, − 7,1) ] ×( − [19/14],[17/28],1 ) = 0 or (x − 2,y + 7,z − 1) ×( − [19/14],[17/28],1 ) = 0.

- Recall that a plane can be represented by two points, →x and →p, on the plane and a vector →N normal to it by the equation (→x − →p) ×→N = 0. We can utilize our three points to find →N.
- Since ( [1/(√2 )], − [1/(√2 )],0 ) and (1,0,0) are points on the plane, we obtain [ ( [1/(√2 )], − [1/(√2 )],0 ) − (1,0,0) ] ×→N = 0 or ( [(1 − √2 )/(√2 )], − [1/(√2 )],0 ) ×→N = [1/(√2 )]( 1 − √2 , − 1,0 ) ×→N = 0.
- Similarly [ ( [1/(√3 )],[1/(√3 )], − [1/(√3 )] ) − (1,0,0) ] ×→N = 0 or ( [(1 − √3 )/(√3 )],[1/(√3 )], − [1/(√3 )] ) ×→N = [1/(√3 )]( 1 − √3 ,1, − 1 ) ×→N = 0.
- We have [1/(√2 )]( 1 − √2 , − 1,0 ) ×→N = 0 and [1/(√3 )]( 1 − √3 ,1, − 1 ) ×→N = 0, letting →N = (a,b,c) gives [1/(√2 )]( 1 − √2 , − 1,0 ) ×(a,b,c) = 0 and [1/(√3 )]( 1 − √3 ,1, − 1 ) ×(a,b,c) = 0.
- Simplifying results in the system of equations

, letting b = 1 yields( 1 − √2 )a + − b = 0 ( 1 − √3 )a + b − c = 0

. Note that choosing b = 1 is arbitrary (except for b = 0).( 1 − √2 )a = 1 ( 1 − √3 )a − c = − 1 - Solving our system of equations results in a = [1/(1 − √2 )], b = 1 and c = [(2 − √2 − √3 )/(1 − √2 )]. Our vector normal to the plane is →N = ( [1/(1 − √2 )],1,[(2 − √2 − √3 )/(1 − √2 )] ).
- The standard form of our plane is then [ ( x,y,z ) − (1,0,0) ] ×( [1/(1 − √2 )],1,[(2 − √2 − √3 )/(1 − √2 )] ) = 0 or (x − 1,y,z) ×( [1/(1 − √2 )],1,[(2 − √2 − √3 )/(1 − √2 )] ) = 0.

- The distance from a point →q to a plane is given by [((→p − →q) ×→N)/(|| →N ||)] where →p is a point in the plane and →N is a normal vector to the plane.
- Substituting gives [((→p − →q) ×→N)/(|| →N ||)] = [([ ( − 1,5,2) − (3,7,0) ] ×(1,1,1))/(|| ( − 1,1,1) ||)] = [(( − 4, − 2,2) ×( − 1,1,1))/(√3 )].

- Recall that a normal vector to the plane ax + by + cz = d is (a,b,c). So →N = ( [1/2],4, − [1/4] ) = [1/4](2,16, − 1).
- A point on the plane [1/2]x + 4y − [1/4]z = 1 satisfies the equation, for instance →p = ( − 4,1,4). We can now find the distance using [((→p − →q) ×→N)/(|| →N ||)].
- Substituting yields [((→p − →q) ×→N)/(|| →N ||)] = [([ ( − 4,1,4) − (0,0,1) ] ×[1/4](2,16, − 1))/(|| [1/4](2,16, − 1) ||)] = [(( − 4,1,3) ×(2,16, − 1))/(√{261} )]. Note that || [1/4](2,16, − 1) || = | [1/4] ||| (2,16, − 1) ||

- We can find the normal vector to a plane →N using three points by using (→x − →p) ×→N = 0.
- We have (1,3,4) − ( − 1,1,1) = (2,2,3) and ( − 2,1,2) − ( − 1,1,1) = ( − 3,0,1). We can form a system of equations
(2,2,3) ×→N = 0 ( − 3,0,1) ×→N = 0 - Letting →N = (a,b,c) we now have

