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

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

### Planes

- 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

- 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

- 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]} )]

- 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.

*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

### 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
- Planes 0:18
- Definition 1
- Example 1
- Example 2
- General Definitions and Properties: 2 Vectors are Said to Be Paralleled If
- Example 3
- Example 4

### Multivariable Calculus

### Transcription: Planes

*Hello and welcome back to educator.com, welcome back to multivariable Calculus.*0000

*Today we are going to start our discussion of planes.*0005

*We are going to spend a couple of lessons on planes, and then we will move on to actual Calculus.*0006

*Differentiation of vectors.*0012

*So, with planes, let us go ahead and get started.*0014

*Let us recall that when we talk, as far as notation is concerned, when we say something like the vector x, we are talking about its components.*0018

*So, in 3 space we have 3 numbers representing that vector -- x1,x2, and x3.*0028

*I know that you know what that is but I just want to recall that because a lot of the definitions we give, we give in vector form, like this.*0037

*When we actually do some of the calculations, we of course have to work with some components, a specific example.*0044

*Okay, let me go ahead and draw out a three dimensional coordinate system here.*0052

*This is the z-axis, this is the y-axis, this is the x-axis.*0058

*What we want to do is find the equation of a plane in a given space.*0060

*Again, we are using geometry for something that we do know, but the definition that we give is going to be valid in any number of dimensions.*0068

*Let us go ahead and draw a plane here, let us see if we can make this look reasonably good.*0076

*It is the sum... random plane, and let us say we have some point p in there... then, another x... then let us go ahead and draw a vector here.*0083

*I am going to call this vector n.*0096

*This here is going to be point p, it is again just a vector.*0099

*This is just going to be point x, I will go ahead and draw a little arrow here, so this will be vector x.*0104

*Now, what we want to do is, let us go ahead and write down the definition and then we will work from the picture.*0114

*So, the definition.*0122

*We define the plane through the point p, and perpendicular to... actually I am going to use the symbol for perpendicular... and perpendicular to the vector n, is the set of all points x such that the located vector px is perpendicular to n.*0124

*So, let us go ahead and write what that means symbolically, and then we will go to the picture and talk about what this means.*0174

*We have x - p · n = 0.*0182

*We have this plane, and we have this random vector n.*0192

*We want to find an equation of this vector n that passes through the point p, and is actually perpendicular to this vector n.*0195

*In other words, every vector in this plane is going to be perpendicular to, or orthogonal to n.*0203

*Well a vector n that starts at the origin is the same as any vector... we can take that vector and sort of move it anywhere we want... it is still the same vector.*0210

*Imagine that I have taken this vector and I have moved it over here, now it is there.*0220

*What we are looking for is, now let me draw a vector from p to x, so now if I have this point p, and this vector n... in this case it is going to be the vector pn, but it is still just n.*0227

*All I have done is move the vector.*0246

*We can do that, that is the whole point of vectors.*0248

*Anywhere in space is the equivalent to the vector that begins in the origin and is in the same direction.*0252

*Now, this point x, all of the x's in this plane, if I draw vectors from p to any point x, all of those vectors px are going to be perpendicular to this vector n.*0257

*Well, perpendicularity is orthogonality which means that the dot product = 0.*0270

*So, I literally write... so px perpendicular or orthogonal to n... well px, remember we take the head - the tail, so that will be x - p, that is this vector, dotted with n, which is this vector, and that is going to equal 0.*0275

*What I am actually looking for, these are the variables, this is sort of why I brought this up.*0294

*This right here is the equation for the plane that passes through the point p and is perpendicular to n.*0302

*The set of all points, this is the equation for it.*0311

*Now let us go ahead and do some math so we can do the equivalent versions of this.*0314

*You are free to use any version you want, whichever is most appropriate for a given problem.*0318

*This, you can distribute the dot product, right?*0325

*This you have x · n - t · n = 0.*0328

*Or, we can go ahead and move this over to the right because the dot product is just a number.*0335

*This becomes x · n = p · n.*0342

*In this case we know what n is and we know what p is, this x is the variables, the x, the y, the z, the t, however many dimensions you are working in.*0349

*If it is 2, this is x,y, if it is three, it is x,y,z, or x1,x2,x3.*0356

*This gives the actual equation of the plane, so any one of these is fine, each are the same representation.*0365

