WEBVTT
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This is Giancoli Answers
with Mr. Dychko.
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The way I like to think about
subtracting vectors is
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adding the opposite
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because then you can use the head
to tail method of adding vectors.
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So we have vector *A* which
is redrawn here
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and on the head of vector *A*,
you put the tail of
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the opposite of *C* so the opposite
of *C* is this vector
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in the other direction, straight up
in this case
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so we have the green one shown here.
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And then the resultant goes from
the very beginning to the very end
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so that's drawn in red here.
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So we'll find the length of the resultant
by finding its components first
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and then use Pythagoras and then
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use inverse tangent to get its angle
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and that's the usual routine for vectors.
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So the resultant *x*-component is
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the *x*-component of vector *A* because
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vector negative *C* has no *x*-component
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so we have 44 times *cos* 28
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so length of vector *A* times *cos*
of its angle here
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gives the *x*-component of
the resultant
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and that's 38.8497
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and then the *y*-component of
the resultant will be
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the *y*-component of *A*
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that's this part here
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and that's found by taking the hypotenuse
or length of *A*—44—
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and multiply by *sin* of 28;
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*sin* is the opposite divided by hypotenuse
so when you multiply by the hypotenuse
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and then times by *sin* 28,
you get the opposite
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and then add to that 31 and we get 51.657;
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I don't know, just in case, it's helpful
to explain that a bit better
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we have *sin* 28, by definition, is opposite
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which in this case is *y*-component of *A*
divided by hypotenuse which is *A*
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and then multiply both sides by *A* here
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and then *A*'s cancel and then
switch the sides around,
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you get *A y* is *A* times *sin* 28
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and that's what we have put here
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and then we get 51.657 there
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and their resultant magnitude is the square
root of the sum of the squares
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so square both of these components
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add them, take the square root
and you get 64.6 units
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and then the angle *Θ* for the resultant is
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the opposite which is this length here—
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oops, it's going upwards though—
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so that's resultant *y*-direction
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that length divided by its *x*-component
and so we get
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inverse tangent of 51.657 divided by
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divided by the *x*-component of 38.8497
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and that gives 53.1 degrees
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and from the drawing, we can see that that's
above the positive *x*-axis.