, if a = 1 then c = 3 and b = − [11/2].2a + 2b + 3c = 0 − 3a + 1c = 0 - We now have a normal vector →N = ( 1, − [11/2],3 ) and a point →p = ( − 1,1,1) on the plane. To find the distance from →q we use [((→p − →q) ×→N)/(|| →N ||)].
- Substituting yields [((→p − →q) ×→N)/(|| →N ||)] = [([ ( − 1,1,1) − ( [2/(√5 )],0,[1/(√5 )] ) ] ×( 1, − [11/2],3 ))/(|| ( 1, − [11/2],3 ) ||)] = [(( [( − 2 − √5 )/(√5 )],1,[( − 1 + √5 )/(√5 )] ) ×( 1, − [11/2],3 ))/(√{[161/4]} )]

- Parallel vectors lie on the same plane. Also note that there are no z - coordinates on these vectors, hence our plane is the xy - plane.
- Any vector along the z - axis is normal to the xy - plane, for instance (0,0,1). We can now find the distance from →q to the plane using [((→p − →q) ×→N)/(|| →N ||)].
- Since →N = (0,0,1) is a unit vector we compute (p − q) ×N = [ ( 1,[1/2],0 ) − (11, − 2,7) ] ×(0,0,1) = ( − 10, − [3/2], − 7 ) ×(0,0,1) = 7.

- We can still utilize [((→p − →q) ×→N)/(|| →N ||)] to find the distance between →q and (0,1,0,1) noting that →N = ( − 4,1, − 2,3).
- Substituting yields [((→p − →q) ×→N)/(|| →N ||)] = [([ (0,1,0,1) − (1,1,1,0) ] ×( − 4,1, − 2,3))/(|| ( − 4,1, − 2,3) ||)] = [(( − 1,0, − 1,1) ×( − 4,1, − 2,3))/(√{30} )].

- To find a vector parallel to the line of intersection of two planes it suffices to find two points on the line of intersection and obtain their difference to construct a vector.
- We can find two points on the line of intersection by fixing z = 1 and z = − 1 (any z0 value will do) and finding the corresponding point from the plane.
- For z = 1 we have

orx + y − 1 = 6 − x + y + 1 = 5

solving the system of equations yields the point ( [3/2],[11/2],1 ).x + y = 7 − x + y = 4 - For z = − 1 we have

orx + y + 1 = 6 − x + y − 1 = 5

solving the system of equations yields the point ( − [1/2],[11/2], − 1 ).x + y = 5 − x + y = 6

- To find a vector parallel to the line of intersection of two planes it suffices to find two points on the line of intersection and obtain their difference to construct a vector.
- We can find two points on the line of intersection by fixing z = 1 and z = − 1 (any z0 value will do) and finding the corresponding point from the plane.
- For z = 1 we have

or3x + 2y − 4 = 11 x − 4y + 2 = 3

solving the system of equations yields the point ( [31/7],[6/7],1 ).3x + 2y = 15 x − 4y = 1 - For z = − 1 we have

or3x + 2y + 4 = 11 x − 4y − 2 = 3

solving the system of equations yields the point ( [41/21], − [4/7], − 1 ).3x + 2y = 7 x − 4y = 5

- To find a vector parallel to the line of intersection of two planes it suffices to find two points on the line of intersection and obtain their difference to construct a vector.
- We can find two points on the line of intersection by fixing z = 1 and z = − 1 (any z0 value will do) and finding the corresponding point from the plane.
- For z = 1 we have

or[1/4]x − [3/4]y + 1 = − 4 [1/3]x − [2/3]y − 1 = − 2

orx − 3y + 4 = − 16 x − 2y − 3 = − 6

solving the system of equations yields the point ( 31,17,1 ).x − 3y = − 20 x − 2y = − 3 - For z = − 1 we have

or[1/4]x − [3/4]y − 1 = − 4 [1/3]x − [2/3]y + 1 = − 2

orx − 3y − 4 = − 16 x − 2y + 3 = − 6

solving the system of equations yields the point ( − 3,3, − 1 ).x − 3y = − 12 x − 2y = − 9

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### More on Planes

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- More on Planes 0:25
- Example 1
- Distance From Some Point in Space to a Given Plane: Derivation
- Final Formula for Distance
- Example 2
- Example 3: Part 1
- Example 3: Part 2

### Multivariable Calculus

### Transcription: More on Planes

*Welcome back to educator.com, welcome back to Multivariable Calculus.*0000

*Last lesson we talked about planes, and today we are going to talk a little bit more about planes.*0005

