00:01
In this problem, we have been given these three vectors, a, b, and c, and we have the magnitude of this vector a, that's 1 550 meters, and we need to determine the magnitudes of vector b and c.
00:16
And we are also given that the resultant of these three vectors, that's a vector plus b vector plus c vector, they add up to give rise to a null vector.
00:28
That means zero vector.
00:29
So first we will convert vectors into their components and write in terms of icap and jcap together.
00:37
So component of vector a will be its magnitude times cost 24 .7 degree along x -axis and along positive why its component will be 1550 sine 24 .7 degree.
00:53
So here we can get a vector in terms of i cap and j cap together.
01:01
So first cost of 24 .7 times 1 550.
01:07
That comes out to be 1408 .2 icap plus we take the value of sign 24 .7 and we multiply that with 1550.
01:23
The result comes out to be 647 .7 and this is along positive y.
01:31
So this is very very a.
01:33
And now in the similar manner we project vector b and vector c.
01:38
So vector b, if we project along negative y -axis, it will be, let's say the magnitude is b.
01:45
So b cost 41 degree, and along positive axis it would be b sine 41 degree.
01:53
So let's keep a value of cost 41 and sine 41.
01:58
So it will be minus .75 b jcap and the value of sign 41 degree that would be .66 b icap.
02:18
And in the same manner we can project and write vector c in terms of the rectangular components.
02:24
So along negative x -axis it would be c cost 34 .8 degree and along positive y it would be c sine 34 .8 degree.
02:38
So here we figure out the value of cost 34 .8 and sign 34 .8 so long negative y negative x it would be minus point 8c i cap and along positive y -axis.
02:56
We take the value of sign 34 .8 that comes up to be 0 .57 times c .j cap.
03:07
So here we add these vectors.
03:10
So while we add them together, we get 1408 .2 plus 0 .66b minus 0 .8c...