*Very important vocabulary here.*0372

*We say the vector n is normal to the plane.*0378

*This idea of normality is actually going to be very important as we go on and start to discover and discuss vector functions and parametric lines and things like that.*0392

*We will have tangent vectors, and we will have normal vectors, and they will play a very prominent role, especially when we do Green's Theorem or Stoke's Theorem later on.*0402

*We are often going to speak of normal, and normal basically just means perpendicular to a plane, or perpendicular to a line.*0410

*So, things like that, the vocabulary of normality will come up quite often.*0420

*Let us do an example here, Example 1.*0424

*Let us let p = the vector (-2,1,-1) and n = the vector (-1,1,4).*0432

*We want to find the equation of the plane that passes through the point and is orthogonal to n.*0449

*Let us let x = (x,y,z).*0460

*We know, let us use this particular form, x · n = p · n, and you are welcome to use any of the 3 forms.*0468

*The first one is the actual definition because it says that something · something = 0, which is the very definition of orthogonality, so that is always a nice one to use.*0478

*But, any variation of it is fine.*0486

*Now, let us do x · n.*0491

*Well x · = is -x + y + 4z, I have just taken x · n, this is this · that, = p · n.*0493

*Well, p · n = -2 × - 1 is 2, 1 × 1 is 1, and -1 × 4 is -4.*0510

*So what we get is -x + y + 4z = 2 +1 is 3, 3 - 4 is -1, there you go, that is it, see what we have done.*0522

*In 3 space, for x,y,z, we were able to take this definition, just run the dot product, and we end up with what is a classical representation of.. it is the equation of a plane in space.*0533

*You have x, you have y, you have z, you have ax, by,cz = d.*0546

*It is the equation of a plane that you have seen before in your high school and possibly in your calculus courses.*0550

*This definition, the vector definition is the general definition, it is true in any number of dimensions.*0557

*In dimensions above 3, well you can still call it a plane, it is more traditional to call it a hyper-plane but that is it.*0563

*You can have a 10-dimensional space, but that is still going to be some plane that is orthogonal to some vector.*0570

*This is how it is represented.*0581

*This is an algebraic general definition for that concept.*0583

*So, a couple of things to note.*0589

*Note, if you run this... well you know what, I am not going to write this part down... I will go ahead and tell you.*0591

*If you were to go ahead and run this calculation in 2-space, what you go ahead and end up with is an equation of a line.*0602

*So that is it.*0607

*In some sense, this is the most general thing you can get.*0608

*In 2 space, you are going to get a line, in 3 space you are going to get the equation of a plane, and in 4 space you are going to get the equation of a hyper plane.*0611

*Again, it is the same thing.*0618

*Now, take a quick look at the coefficients of this thing.*0622

*I will go ahead and write, "note" (-1,1,4).*0626

*Note: the coefficients of the standard form of the plane equation.*0635

*When I take the vector (-1,1,4), that is a vector... as a vector... it is... it, or any multiple of it, is perpendicular to the plane.*0656

*Expressed in standard form.*0697

*So, if you start it actually with an equation of a plane that you are accustomed to seeing, this ax+by+cz=d, if you just take a,b,c, as a vector, that is your normal vector or any multiple of it.*0706

*That is the whole idea.*0720

*Let me write that last one down.*0723

*Given ax + by + cz = d, then a,b,c is perpendicular to the plane, or any multiple of it, that is the whole idea.*0725

*We started off with the general definition we recovered the standard form, in 3 space, but if you are given it in 3 space, then take the coefficients and you have got yourself your normal vector.*0754

*Very nice actually, it comes in very handy.*0762

*Let us do a couple of examples here.*0766

*So, example 2, we want to find a point in the plane of example 1.*0772

*In other words, a random point, so now that we have this plane we have this equation for the plane which is -x - y + 4z = -1.*0790

*Find a random point in that plane.*0809

*Okay, this is actually pretty easy.*0812

*Essentially what you have to do is... there are infinite number of points... so, you can choose any two, then solve for the third.*0814

*So, let us just say, let x = 5, and again it does not matter which numbers you choose, and y = 2, then we have, -5 - ... actually let us make y a - 2 so -5 - a minus 2, + 4z = -1.*0831