*We want to become reasonably familiar with the definition, sort of manipulating, doing different types of problems, thinking geometrically, and using already the intuition that we have about geometry to solve different problems regarding this.*0009

*Let us just jump right on in and begin with an example. So, example 1.*0026

*We want to find an equation for the plane passing through s, which is the point (1,2,1), t which is the point (-1,2,4), and u which is the point (1,3,-2).*0035

*In this case, they have not given us a normal vector, they just gave us 3 points in space and they want us to find an equation that passes through those 3 points.*0076

*Well, you know, our equation was based on finding and having some sort of a normal vector.*0088

*We are going to use these to find the normal vector.*0093

*Let us just draw this out real quickly just to see what it looks like, this is s, this is t, this is u.*0096

*So we want to find the plane that passes through those, this is s, this is t, and this is u.*0104

*So, we need a normal vector.*0111

*Let us solve some simultaneous equations, we need basically some vector which is going to be normal to the plane.*0114

*I have 3 points. If I take this vector and if I take this vector right here, so st, and su, both st and su are perpendicular to this normal vector n that we want to find in order to use the equation that we have for the plane.*0124

*We can find this normal vector by actually finding some simultaneous equations since we have some vectors here.*0145

*Let us go ahead and do that.*0153

*We will let the normal vector n = (a,b,c), so the (a,b,c) is what we are going to be looking for.*0155

*Now we know that n is perpendicular to st, and we also know that n is also going to be perpendicular to su, so let us just write out some equations.*0165

*st, the vector st is going to equal t - s, right? that is how we get a located vector.*0180

*A located vector means that we are going to get a vector that is somewhere different than the origin.*0190

*That is going to equal to... when we do t-s... we can put (-2,0,3).*0195

*Let us go ahead and do su. The vector su is going to be u - s, and that is going to equal (0,1,-3).*0204

*We know that since n is perpendicular to this, we know that n · this = 0.*0216

*n · t - s = 0, and we also... so let me go ahead and solve that one, when I do n · t - s, I am going to write it as... actually you know what, let me know, no, a little bit too much here.*0233

*Let me go ahead and write it here, n · t - s = 0, and I also know that n · u - s is equal to 0.*0264

*Now, let me go over here.*0281

*That is going to be a,b,c · (-2,0,3) = 0.*0286

*This one gives me -2a + 0b + 3c = 0.*0289

*I also know that n · u - s = 0, that is going to be the vector a,b,c, that is my vector n.*0309

*u - s = 0, 1,-3 = 0, so when I do this dot product, I get a × 0 is 0a + b - 3c = 0.*0320

*I have these two equations that I can actually end up solving, so I end up with b - 3... so I have, let us see, actually let me write these a little bit better, - 2a + 3c = 0.*0345

*I also have b - 3c = 0.*0364

*We will let c = 1, we will just choose some value and go ahead and solve for a and b.*0370

*So if we let c = 1, then b is going to end up equaling 3, and a will end up equaling 3/2.*0377

*If I put c in for 1 here, b is going to end up being 3, and if I put c in 1 here, -2a is going to = -3, so a = 3/2.*0394

*There we go, we have our a,b,c.*0405

*Now n = the vector (3/2,3/1) and now I can go ahead and write my equation for the plane.*0406

*It is going to be x · n = ... and I choose any point, the s, the t, or the u, that the plane passes through, it does not matter.*0421

*I will go ahead and take s, so s · n.*0436

*So x = x,y,z, and n = (3/2,3,1) = s which is (1,2,1) × (3/2,3,1).*0445

*When I actually run this, I get the following.*0464

*I get 3/2x + 3y + 1z = 3/2 + 6 + 1... I will go ahead and multiply everything by 2 to get rid of the 2 here.*0469

*So I will get 3x + 6y + 2z = 3 + 12 + 2.*0494

*I get 3x + 6y + 2z = 17.*0504

*I will rewrite this a little bit better because I do not want all of these random lines to be all over the place.*0513

*3x + 6y + 2 × = 3 + 12, 17, there we go that is it.*0521

*In this case we started off with 3 points in space, and we ended up with defining the plane that passes through those 3 points.*0532

