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# Calculating dot and cross products with unit vector notation

Calculating the dot and cross products when vectors are presented in their x, y, and z (or i,j, and k) components. Created by Sal Khan.

Video transcript

So far, when I've told you about
the dot and the cross products, I've given you the
definition as the magnitude times either the cosine
or the sine of the angle between them. But what if you're not given
the vectors visually? And what if you're not given
the angle between them? How do you calculate the dot
and the cross products? Well, let me give you
the definition that I giving you already. So let's say I have a
dot b dot product. That's the magnitude of a times
the magnitude of b times cosine of the angle
between them. a cross b is equal to the
magnitude of a times the magnitude of b times sine of
the angle between them-- so the perpendicular projections
of them-- times the normal vector that's perpendicular
to both of them. The normal unit vector, and you
figure out which of the two perpendicular vectors
it is by using the right hand rule. But what if we don't have
the thetas; the angles between them? What if, for example, I were
to tell you that the vector a,-- if I were to give it to you
in engineering notation. In engineering notation,
you're essentially just breaking down the vector into
its x, y and z components. So let's say vector a is 5i--
i is just the unit vector in the x direction, minus
6j, plus 3k. i,j and k are just the unit
vectors of the x, y and z directions. And the 5 is how much it goes
in the x direction. The minus 6 is how much it
goes in the y direction. And the 3 is how much it goes
in the z direction. You could try to graph it. And actually, I'm trying to look
for a graphing calculator that'll do this, so I can show
you it all in videos to give you more intuition. But lets say this is
all you're given. And let's say that b-- I'm just
making these numbers up-- let's say it's minus 2i-- and,
of course, we're working in three dimensions right now--
plus 7j, plus 4k. You could graph it. But obviously, if you were given
a problem, and if you were actually trying to model
vectors on a computer simulation, this is the
way you would do it. You'd break it up into the x, y,
and z components because of the add vectors. You just have to add the
respective components. But how do you multiply them
either taking the cross or the dot product? Well it actually turns out I'm
not going to prove it here but I'll just show you
how to do it. The dot product is very
easy when you have it given in this notation. And actually another way of
writing this notation, sometimes it's in bracket
notation. Sometimes they would rewrite
this as 5 minus 6, 3. Or it's just the magnitudes of
the x,y and z direction. I just want to make sure you're
comfortable with all of these various notations. You could have written b
as minus is 2, 7, 4. These are all the same things. You shouldn't get daunted if
you see one or the other. But anyway, so how do
I take a dot b? This, I think you'll find
fairly pleasant. All you do is you multiply the
i components, add that to the j components multiplied, and
then add that to the k components multiplied
together. So it would be 5 times minus 2
plus minus 6 times 7 plus 3 times 4, so it equals minus
10 minus 42 plus 12. So this is minus 52 plus 12,
so it equals minus 40. That's it. It's just a number. And I'd actually be curious
to graph this on a three dimensional grapher to see
why it's minus 40. They must be going in
opposite directions. And their projections onto each
other go into opposite directions. And that's why we get
a minus number. The purpose of this-- I don't
want to get too much into the intuition-- this is just how to
calculate, but it's fairly straightforward. You just multiply the
x components. Add that to the y components
multiplied and add that to the z components multiplied. So whenever I am given something
in engineering or bracket notation and I have to
find the dot product, it's very, almost soothing, and
not so error prone. But, as you will see, taking the
cross product of these two vectors when given in this
notation isn't so straightforward. And I want you to keep in mind,
another way you could have done it, you could have
figured out the magnitude of each of these vectors and then
you could have used some fancy trigonometry to figure out the
thetas and then used this definition. But I think you appreciate the
fact that this is a much simpler way of doing it. So the dot product
is a lot of fun. Now let's see if we could
take the cross product. And once again, I'm not
going to prove it. I'm just going to show
you how to do it. In a future video, I'm sure I'll
get a request to do it eventually, and I'll prove it. But the cross product, this
is more involved. And I never look forward to
taking the cross product of two vectors in engineering
notation. a cross b. It equals. So this is an application
of matrices. So what you do is you take the
determinant-- I'll draw a big determinant line-- on the top
line of the determinant. This is really just
a way to make you memorize how to do it. It doesn't give you much
intuition, but the intuition is given by the actual
definition. How much of the vectors are
perpendicular to each other. Multiply those magnitudes. Right hand rule figures
out what direction you're pointing in. But the way to do it if you're
given engineering notation, you write the i, j, k unit
vectors the top row. i, j, k. Then you write the first vector
in the cross product, because order matters. So it's 5 minus 6, 3. Then you take the second vector
which is b, which is minus 2, 7, 4. So you take the determinant
of the 3 by 3 matrix, and how do I do that? Well that's equal to the
subdeterminant for i. So the subdeterminant for i, if
you get rid of this column and this row, the determinant
that's left over, so that's minus 6, 3, 7, 4 times i--
you might want to review determinants if you don't
remember how to do this, but maybe me working through it
will just jog your memory. And then remember, it's
plus, minus, plus. So then minus the subdeterminant
for j. What's the subdeterminant
for j? You cross out j's
row and columns. You have 5, 3, minus 2, 4. We just crossed j's
row and column. And whatever's left over, those
are the numbers in its subdeterminant. That's what I call it. j plus-- I want to do them all
on one line because it would have been a little bit
neater-- plus the subdeterminant for k. Cross out the row and
the column for k. We're left with 5 minus 6,
minus 2 and 7 times k. And now let's calculate them. And let me make some
space, because I've written this too big. I don't think we need
this anymore. So what do we get? Let's take this up here. So these 2 by 2 determinants
are pretty easy. This is minus 6 times
4 minus 7 times 3. I always make careless
mistakes here. Minus 24 minus 21 times i minus
5 times 4 is 20, minus minus 2 times 3, so minus minus
6 j, plus 5 times 7, 35 minus minus 2 times minus 6. So it's minus positive 12k. We could simplify this, which
equals minus 24 minus 21. It is minus 35-- I didn't have
to put a parentheses-- i, and then what's 20 minus minus 6? Well that's 20 plus
plus 6, so 26. And then we have a
minus out here. So minus 26j. And that was 35 minus
12, that's 23. Plus 23k. So that's the cross product. And if you were to graph this in
three dimensions, you will see-- and this is what's
interesting-- you will see that vector, if my math is
correct, minus 35i, minus 26j, plus 23k, is perpendicular
to both of these vectors. I think I'll leave you there for
now, and I will see you in the next video. And hopefully, I can track down
a vector graphic program. Because I think it'll be fun to
both calculate the dot and the cross products using the
methods I just showed you and then to graph them. And to show that it
really does work. That this vector really is
perpendicular to both of these and pointing in the direction as
you would predict using the right hand rule. I'll see you in the next video.