*So we have -5 + 2 is -3, when we move it over here, -1 + 3 is 2, so we end up with 4z = 2, so we get z = 1/2.*0856

*If we take the point (5,-2,1/2), this is a random point in the plane.*0871

*That is it, choose any two and solve for the third, nice and easy.*0886

*Now, just some general definitions of properties.*0890

*So, 2 vectors a and b, are said to be parallel, and I will often use the symbol just like that for parallel, if there is a number c such that c × a = b.*0897

*As you see, parallel is a geometric notion that you know of.*0926

*You know that it is two lines going like this, or two planes going like this.*0930

*2 lines going like this, 2 vectors going like that, it is a geometric notion.*0935

*We need an algebraic definition in order to do some math.*0940

*Here is your algebraic definition.*0942

*If I have a vector, and I have another vector, or if I have a vector and I multiply it by some multiple, smaller than 1, bigger than 1, negative, it does not matter, and I end up with the other vector, well those two vectors are parallel.*0943

*This is an algebraic definition of parallelity, or parallel-ness, or whatever you want to say.*0957

*Two planes are said to be parallel if their normal vectors are parallel.*0963

*If I want to deal with these 2 planes and I want to find out if they are parallel, I have to deal with their normal vectors.*0983

*If I have a plane and a plane, well their normal vectors are going to be, one is here, and one is there, those are going to be parallel.*0990

*Now we fall back again on this definition, the parallelity of vectors.*0995

*So, let us do another example here.*1004

*Example 3.*1011

*Find the angle θ between the planes 2x + y - z = 0, and x - 2y + z = 2.*1014

*So, we just said that 2 planes are parallel if their normals are parallel.*1042

*If we want to find the angle between 2 planes, so clearly they are not parallel, they are intersecting somewhere, there is an angle between those 2 planes like that.*1049

*The angle between the planes is the angle between the normal vectors.*1058

*So, we find a couple of normal vectors and we do exactly what we do with that definition of dot product and cos(θ) and we find the angle between them.*1063

*Now we said if we are given a plane in this form, we can find two normal vectors.*1070

*Let us take this one and call this plane number 1, and this is plane number 2, therefore... actually, you know what, let me draw a picture of this.*1077

*I think it is important to actually see what it is I just described.*1090

*So this is one plane... and let me see if I can get this write, it is always a little challenging in drawing these.*1095

*That way, and that way, and maybe that way... so here is a plane and there is some angle between them.*1102

*Well, the angle between them is going to be the angle of the normal vectors.*1113

*That is 1 normal vector, and that is say another normal vector, it is going to be the angle between the normal vectors.*1121

*That is what is happening.*1129

*So, the normal of 1 is going to equal the coefficient (2,1,-1).*1131

*Normal 2 is going to equal (1,-2,1).*1140

*Now we say that cos(θ) is going to be n1 · n2 / norm(n1) × norm(n2).*1147

*Well n1 · n2 = 2-2-1, and the norm of 1 is (4,5,6... this is going to be sqrt(6)).*1165

*It is going to = 2 - 2, -1/6 therefore θ = arcos(-1/6) and it gives us an angle of 99.6 degrees.*1183

*So these 2 planes, the angle between them is 99.6 degrees, that is it, we worked with the normal vector.*1201

*So let us see what else we can do here.*1210

*Possibly another example.*1214

*This is going to be example number 4.*1218

*Example number 4.*1225

*Now, let q = (1,1,2), and again we are just doing examples to get familiar with the notion of planes, of using the formulas, to use our intuition, our geometric intuition regarding lines and planes, and how they are.*1228

*You are going to be doing actually a fair amount of that.*1248

*So q, and let t = (1,-1,3) and n = (2,1,3).*1252

*Okay, now, here we go.*1264

*We want to find the point of intersection of the line t in the direction of n, and the plane through q perpendicular to n.*1268

*Okay, so we have these 3 points, these 3 vectors, we want to find the point of intersection of the line that passes through t and is moving in the direction of n, and the plane that passes through q and is perpendicular to n.*1313

*This is often this case, you will often be given these random points and you have to draw it out, draw it out in 2 space or 3 space so that you can see what it is that is going on, so that you can decide what equations to use, what to set equal to what.*1329

*We have this plane, here is a plane, and I will go ahead and put q here, so this is q.*1342