*We needed a normal vector.*0538

*In this particular case, we just took the vectors from one point to the other, or from that same point t the other point, and we let n = some (a,b,c) and we ended up solving some simultaneous equations.*0540

*There is another way to do this, any time you are given two vectors in a plane there is a very quick and easy way to find a vector that is normal to both of those.*0552

*It is something called the cross product or the vector product.*0564

*Now, we could introduce it now, many people often do after they discuss the scalar product or the dot product, however we are probably not be going to use the cross product for a while, not until later on in the course.*0566

*I did not want to introduce something simply for the sake of having it, when we are not going to not use it and then just reintroduce it later.*0580

*So at this point I think it is nice to use other tools at our disposal, other geometric intuitions, you know solving simultaneous linear equations.*0587

*Our tool box is getting bigger, we have a lot of things at our disposal.*0595

*Including everything that we learned before, it is not just what we are learning now that we are applying.*0598

*I think that is nice that we were able to use this without having to fall back on this notion of the cross product.*0605

*We will deal with it when we get to it.*0609

*Okay, now we are going to work on a very important notion.*0612

*What we want to do is we want to say, we want to be able to find the distance from some point in space to a given plane.*0618

*I could go ahead and write out the formula for this, and that will be fine.*0626

*For all practical purposes, we are going to be doing a lot of that, especially if the formulas seem to make a little bit of sense.*0634

*This formula, I am going to go through a bit more of a derivation for, simply because I want you to understand where it comes from.*0640

*In this case, where it comes from will make more sense.*0646

*We will not often derive a lot of the formulas in this course, because we want to concentrate on just problem-solving and intuition, but this is a derivation that I think you should see.*0650

*Now, let x - p · n = 0 be a plane through p perpendicular to n.*0661

*That is the standard definition of a plane.*0684

*Now, let q be an arbitrary point in the space, whatever that space happens to be.*0689

*Now, we want a formula for the distance, the length, a number, for the distance from q to the plane.*0709

*Okay, let us draw a picture and get a sense of what is going on and we can use the things we discussed over the last few lessons to come up with a formula here.*0732

*Let us go ahead and draw a plane.*0740

*We have some plane, and let me see, we have some point p here.*0748

*A point p, it passes through that, and it is perpendicular to some vector n.*0756

*I will go ahead and write n here, it is perpendicular to that.*0764

*Then I have some arbitrary point q, so I will go ahead and put q here, well this is the point q.*0765

*We want to find the distance from this to the plane, well, we know what distance is, when we find the distance we drop a perpendicular to that plane.*0774

*We do not go this way or that way, we go straight down to the plane.*0780

*So, if the point is here, not that, or that, it is just straight down. This distance is what we are looking for.*0785

*That is it, so we are looking for this distance.*0797

*So we want a formula, we want this length.*0802

*Let us take a look at what we know.*0808

*We have p, right? that is the point that the plane passes through.*0810

*We have n, that is the normal vector, and we have q.*0819

*So we actually have a lot here. Let us take a look at what we have got.*0823

*This length is what we actually want, so here is how I am going to do this.*0830

*Remember we discussed something called a projection? Where if I had one vector, and if I had another vector, the projection of this vector onto this vector is just the vertical... it is like shining a light this way.*0832

*We had some formula for that, remember what that was?*0847

*I will get to that in a minute, but here is what I am going to do.*0850

*I actually am going to take the vector qp, and this right here is also another vector.*0855

*Let us go ahead and call this point, where it actually hits at, q'.*0864

*Now, if I take qp and I project it onto the vector q'... if I project it onto this vector... now let me draw it separately, I have qp, q, q', and qp.*0872

*If I take this vector and project it onto this vector, in other words, if I project it this way, I am going to end up getting this distance.*0895

*That is what this distance was, remember when we did a projection, no matter how long this vector was, if I just project it onto this, it is actually just going to give me the shadow.*0908

*The shadow of this vector on here, because these are actually in the same... this point and this point are on the plane.*0917

*This distance is the projection of this vector onto this vector.*0927

*That is what we are going to do, we are going to find the projection and then we are going to use that to give us our length.*0931

*Let us go ahead and write what this is.*0938

*First of all, let us go ahead and write what qp is, so qp, qq', alright, let us write that over here.*0941

