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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, among
other things, you've

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been learning about
computing components

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of one vector in the
direction of another vector.

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So I have a
straightforward problem

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about that for you here.

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So we've got two vectors.

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The vector 2i minus 2j plus k.

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And we've got the
vector i plus j plus k.

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And so what I'd like
you to do for me

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is to compute the component
of this first vector

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in the direction of
the second vector.

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So why don't you pause
the video, take some time

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to work that out, come back,
and we can work on it together.

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Welcome back.

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So hopefully you had some
luck with this problem.

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Now this problem is
pretty straightforward.

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Really all you have
to do is remember what

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the definition of component is.

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And after that,
it's smooth sailing.

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So in particular, the
component of one vector

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in the direction of
another is the length

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of what you get when you project
this vector onto that one,

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well, plus a sign, right?

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So if the projection is
in the same direction,

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then it's positive, or if it's
in the opposite direction,

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it's negative.

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So let me draw a picture
of what I mean by that.

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So if you have a vector v and
you have another vector w,

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then the projection
of v onto w is

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what you get if you make a,
drop a perpendicular line there,

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and then it's just
this vector here.

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So that's the projection.

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And then the component is the
length of that projection.

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Again, with the
sign if necessary.

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And so we can see
since this is going

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to be a right
triangle here, we can

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see that this vector
has length that's

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just given by the length of
v-- so the length-- rather,

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the component of
v in direction w

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is-- so it's just the
length of v, right-- that's

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the length of the hypotenuse--
times the cosine of the angle

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between them, so it's
times cosine of theta.

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Now this length of v
times cosine theta,

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this should remind
you of something.

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This looks very much
like this formula we

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had for the dot product, right?

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So another way of
writing this is

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to say that this is
equal to v dot w--

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so v dot w is the length
of v times the length of w

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times the cosine of the angle.

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And so here we just
have the length of v

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times the cosine of the angle.

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So we have to divide this
through by the length of s.

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So this is what
the component is,

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and this is the
simple formula for it,

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if you're given v and w
in some easy-to-use form.

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And indeed in this
problem, we're

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given v and w just in their
nice coordinate forms.

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So we're given that
our vector v that we

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want the component of
is 2i minus 2j plus k.

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And the direction w that we're
looking at is i plus j plus k.

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So in our case, we just have to
compute these expressions v dot

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w and the length of w in order
to put them into this formula,

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and then we'll be done.

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So in our case, v dot w--
well, that's straightforward

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because we're given v
and w in coordinates--

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so this is just 2 times 1 plus
minus 2 times 1 plus 1 times 1.

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So that's, OK, 2
minus 2 plus 1 is 1.

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And the length of
w-- well, you know,

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it's just your usual
length formula--

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is 1 squared plus 1 squared plus
1 squared, the whole thing is

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square rooted, which is equal
to the square root of 3.

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So we've got v dot w and
we have the length of w,

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and so then we just need to put
them right into this formula.

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That the component of
v in the direction of w

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is given by this expression.

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So the component of-- I'm
not going to write it out

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with i's, j's, and k's,
I'm going to write 2,

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minus 2, 1-- in
direction [1, 1,  1]

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is, well we just have to divide
the dot product by the length

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of the direction vector.

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So that's 1 divided by
the square root of 3.

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So that's that, I'll end there.