00:01
Hi there, so for this problem, we are told that given the pair of bettors, so the vector a, it has an x component that is equal to 6 in the x direction and minus 2 in the y direction.
00:17
Now we have also the vector b, which is equal to minus 4 in the x direction and 7 in the y direction.
00:29
So with that said, we are asked to use the definition of a scalar product to determine the following.
00:36
So for part a, we are asked to obtain the scalar product.
00:44
Now, the scalar product between two vectors is just simply the product between the each component.
00:56
So component by component.
00:58
So in this case, it's just simply the x component.
01:02
Of the vector a with the x component of the vector b, so that will be 6 times minus 4.
01:09
And this plus the white component of the vector a with the white component of the vector b.
01:17
So that will be minus 2 times 7.
01:21
So this will give us 6 times 4, so that will be minus 24.
01:26
And this minus 14.
01:30
So from this, we obtain an scholar product, of minus 38.
01:39
So that's a solution for par a of this problem.
01:43
Now for par b, we are asked about the angle between the vectors.
01:50
And for that, we use the definition that the dot product between two vectors is equal to the magnitude of the first vector times the magnitude of the second vector, cosine of the angle between these vectors.
02:04
So if we want to solve for the first for the cosine of the angle, that will be dot product divided by the magnitude of the vector a times the magnitude of the vector b.
02:17
So the first thing that we need to determine is the magnitude of the vector a is just simply the square root of the sum of each component to the square.
02:26
So that will be 6 to the square, plus minus 2 to the square.
02:34
And we take the square root of that.
02:37
So if we use our calculator, we obtain a value of 6 point.
02:44
Well, let me just express this as the square root of 40.
02:50
And the magnitude of the vector b that will be the square root of minus 4 to the square plus 7 to the square.
03:03
So we will have in here that this is equal to the square root of 65.
03:12
So with that set, if we substitute the values in here, because we already know the value for that product, so that will be minus 38, divided by the square root of 40 times the square root of 65.
03:30
So now to obtain the angle theta, we apply the consign of minus 1 to both sides of this, expression, so we will obtain that this corresponds to the angle theta...