*The vector qp is going to equal p - q, that is the definition of a located vector.*0956

*Now we are going to project it onto this vector qq'.*0964

*What is interesting is we know what q is, we do not really know what q' is, however we know that we have this vector n that is perpendicular to the plane.*0968

*Well this vector n, well this is perpendicular, and this is perpendicular, so for all practical purposes, what I can do is... these are parallel.*0980

*We said parallel vectors in space are the same, you can use one for the other.*0990

*So, I am not only going to project this onto this, it is the same thing as projecting this vector onto n itself, and we have n already.*0995

*Again, it is just like taking this vector and moving it, over here until it is on top.*1006

*That is the whole idea behind parallel vectors, as long as they are parallel I can use any vector I want.*1014

*Since this is parallel to this, instead of projecting this onto this, it is the same as projecting this vector onto n.*1017

*Okay, so this is what we are going to do, we are going to find the projection of qp onto the vector n.*1023

*That is what we are going to do. Let us go ahead and do that.*1038

*Now let us write the formula. Let us recall what the formula is generally.*1041

*If I have... so the projection of some vector a onto another vector b is a · b/b · b, that is a number × vector b.*1048

*Now what we are doing is we are projecting the vector qp onto the vector n.*1062

*Well, that is going to equal the vector qp · vector n/n · n × n.*1071

*Well qp is p - q · n/n · n × n.*1087

*This number right here, c, this was the number that we almost want, it is not quite what it is that we want.*1107

*Recall that when we project one vector on top of another, we end up with this c, that is what we get.*1116

*However, this number c, based on the definition of the projection, is not a length, it is a scale factor.*1125

*It is saying take the length of this vector and scale it back.*1133

*This gives us a scale factor.*1137

*The actually length of this thing which is what we want, a positive number because a distance is actually a positive number, it is actually a scale factor × the norm of the vector.*1140

*Again, the norm is the length of the vector.*1153

*Let me write this down. C is not the distance but a scale factor.*1156

*It is saying scale back this vector back to here.*1170

*C × norm(n) is the distance.*1175

*In other words, if this vector is a certain length, the scale factor times that length gives me this length.*1185

*Let us write this out.*1194

*We have p - q... oops, let me use lower-case... p - q · n... oops, I want this to be as clear as possible... · n/n · n ×... that is c... that is what this thing is.*1198

*Well we are going to multiply that thing by the norm of n.*1228

*Well the norm of n, you remember the definition of a norm, you dot the vector with itself and you take the square root which will represent a fractional coefficient.*1230

*(n · n) ^{1/2}, (n · n)^{1}, this cancels with that, and this leaves you with p - q · n/n · n... nope, I am not going to do that, I want these to be very clear.*1239

*(n · n) ^{1/2}, well n · n raised to the 1/2 power is the norm of n.*1270

*I am going to write that as my final formula.*1280

*It is going to be so... distance, length, distance = p - q · n / norm(n).*1285

*This is the equation that we want.*1311

*Given a plane that passes through the point p, and a random point in that space q, the distance from q to the plane is equal to the point p - the point q dotted with the vector n, dotted with the normal vector, and all of that divided by the norm of n.*1316

*This is all based on standard projection. That is all it is.*1339

*This is true in all number of dimensions.*1345

*You are probably thinking to yourself, what does distance mean in 5 dimensions, well that is the whole idea.*1348

*Distance is an algebraic property, not a geometric property.*1354

*We think of it in terms of a geometric property because we live in 3-dimensional space, so I know what the distance between me and the wall might be, but that distance is not just a physical number in terms of 3 dimensions, it is a number that exists in any number of dimensions.*1359

*This is how it is. If I have some plane in 10 dimensions, and I have a point in that, you know, 10-dimensional space, this formula will give me the distance from the point to the plane.*1376

*Let us just go ahead and do an example here. Example 2.*1388

*We will let p = (1,2,1) and we will let the point q = (3,-1,-1), and we will let n = (3,4,-2).*1399

*Now, we want to find the distance from q to the plane passing through p and perpendicular to n.*1420

*That is it. Find the distance from q to the plane passing through p and perpendicular to n.*1448

*Okay, we have to draw this out. We have a nice little formula, perfectly algebraic.*1454

*Let us go ahead and write what we have here.*1459

*We have the length = p ... oh, you know what, this is actually a good time to stop and talk about general technique in solving problems.*1462

