WEBVTT

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So far, when I've told you about
the dot and the cross

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products, I've given you the
definition as the magnitude

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times either the cosine
or the sine of the

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angle between them.

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But what if you're not given
the vectors visually?

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And what if you're not given
the angle between them?

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How do you calculate the dot
and the cross products?

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Well, let me give you
the definition that

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I giving you already.

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So let's say I have a
dot b dot product.

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That's the magnitude of a times
the magnitude of b times

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cosine of the angle
between them.

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a cross b is equal to the
magnitude of a times the

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magnitude of b times sine of
the angle between them-- so

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the perpendicular projections
of them-- times the normal

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vector that's perpendicular
to both of them.

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The normal unit vector, and you
figure out which of the

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two perpendicular vectors
it is by using

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the right hand rule.

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But what if we don't have
the thetas; the

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angles between them?

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What if, for example, I were
to tell you that the vector

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a,-- if I were to give it to you
in engineering notation.

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In engineering notation,
you're essentially just

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breaking down the vector into
its x, y and z components.

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So let's say vector a is 5i--
i is just the unit vector in

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the x direction, minus
6j, plus 3k.

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i,j and k are just the unit
vectors of the x, y and z

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directions.

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And the 5 is how much it goes
in the x direction.

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The minus 6 is how much it
goes in the y direction.

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And the 3 is how much it goes
in the z direction.

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You could try to graph it.

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And actually, I'm trying to look
for a graphing calculator

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that'll do this, so I can show
you it all in videos to give

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you more intuition.

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But lets say this is
all you're given.

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And let's say that b-- I'm just
making these numbers up--

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let's say it's minus 2i-- and,
of course, we're working in

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three dimensions right now--
plus 7j, plus 4k.

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You could graph it.

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But obviously, if you were given
a problem, and if you

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were actually trying to model
vectors on a computer

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simulation, this is the
way you would do it.

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You'd break it up into the x, y,
and z components because of

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the add vectors.

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You just have to add the
respective components.

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But how do you multiply them
either taking the cross or the

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dot product?

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Well it actually turns out I'm
not going to prove it here but

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I'll just show you
how to do it.

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The dot product is very
easy when you have it

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given in this notation.

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And actually another way of
writing this notation,

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sometimes it's in bracket
notation.

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Sometimes they would rewrite
this as 5 minus 6, 3.

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Or it's just the magnitudes of
the x,y and z direction.

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I just want to make sure you're
comfortable with all of

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these various notations.

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You could have written b
as minus is 2, 7, 4.

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These are all the same things.

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You shouldn't get daunted if
you see one or the other.

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But anyway, so how do
I take a dot b?

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This, I think you'll find
fairly pleasant.

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All you do is you multiply the
i components, add that to the

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j components multiplied, and
then add that to the k

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components multiplied
together.

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So it would be 5 times minus 2
plus minus 6 times 7 plus 3

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times 4, so it equals minus
10 minus 42 plus 12.

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So this is minus 52 plus 12,
so it equals minus 40.

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That's it.

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It's just a number.

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And I'd actually be curious
to graph this on a three

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dimensional grapher to see
why it's minus 40.

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They must be going in
opposite directions.

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And their projections onto each
other go into opposite

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directions.

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And that's why we get
a minus number.

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The purpose of this-- I don't
want to get too much into the

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intuition-- this is just how to
calculate, but it's fairly

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straightforward.

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You just multiply the
x components.

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Add that to the y components
multiplied and add that to the

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z components multiplied.

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So whenever I am given something
in engineering or

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bracket notation and I have to
find the dot product, it's

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very, almost soothing, and
not so error prone.

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But, as you will see, taking the
cross product of these two

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vectors when given in this
notation isn't so

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straightforward.

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And I want you to keep in mind,
another way you could

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have done it, you could have
figured out the magnitude of

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each of these vectors and then
you could have used some fancy

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trigonometry to figure out the
thetas and then used this

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definition.

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But I think you appreciate the
fact that this is a much

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simpler way of doing it.

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So the dot product
is a lot of fun.

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Now let's see if we could
take the cross product.

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And once again, I'm not
going to prove it.

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I'm just going to show
you how to do it.

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In a future video, I'm sure I'll
get a request to do it

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eventually, and I'll prove it.

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But the cross product, this
is more involved.

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And I never look forward to
taking the cross product of

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two vectors in engineering
notation.

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a cross b.

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It equals.

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So this is an application
of matrices.

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So what you do is you take the
determinant-- I'll draw a big

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determinant line-- on the top
line of the determinant.

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This is really just
a way to make you

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memorize how to do it.

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It doesn't give you much
intuition, but the intuition

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is given by the actual
definition.

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How much of the vectors are
perpendicular to each other.

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Multiply those magnitudes.

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Right hand rule figures
out what direction

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you're pointing in.

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But the way to do it if you're
given engineering notation,

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you write the i, j, k unit
vectors the top row.

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i, j, k.

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Then you write the first vector
in the cross product,

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because order matters.

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So it's 5 minus 6, 3.

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Then you take the second vector
which is b, which is

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minus 2, 7, 4.

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So you take the determinant
of the 3 by 3 matrix,

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and how do I do that?

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Well that's equal to the
subdeterminant for i.

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So the subdeterminant for i, if
you get rid of this column

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and this row, the determinant
that's left over, so that's

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minus 6, 3, 7, 4 times i--
you might want to review

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determinants if you don't
remember how to do this, but

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maybe me working through it
will just jog your memory.

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And then remember, it's
plus, minus, plus.

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So then minus the subdeterminant
for j.

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What's the subdeterminant
for j?

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You cross out j's
row and columns.

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You have 5, 3, minus 2, 4.

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We just crossed j's
row and column.

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And whatever's left over, those
are the numbers in its

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subdeterminant.

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That's what I call it.

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j plus-- I want to do them all
on one line because it would

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have been a little bit
neater-- plus the

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subdeterminant for k.

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Cross out the row and
the column for k.

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We're left with 5 minus 6,
minus 2 and 7 times k.

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And now let's calculate them.

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And let me make some
space, because I've

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written this too big.

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I don't think we need
this anymore.

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So what do we get?

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Let's take this up here.

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So these 2 by 2 determinants
are pretty easy.

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This is minus 6 times
4 minus 7 times 3.

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I always make careless
mistakes here.

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Minus 24 minus 21 times i minus
5 times 4 is 20, minus

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minus 2 times 3, so minus minus
6 j, plus 5 times 7, 35

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minus minus 2 times minus 6.

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So it's minus positive 12k.

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We could simplify this, which
equals minus 24 minus 21.

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It is minus 35-- I didn't have
to put a parentheses-- i, and

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then what's 20 minus minus 6?

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Well that's 20 plus
plus 6, so 26.

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And then we have a
minus out here.

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So minus 26j.

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And that was 35 minus
12, that's 23.

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Plus 23k.

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So that's the cross product.

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And if you were to graph this in
three dimensions, you will

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see-- and this is what's
interesting-- you will see

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that vector, if my math is
correct, minus 35i, minus 26j,

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plus 23k, is perpendicular
to both of these vectors.

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I think I'll leave you there for
now, and I will see you in

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the next video.

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And hopefully, I can track down
a vector graphic program.

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Because I think it'll be fun to
both calculate the dot and

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the cross products using the
methods I just showed you and

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then to graph them.

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And to show that it
really does work.

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That this vector really is
perpendicular to both of these

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and pointing in the direction as
you would predict using the

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right hand rule.

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I'll see you in the next video.

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