Video Transcript
If the angle between vectors 𝐀 and
𝐁 is 𝜃 one and the angle between the two vectors negative 𝐀 and negative 𝐁 is 𝜃
two, what is the relationship between 𝜃 one and 𝜃 two?
To answer this question, we’ll need
to be able to relate the angle between two vectors to the vectors themselves. In general, when thinking about two
vectors and the angle between them, the easiest relationship to deal with is the dot
product. Recall that we can express the dot
product of two vectors 𝐮 and 𝐯 as the magnitude of 𝐮 times the magnitude of 𝐯
times the cos of 𝜃, the angle between them. This formula relates the vectors 𝐮
and 𝐯 to the angle between them, which is exactly what we want. Moreover, calculating the magnitude
of a vector is a fairly straightforward process. And we can directly find the dot
product between two vectors by multiplying corresponding components and then adding
up each of those products.
So if we rearrange this formula by
dividing by the magnitude of 𝐮 times the magnitude of 𝐯, we have one side of the
equation that depends only on the angle between the vectors and the other side of
the equation that can be directly calculated without knowing that angle
beforehand. To see this more clearly, if we
take the inverse cosine of both sides of this equation, we now have a calculation
that can be done with only our initial knowledge of the two vectors and that gives
us back the angle between them.
Let’s now use these formulas to
express 𝜃 one and 𝜃 two in terms of 𝐀 and 𝐁 and negative 𝐀 and negative 𝐁. 𝜃 one is the inverse cos of 𝐀 dot
𝐁 divided by the magnitude of 𝐀 times the magnitude of 𝐁. 𝜃 two is the inverse cos of
negative 𝐀 dot negative 𝐁 divided by the magnitude of negative 𝐀 times the
magnitude of negative 𝐁. To proceed, we’ll need to relate
the two quantities in the parentheses. Specifically, we’ll need to relate
𝐀 dot 𝐁 to negative 𝐀 dot negative 𝐁 and the magnitude of 𝐀 and the magnitude
of 𝐁 to the magnitude of negative 𝐀 and the magnitude of negative 𝐁. As a clue to the right direction,
we observe that negative 𝐀 is negative one times 𝐀. In other words, negative 𝐀 is a
scalar multiple of 𝐀.
With this in mind, let’s recall two
useful identities for taking dot products and magnitudes involving the scalar
multiple of a vector. If 𝑎 is a scalar, then the dot
product of the two vectors 𝑎𝐮 and 𝐯 is equal to 𝑎 times the dot product of the
two vectors 𝐮 and 𝐯. Also, the magnitude of the vector
𝑎𝐮 is equal to the absolute value of 𝑎 times the magnitude of the vector 𝐮. This last relationship makes
intuitive geometric sense. If we double a vector, we double
its length. If we have the vector, we have its
length.
We use the absolute value of 𝑎
because, by definition, the negative of a vector is a vector of the same length but
opposite direction. But when we calculate the magnitude
of a vector, all we care about is its length, not its direction. So we see immediately that the
magnitude of negative 𝐮 is the absolute value of negative one times the magnitude
of 𝐮. But the absolute value of negative
one is just one. So this is just equal to the
magnitude of 𝐮. Also, our relationship for the dot
product actually applies to a scalar multiple of both vectors. That is, 𝑎𝐮 dot 𝑏𝐯 is 𝑎𝑏
times 𝐮 dot 𝐯.
So negative 𝐮 dot negative 𝐯 is
negative one, the scalar from negative 𝐮, times negative one, the scalar from
negative 𝐯, times 𝐮 dot 𝐯. But negative one times negative one
is just one. So this is equal to 𝐮 dot 𝐯. We now have exactly the
relationships that we need. Negative 𝐮 dot negative 𝐯 is
equal to 𝐮 dot 𝐯. So negative 𝐀 dot negative 𝐁 is
equal to 𝐀 dot 𝐁. Also, the magnitude of negative 𝐮
is equal to the magnitude of 𝐮. So the magnitude of negative 𝐀 is
the magnitude of 𝐀, and the magnitude of negative 𝐁 is the magnitude of 𝐁.
But combining these results, the
two expressions in the parentheses are exactly equal. So 𝜃 two is also equal to the
inverse cos of 𝐀 dot 𝐁 divided by the magnitude of 𝐀 times the magnitude of
𝐁. But this is the same expression
that 𝜃 one is equal to. So 𝜃 one and 𝜃 two are equal to
the same expression, which means they must be equal to each other. So the relationship between 𝜃 one
and 𝜃 two is that they are in fact exactly the same.
Now, we arrived at this result
algebraically, but we actually could’ve arrived at the exact same result using
purely geometric methods. Here, we’ve drawn a picture
representing 𝐀 and 𝐁 with the angle between them. Since we don’t know the particular
identity of 𝐀, 𝐁, or 𝜃 one, this diagram is just meant to show qualitative
relationships. So now we want to include negative
𝐀, negative 𝐁, and 𝜃 two onto this diagram. The qualitative relationship
between 𝐀 and negative 𝐀 is that negative 𝐀 has the exact same length but points
in exactly the opposite direction. Similarly, negative 𝐁 points in
exactly the opposite direction from 𝐁 with exactly the same length. And here’s 𝜃 two, the angle
between negative 𝐀 and negative 𝐁.
Now, here is our key
observation. All scalar multiples of a vector,
including the negative ones, lie on the same straight line. Now, since this is true for all
vectors, it’s not only true for 𝐀 and negative 𝐀; it’s also true for 𝐁 and
negative 𝐁. But now look at what we have. We have two intersecting straight
lines and 𝜃 one and 𝜃 two form a pair of opposite angles. At this point, we either recall or
derive the basic fact from Euclidean geometry that when two straight lines
intersect, opposite angles are congruent. So by purely geometric arguments,
we’re led to the conclusion that 𝜃 two is equal to 𝜃 one.