*When you are solving a problem, and if you have a formula, even if you are familiar with that formula, it is always a good idea to write down that formula first*1475

*It is just a good habit to get into so that you know what it is you are working with.*1483

*If you happen to be looking at the formula on a separate page, and sort of running back and forth and putting numbers into the formula, that is one thing, but if you actually write it out, you get in some good mathematical habits, writing out formulas, making sure they are on the page, very systematic.*1489

*The idea of mathematics, and the idea of being correct is about being systematic.*1501

*Do not do things in your head, make sure you write everything out.*1506

*So it is equal to p - q · n/the norm of n.*1511

*What is p - q? Well p - q is going to equal (-2,3,2), that is p - q.*1523

*Then n is (3,4,-2), so write it all out, no big deal, you do not need to do any of this in your head, you want to be correct.*1540

*The norm of n is going to be, 3 ×3 is 9 + 16 + 4 all under the radical.*1548

*That equals -6 + 12 - 4 all over sqrt(29), that is going to equal -6 + 12 is 6, 6 - 4 is 2, so all of that = 2/sqrt(29).*1560

*There you go. That is your distance, strictly algebraic property.*1581

*Once again, if I have a plane in the given space that passes through a point p and is perpendicular to n, and if I have a q a random point in that space, the distance from the point to that plane is given by p - q · n/norm(n).*1587

*Clearly n, the normal vector is very important. It will continue to be important as we go on.*1608

*Let us do another example, just more practice with planes, more practice with some of the other resources we have at our disposal.*1617

*Let us see what we have got. Example 3.*1626

*Find a vector parallel to the line of intersection of the planes 3x - 2y + z = 1, and 2x + 2y - 2z... let me make my 2's and z's a little bit different here... -2z = 2.*1632

*So, we have 2 planes, and they intersect in a line.*1685

*If they intersect in a line, we want to find a vector that is parallel to that vector.*1692

*I am sure there are a number of ways to do this, I mean one of the nice things as you move along into higher and higher mathematics is that there is a whole plethora of ways that you can approach a problem as you gain more tools at your disposal.*1697

*What I am going to do in this case is I am just going to choose 2 random points on that line, and then I am going to sort of take the vector that connects those 2 points.*1710

*It is parallel, it happens to actually be on the line, but it is parallel to the line.*1721

*That is it, that is what I am going to do.*1727

*Find 2 points on this line, we will call them p and q, then take q - p or p - q, again the direction of the vector does not matter.*1731

*Any vector is fine whether it is one direction or the other direction, parallel is this way and that way.*1754

*We are going to end up actually solving 2 equations and 3 unknowns, so here is what I am going to do.*1760

*I am actually going to choose, and again this is a line so there are an infinite number of points that I can choose from... what I am going to do is I am going to choose a z, and I am going to solve for x and y.*1766

*I am going to do the same thing for another point -- choose a z and then solve for x and y.*1781

*Then those are going to be my two points p and q and then I am going to choose the vector in between them.*1786

*So let us just start off, so we will let z=1.*1790

*So when z=1 these equations become, implies so we have 3x - 2y + 1 = 1, and 2x + 2y - 2 = 2.*1792

*So this equation becomes, 3x - 2y = 0, and this equation becomes 2x + 2y = 4.*1820

*Now I can go ahead and solve this, I end up with 5x = 4, and I get x = 4/5.*1840

*Now when i put 4/5 back into here, let us just sort of choose this one, I get 3 × 4/5 which is 12/5 - 2y = 0, I get y = 6/5.*1848

*My point p, my first point, is going to equal (4/5,6/5,1).*1867

*I do not want to deal with fractions, so again, any multiple of a vector, is pretty much the same vector, it is in the same direction.*1880

*I am going to go ahead and take p, and I am going to multiply everything by 5 so that I can work with whole numbers, and this is going to be (4,6,5).*1889

*(4,6,5) is my first point and now I will do the same thing with another point.*1898

*I will choose another z and go through the same process.*1903

*This time, we will let z = 2.*1908

*This implies, we get 3x - 2y + 2 = 1, and we get 2x + 2y - 4 = 2.*1915

*These equations become, 3x - 27 = -1, and 2x + 2y = 6.*1932

*When we add those we get 5x = 5, we get x = 1.*1948

*We put 1 back in here, either equation is fine, so I will just go ahead and put it in this equation, 2 × 1 + 2y = 6, 2y = 4, y = 2.*1956

