00:01
Hi there, so for this problem, we are told for par a of this problem.
00:08
To find the vectors that are shown below in terms of unit vectors i and j.
00:14
Now, to do this, we need to obtain information from the image that is given, this one.
00:22
So, as you can see, the vector a, it has a magnitude that is given, and that magnitude is 3 .6 meters and makes an angle of 70 degrees with respect to the positive x axis.
00:38
So that vector is just simply.
00:41
The magnitude times.
00:45
Since the angle is given with respect to the horizontal axis, then the x component is given by the cosine of the angle.
00:53
So that will be the cosine of 70 degrees.
00:55
And this in the y in the x direction, and this plus the sign of 70 degrees, this will give us the y component of this vector in the j direction.
01:12
So we know that the magnitude is given, that magnitude is 3 .6 meters.
01:20
So using our calculator, we obtain that the vector a is equal to 1 .23 meters in the eds direction and 3 .38 in the j direction of course let me just put this upside so all of this is in meters now for the better b we have something similar however the edge company the angle is given with respect to the negative x a axis.
02:07
And as you can see, the vector is in the third quadrant.
02:12
So both components, the x component and the y component should be negative.
02:17
So we will have 2 .4 meters for its magnitude times minus the cosine of the angle that is 30 degrees in the in the x direction.
02:28
And this minus the sign of 30 degrees in the j direction.
02:36
Now using our calculator we obtain that the vector b is equal to, so the x component is minus 2 .17 meters and the y component is 1 .2 meters.
03:13
Out, so that will be, and this in the minus 1 .2 in the j direction.
03:22
So that's a solution for part a of this problem.
03:26
This is the vector b, this is the vector a.
03:30
Now, for part b of this problem, we are told that using the vectors that you found, we need to find the vector c.
03:42
And the vector c is equal to 3 times the vector a minus 4 times the vector b.
04:00
So what we need to do is to multiply all of the components for the vector a times 3 and all of the components of the vector in b by minus 4.
04:11
And we just add together these to vectors.
04:13
So we will have that the vector c is equal to.
04:20
What we can do in this case is to multiply component by component.
04:24
So we start with the x component of the vector a.
04:26
So that will be 1 .23.
04:30
This in the, well, we know that this is in the x direction.
04:35
And this minus, well, that will convert this into plus because we know that for the vector b, both components are negative.
04:44
So that will be 4 times 2 .17.
04:53
And this in the x direction.
04:57
And this plus 3 times 3.
05:01
The y component of the vector a, which is 3 .38, and this plus 4 times 1 .2.
05:17
Now, simplifying this expression, we will find that this is equal to, so let's start with the ed's component for this betr.
05:28
That is equal to 12 .37 in the ex -derption.
05:36
And for the other component, we have about of 14 .94 in the y direction and this in meters.
05:54
So that's a solution for par b of this problem.
05:57
Now for par c, we are asked about to find the magnitude and direction of the vector c.
06:09
So the magnitude of a vector is just the square root of the square of each component, the sum of the square of each component, just like this.
06:26
So using our calculator, we obtain a value of 19 .360 meters.
06:46
Oh, 396 meters.
06:52
So that's the magnitude.
06:55
And the direction is given by, the tangent of minus 1 of the y component, and the y component in this case is 14 .94 divided by the x component that is 12 .37.
07:18
So using our calculator, we obtain a value of 50 .4 degrees, and this is with respect to the positive x axis.
07:42
This is the direction.
07:45
Now, for part d of this problem, we are asked about define this color product between the vectors a and b...