*Then we have a line, which passes through p, so here is some p.*1352

*It actually moves in the direction of n, so I will go ahead and put n right on this.*1362

*n is like that.*1378

*This line that passes through p, it is in the direction of n.*1381

*This is the line t that is in the direction n, and we will call this the vector n, it can actually be anywhere, I just happened to put it on top of a line.*1393

*It is perpendicular to.. passing through p in the direction of n... and the plane through q and perpendicular to n.*1405

*What we are looking for is basically this point right here, this point of intersection, this x.*1413

*Where does the line actually cross that axis?*1418

*Okay, let us see what we have got here.*1423

*Let us go ahead and find some equations here.*1427

*We know that... let x = the point p + t × vector n... so that is the equation of the line that passes through p in the direction of n.*1431

*Actually, let me specify these a little bit better.*1447

*So we say that x = p + t × n, and now we will do an equation for the plane.*1453

*The plane is going to be x - q, and we are going to have, remember we said this vector.*1470

*So the equation of a plane is x - q, I think I should use a small q here, so x - q · n = 0.*1480

*That is the equation of the plane that passes through q and is perpendicular to n.*1497

*This is the equation of the line that passes through p and is in the direction of n.*1504

*We want the point of intersection, in other words, we want the value x that makes this true.*1507

*Well, the value x that makes this true, just take this x and put it in here, that is all you have got to do.*1515

*So we will take this x, which is p + t × n, I am sorry if I keep going back and forth between upper and lower cases.*1523

*Let me go ahead and be consistent.*1534

*That is x - q · n = 0.*1540

*We took the equation of a line and the equation of a plane and we want the point of intersection.*1548

*Well the point of intersection is a point that is on both, that satisfies both.*1553

*I took the value x and I just stuck it in for this value x here, simple substitution, and now I solve this.*1558

*let me see what I do, let me go ahead and distribute the dot product here.*1565

*This becomes p · n + t × n · n... I can do that... - q · n = 0.*1569

*Now we just do it.*1588

*When I take p · n, I end up with 10.*1590

*When I take n · n, I end up with 14.*1595

*When I end up doing q · n, I end up with -9, equals 0.*1602

*So I end up with 14t = -1, make this a little bit clearer and get rid of some of these whack lines down at the bottom of the page.*1609

*Therefore t = -1/14, so the value of t that makes this true, that gives us this point x is -1/14.*1623

*Now, of course, I have to take this and put it back into here and solve this equation for the actual point x.*1635

*Let us go ahead and actually do that.*1642

*So x, again let me rewrite the equation, is p + t × n.*1646

*This is equal to (1,-1,3) + t, which we said was -1/14) which was (2,1,3).*1656

*That equals 1, -1, and 3, - 2/14 - 1/14... actually you know what, I think I am actually going to leave this plus sign as is, I will just multiply the -1/14 × 2, which is -2/14, -1/1, which is... wait, where am I, yes, which is -1 × 1, which is -1/14, and -1/14 × 3 is -3/14.*1672

*You can do it either way, if you want to pull this negative out that is fine.*1729

*So there you go, this is our answer.*1733

*You can do the addition if you want, this + that, this + that, this + that, it is going to give you some vector.*1734

*This value, this value, this value.*1740

*This value is the sum of those two, this value is the sum of those two in the middle, and this is the sum of those two.*1743

*That is it, so again, we started off with just a standard definition of a plane that passes through a given point, that is perpendicular to some vector, and of course we had what we had previously.*1749

*The equation of a line in any dimensional space is a point p + some parameter t.*1762

*T is just some real number in the direction of n, so take a point and if the vector is in this direction, it is going to be in this direction and that direction it is in a line.*1770

*So, we just use our intuition, what it is we know about geometry, what it is we know about perpendicularity to get comfortable with these straight algebraic definitions in vector form.*1780

*Okay, thank you for joining us here at educator.com*1792

*We will see you next time for a continued discussion of planes.*1794

*Take care, bye-bye.*1798

1 answer

Last reply by: Professor Hovasapian

Tue Jan 29, 2013 3:35 PM

Post by Josh Winfield on January 29, 2013

Coefficients in Example#1 are (-1,1,4). In Example#2 you used (-1,-1,4) therefore the point you found is not in the plane of Example#1