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PxCxlsl_YwY	Q, and R, and I'm wondering, what is this angle here?
PxCxlsl_YwY	OK, so, of course, one solution is to build a
PxCxlsl_YwY	model and then go and measure the angle.
PxCxlsl_YwY	But, we can do better than that. We can just find the angle
PxCxlsl_YwY	using dot product. So, how would we do that?
PxCxlsl_YwY	Well, so, if we look at this formula, we see,
PxCxlsl_YwY	so, let's say that we want to find the angle here.
PxCxlsl_YwY	Well, let's look at the formula for PQdotPR.
PxCxlsl_YwY	Well, we said it should be length PQ times length PR times
PxCxlsl_YwY	the cosine of the angle, OK?
PxCxlsl_YwY	Now, what do we know, and what do we not know?
PxCxlsl_YwY	Well, certainly at this point we don't know the cosine of the
PxCxlsl_YwY	angle. That's what we would like to
PxCxlsl_YwY	find. The lengths,
PxCxlsl_YwY	certainly we can compute. We know how to find these
PxCxlsl_YwY	lengths. And, this dot product we know
PxCxlsl_YwY	how to compute because we have an easy formula here.
PxCxlsl_YwY	OK, so we can compute everything else and then find
PxCxlsl_YwY	theta. So, I'll tell you what we will
PxCxlsl_YwY	do is we will find theta -- -- in this way.
PxCxlsl_YwY	We'll take the dot product of PQ with PR, and then we'll
PxCxlsl_YwY	divide by the lengths.
PxCxlsl_YwY	OK, so let's see. So, we said cosine theta is
PxCxlsl_YwY	PQdotPR over length PQ length PR.
PxCxlsl_YwY	So, let's try to figure out what this vector,
PxCxlsl_YwY	PQ, well, to go from P to Q,
PxCxlsl_YwY	I should go minus one unit along the x direction plus one
PxCxlsl_YwY	unit along the y direction. And, I'm not moving in the z
PxCxlsl_YwY	direction. So, to go from P to Q,
PxCxlsl_YwY	I have to move by *amplt;-1,1,0amp*gt;.
PxCxlsl_YwY	To go from P to R, I go -1 along the x axis and 2
PxCxlsl_YwY	along the z axis. So, PR, I claim, is this.
PxCxlsl_YwY	OK, then, the lengths of these vectors, well,(-1)^2 (1)^2
PxCxlsl_YwY	(0)^2, square root, and then same thing with the
PxCxlsl_YwY	other one. OK, so, the denominator will
PxCxlsl_YwY	become the square root of 2, and there's a square root of 5.
PxCxlsl_YwY	What about the numerator? Well, so, remember,
PxCxlsl_YwY	to do the dot product, we multiply this by this,
PxCxlsl_YwY	and that by that, that by that.
PxCxlsl_YwY	And, we add. Minus 1 times minus 1 makes 1
PxCxlsl_YwY	plus 1 times 0, that's 0.
PxCxlsl_YwY	Zero times 2 is 0 again. So, we will get 1 over square
PxCxlsl_YwY	root of 10. That's the cosine of the angle.
PxCxlsl_YwY	And, of course if we want the actual angle,
PxCxlsl_YwY	well, we have to take a calculator, find the inverse
PxCxlsl_YwY	cosine, and you'll find it's about 71.5°.
PxCxlsl_YwY	Actually, we'll be using mostly radians, but for today,
PxCxlsl_YwY	that's certainly more speaking. OK, any questions about that?
PxCxlsl_YwY	No? OK, so in particular,
PxCxlsl_YwY	I should point out one thing that's really neat about the
PxCxlsl_YwY	answer. I mean, we got this number.
PxCxlsl_YwY	We don't really know what it means exactly because it mixes
PxCxlsl_YwY	together the lengths and the angle.
PxCxlsl_YwY	But, one thing that's interesting here,
PxCxlsl_YwY	it's the sign of the answer, the fact that we got a positive
PxCxlsl_YwY	number. So, if you think about it,
PxCxlsl_YwY	the lengths are always positive.
PxCxlsl_YwY	So, the sign of a dot product is the same as a sign of cosine
PxCxlsl_YwY	theta. So, in fact,
PxCxlsl_YwY	the sign of AdotB is going to be positive if the angle is less
PxCxlsl_YwY	than 90°. So, that means geometrically,
PxCxlsl_YwY	my two vectors are going more or less in the same direction.
PxCxlsl_YwY	They make an acute angle. It's going to be zero if the
PxCxlsl_YwY	angle is exactly 90°, OK, because that's when the
PxCxlsl_YwY	cosine will be zero. And, it will be negative if the
PxCxlsl_YwY	angle is more than 90°. So, that means they go,
PxCxlsl_YwY	however, in opposite directions.
PxCxlsl_YwY	So, that's basically one way to think about what dot product
PxCxlsl_YwY	measures. It measures how much the two
PxCxlsl_YwY	vectors are going along each other.
PxCxlsl_YwY	OK, and that actually leads us to the next application.
PxCxlsl_YwY	So, let's see, did I have a number one there?
PxCxlsl_YwY	Yes. So, if I had a number one,
PxCxlsl_YwY	I must have number two. The second application is to
PxCxlsl_YwY	detect orthogonality. It's to figure out when two
PxCxlsl_YwY	things are perpendicular. OK, so orthogonality is just a
PxCxlsl_YwY	complicated word from Greek to say things are perpendicular.
PxCxlsl_YwY	So, let's just take an example. Let's say I give you the
PxCxlsl_YwY	equation x 2y 3z = 0. OK, so that defines a certain
PxCxlsl_YwY	set of points in space, and what do you think the set
PxCxlsl_YwY	of solutions look like if I give you this equation?
PxCxlsl_YwY	So far I see one, two, three answers,
PxCxlsl_YwY	OK. So, I see various competing
PxCxlsl_YwY	answers, but, yeah, I see a lot of people
PxCxlsl_YwY	voting for answer number four. I see also some I don't knows,
PxCxlsl_YwY	and some other things. But, the majority vote seems to
PxCxlsl_YwY	be a plane. And, indeed that's the correct
PxCxlsl_YwY	answer. So, how do we see that it's a
PxCxlsl_YwY	plane?
PxCxlsl_YwY	So, I should say, this is the equation of a
PxCxlsl_YwY	plane. So, there's many ways to see
PxCxlsl_YwY	that, and I'm not going to give you all of them.
PxCxlsl_YwY	But, here's one way to think about it.
PxCxlsl_YwY	So, let's think geometrically about how to express this
PxCxlsl_YwY	condition in terms of vectors. So, let's take the origin O,
PxCxlsl_YwY	by convention is the point (0,0,0).
PxCxlsl_YwY	And, let's take a point, P, that will satisfy this
PxCxlsl_YwY	equation on it, so, at coordinates x,
PxCxlsl_YwY	y, z. So, what does this condition
PxCxlsl_YwY	here mean? Well, it means the following

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