00:01
In this exercise, we have these three vectors shown in the screen, and in question a, we have to find the magnitude in the direction of vector r, which is equal to the sum of the three vectors a, b, and c.
00:18
So in order to find this, first, i'm going to find a, b, and c in terms of their components.
00:24
Notice that a only has a component in the negative y direction, so its y component is minus 8 meters.
00:36
So we have a equals minus 8 jm, j is a unvector in the y direction.
00:44
Notice that both the x and y components of b point in the positive direction.
00:52
And i'm going to define right triangle.
00:57
Such that the legs of the triangle are b y and b x and the hypotenuse is just the magnitude of b so b we have an x component of b times the sign of 30 and a y component of b times a cosign of 30 so b is equal to 15 times a sign of 30 that's just one half plus 15 times the cosine of 30 that's a square root of 3 over 2 so b is equal to 7 .5 i plus 13 j this is vector b and finally vector c notice that vector c has negative x and y components so again we can define a right triangle here.
02:09
So i said the legs are the magnitude of cy and the magnitude of cx.
02:15
So the cx component of c is minus the magnitude of c times the cosine of 25 and the y component is minus c times the sign of 25.
02:34
So we get that c is equal to minus 12 times a core sign of 25 minus 12 times the sign of 25 j so you get the c is equal to minus 10 .9 i minus 5 .07 j and r is just a sum of the three vectors so it's a plus b plus c so we have to sum them component by component in the x component a x is zero bx is 7 .5 and cx is minus 10 .9.
03:17
In the y component, we have that ay is minus 8.
03:23
By is 13 and cy is minus 5 .07.
03:30
So r is equal to minus 3 .4i minus 0 .07j.
03:40
And we can find the angle.
03:43
You notice that r is pointing in the negative x in line direction.
03:48
So it's pointing in a third quadrant like this.
03:55
And we want to find this angle here.
04:02
The angle that it makes with a negative x direction.
04:06
So notice that theta is just the arc tangent of the magnitude of our width.
04:16
Y the y component of r divided by the magnitude of r x which is the x component of r so this is the arc tangent of 0 .07 divided by 3 .4 and this is equal to 1 .2 degrees and then we can we have to show it on the diagram that we have to show this vector on the diagram so basically i'm going to take this diagram here i'm going to just erase some things to get the diagram a little tighter here.
05:01
Okay, so now we can make the vector addition.
05:08
First, i'm going to add a to b, and then i'm going to add the result to c.
05:14
So notice that this here is the sum of a and b, this vector in green.
05:27
I just use the parallelogram rule, and then we have to use the parallelogram rule in order to sum this vector with c.
05:36
So we have this vector here.
05:47
Notice that it's very close to the x -axis as it should be since the y coordinate is just minus 0 .07 and it points to the third quadrant as it should.
06:01
This is r.
06:03
Then in question b we have to calculate the vector s which is equal to c minus a minus b again we have to do it component by component so the x component of c is minus 10 .9 the x component of a 0 and the x component of b is 7 .5 then the y component of c is 5 .07 the y component of b is 3 .07 the y component of b is 3 .5 .7 the y component of b is 3 .5 and the y component of x is minus 8.
06:45
So s is equal to minus 18 .4i plus 10 .07 j...