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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

Planes

Find a vector parallel to the line of intersection of the planes 3x + 2y − 4z = 11 and x − 4y + 2z = 3.
  • 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
    3x + 2y − 4 = 11
    x − 4y + 2 = 3
    or
    3x + 2y = 15
    x − 4y = 1
    solving the system of equations yields the point ( [31/7],[6/7],1 ).
  • For z = − 1 we have
    3x + 2y + 4 = 11
    x − 4y − 2 = 3
    or
    3x + 2y = 7
    x − 4y = 5
    solving the system of equations yields the point ( [41/21], − [4/7], − 1 ).
Our vector parallel to the line of intersection of the planes is therefore ( [31/7],[6/7],1 ) − ( [41/21], − [4/7], − 1 ) = ( [52/21],[10/7],2 ).
Find a vector parallel to the line of intersection of the planes [1/4]x − [3/4]y + z = − 4 and [1/3]x − [2/3]y − z = − 2.
  • 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
    [1/4]x − [3/4]y + 1 = − 4
    [1/3]x − [2/3]y − 1 = − 2
    or
    x − 3y + 4 = − 16
    x − 2y − 3 = − 6
    or
    x − 3y = − 20
    x − 2y = − 3
    solving the system of equations yields the point ( 31,17,1 ).
  • For z = − 1 we have
    [1/4]x − [3/4]y − 1 = − 4
    [1/3]x − [2/3]y + 1 = − 2
    or
    x − 3y − 4 = − 16
    x − 2y + 3 = − 6
    or
    x − 3y = − 12
    x − 2y = − 9
    solving the system of equations yields the point ( − 3,3, − 1 ).
Our vector parallel to the line of intersection of the planes is therefore ( 31,17,1 ) − ( − 3,3, − 1 ) = ( 34,14,2 ).
Find the distance from q = (11, − 2,7) and the plane containing parallel vectors ( [1/2],[1/4],0 ) and ( 1,[1/2],0 ).
  • 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.
Hence the distance from q to the xy - plane is 7. Note that we could have figured this out from the start, as the only distance from q is the z - coordinate 7. See image:
Find the distance from q = (1,1,1,0) and the hyperplane perpendicular to ( − 4,1, − 2,3) containing the point (0,1,0,1).
  • 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} )].
Hence the distance from q to the hyperplane is [1/(√{30} )](4 + 0 + 2 + 3) = [9/(√{30} )].
Find a vector parallel to the line of intersection of the planes x + y − z = 6 and − x + y + z = 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
    x + y − 1 = 6
    − x + y + 1 = 5
    or
    x + y = 7
    − x + y = 4
    solving the system of equations yields the point ( [3/2],[11/2],1 ).
  • For z = − 1 we have
    x + y + 1 = 6
    − x + y − 1 = 5
    or
    x + y = 5
    − x + y = 6
    solving the system of equations yields the point ( − [1/2],[11/2], − 1 ).
Our vector parallel to the line of intersection of the planes is therefore ( [3/2],[11/2],1 ) − ( − [1/2],[11/2], − 1 ) = ( 2,0,2 ).
Find the distance from q = (3,7,0) to the plane perpendicular to N = ( − 1,1,1) and containing the point p = ( − 1,5,2).
  • 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 )].
Hence the distance from q and the plane is [2/(√3 )]( − 2, − 1,1) ×( − 1,1,1) = [2/(√3 )].
Find the distance from q = (0,0,1) to the plane [1/2]x + 4y − [1/4]z = 1.
  • 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) ||
Hence the distance from q and the plane is [1/(3√{29} )]( − 4,1,3) ×(2,16, − 1) = [5/(3√{29} )].
Find the distance from q = ( [2/(√5 )],0,[1/(√5 )] ) to the plane containing points (1,3,4), ( − 2,1,2) and ( − 1,1,1).
  • 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
    2a + 2b + 3c = 0
    − 3a + 1c = 0
    , if a = 1 then c = 3 and b = − [11/2].
  • 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]} )]
Hence the distance from q and the plane is [4/(√{161} )]( [( − 2 − √5 )/(√5 )],1,[( − 1 + √5 )/(√5 )] ) ×( 1, − [11/2],3 ) = [4/(√{161} )]( [( − 10 − 7√5 )/(2√5 )] ) = [( − 20 − 14√5 )/(√{805} )]
Find the standard equation of a plane containing the points (5,3, − 1), (2, − 7,1) and (4,9, − 6).
  • 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
    3a + 10b − 2c = 0
    2a + 16b − 7c = 0
    , letting c = 1 yields
    3a + 10b = 2
    2a + 16b = 7
    . Note that choosing c = 1 is arbitrary (except for c = 0).
  • 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.
Simplifying yields − 38x + 17y + 28z = − 167.
Find the standard equation of a plane containing the points ( [1/(√2 )], − [1/(√2 )],0 ), ( [1/(√3 )],[1/(√3 )], − [1/(√3 )] ) and (1,0,0).
  • 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
    ( 1 − √2 )a + − b = 0
    ( 1 − √3 )a + b − c = 0
    , letting b = 1 yields
    ( 1 − √2 )a = 1
    ( 1 − √3 )a − c = − 1
    . Note that choosing b = 1 is arbitrary (except for b = 0).
  • 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.
Simplifying yields x + ( 1 − √2 )y + ( 2 − √2 − √3 )z = 1.

*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

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.com1792

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

Take care, bye-bye.1798