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Lecture Comments (20)

1 answer

Last reply by: Professor Hovasapian
Wed Sep 2, 2015 11:54 PM

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 qq´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

More on Planes

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.
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 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 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 ).

*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

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 first1475

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