*Now our point q = x=1, y=2, z=2, so now I have q is (1,2,2).*1975

*On the previous page we said that p = (4,6,5), therefore q - p = 1-4 is -3, 2 - 6 is -3... wait 2 -6 is not -3, it is -4... 2 - 5 is... this is -3, so there you go.*1990

*This is a vector that is parallel to the line which is the intersection of those 2 planes.*2028

*I simply chose 2 points on that line, and I took the vector.*2035

*I can choose (3,4,3) if I want just multiply by -1, it does not matter, again the direction does not matter, I have an infinite number of vectors to choose from.*2042

*That is it.*2048

*Okay, thank you very much for joining us here at educator.com and Multivariable Calculus.*2051

*Next time we will begin our discussion of differentiation of vectors, take care.*2055

0 answers

Post by Indu Sreedhar on May 12, 2017

how the b-3c=1? it should be b-3c=0 right?

0 answers

Post by Indu Sreedhar on May 12, 2017

how the second equation became b-3c =1 I thought it should be b-3c=0

3 answers

Last reply by: Professor Hovasapian

Wed Jan 3, 2018 4:14 AM

Post by Hen McGibbons on August 22, 2015

I have another question. I recall from a previous video lecture, that you assumed 2 of the variables. So for example, you let x= 5 and y = 2. I went back through all the video lectures but I couldn't find the one I am thinking about. Why would you be able to assume 2 of the variables like that? To me, this is even more confusing than being able to assume 1 variable (letting z = 2 for examples.)

1 answer

Last reply by: Professor Hovasapian

Sat Aug 29, 2015 9:14 PM

Post by Hen McGibbons on August 22, 2015

i can't wrap my head around letting z = 1 and then letting z = 2 in Example 3. I've also seen you use this method in several other examples in the videos before this. Why can you just assume that z=1 will work for both equations of a plane?

0 answers

Post by Shelly Wang on May 24, 2014

For Example 3, I used x=0 and received an answer of y=-2 and z=-3 yielding a p=(0,-2,-3). Then, I used x=1 and received an answer of y=2 and z=2 yielding a q=(1, 2, 2).

After doing q-p, my final answer for the vector is (1, 4, 5) which does not equal c(-3,-4,-3) where c is any real number.

Is my answer correct?

2 answers

Last reply by: Professor Hovasapian

Wed Oct 30, 2013 2:41 AM

Post by Christian Fischer on October 26, 2013

A quick question regarding planes: When you in your last example can choose any value of z (you choose 1 and 2) does that imply that planes just like vectors are independent of the coordinate system, so the reason you can choose ANY value of z is because you can lay the plane any place in the z axis? So a plane is not fixed in space until the point where you choose values of x,y and z?

2 answers

Last reply by: Professor Hovasapian

Wed Oct 30, 2013 2:26 AM

Post by Christian Fischer on October 26, 2013

Hi professor, thank you for a great video. I don't understand why you project vector qp onto qq`, and not instead project qq`onto qp`since qqÂ´has the length of the distance between the two points and thus it's shadow projected onto qp would have that same length?

Have a great day,

Christian

1 answer

Last reply by: Professor Hovasapian

Fri Sep 20, 2013 5:28 PM

Post by yaqub ali on September 20, 2013

professor, why is it when solving for the plane running through the three points. why did you randomly let C=1

could you have randomly chosen B=1

besides what if C=1 isn't on the plane. how do you make sure the point you randomly chose is on the plane??

2 answers

Last reply by: Justin Malaer

Wed Feb 13, 2013 8:57 PM

Post by Justin Malaer on February 1, 2013

To find the vector n orthogonal to the plane, could you just take the cross product of vector ST cross vector SU? Because then you would have a vector orthogonal to the plane if I'm correct.

3 answers

Last reply by: Professor Hovasapian

Sat Dec 29, 2012 5:24 PM

Post by Mohammed Alhumaidi on July 14, 2012

Isn't b = 4 when C = 1 ?

b - 3C = 1

b - 3 = 1

b = 1 + 3

b = 4