00:01
In this problem, we have been given these three vectors that are in this plane of xy.
00:06
Vector a having magnitude 100 newton, b having magnitude 80 newtons, and c vector having magnitude 40 newtons, and the respective angles are shown.
00:17
So first we need to figure out the components of each vector.
00:22
So we will use the idea that with respect to any vector, whatever direction it makes an angle theta, its component will be a cost theta and perpendicular to that its component will be a sine theta that's what we can get using the normal idea of the trigonometry so from this vector a the component will be one along the positive x direction so that will be 100 cost 30 degree and its component along positive y direction that will be 100 sine 30 degree so we can definitely conclude that for vector a let's write here the x component and here we will write the y component.
01:07
So for vector a, the x component will be 100 cost 30 degree.
01:14
So when we find their value, we get the component as 86 .6 newtons and the y component will be 50 newtons.
01:23
Similarly, vector b, the x component, here we can see to resolve vector b in its rectangular components it will be 80 cost 30 degree along positive y because it's making angle theta with y axis and 80 sign 30 degree will be along the negative x so along negative x direction it will be minus 40 newton and along positive y direction it will be 292 let's just take this 80 times plus 30 degree that will be 60 0 .28 newtons approximately we take this as 69 .3 newtons plus sign indicates that they are along the positive x or y direction and negative indicates they are along negative accent y direction so vector c will have one of the component along x and the other along negative y so both along negative x and negative y respectively so along negative x direction the component will be 40 cost 53 degree and along negative y direction it will be 40 sign 53 degree so we just figure out the value of cost 53 and we multiply that with 40 to get the x component as minus 24 .1 newton and the y component can be obtained by taking sign 53 and multiplying that with 40 and that comes out to be 31 .9 newton of course this will also be negative so we have figured out the x and y components of each of these vectors and now we can represent these vectors in terms of i cap and j cap so vector a will be 86 .6 icap plus 50 j cap vector b will be minus 40 i cap plus 69 .3 j cap and vector c that will be minus 24 .1 i cap minus 31 .9 jcap.
03:39
And now we have to figure out the magnitude as well as the direction of the vector d which is expressed as half of vector a minus three times of vector b plus vector c.
03:57
So here we do half of vector a, so that will be 43 .3 i cap plus 25 j cap.
04:09
We subtract with three times of vector b.
04:13
So that will be minus 120 i cap plus three times of 69 .3.
04:20
That will be 207 .9 j cap.
04:25
And we add to this vector c, that's minus 24 .1 i cap minus 31 .9 j cap.
04:34
So here we add or subtract the i components together.
04:38
So that will be 43 .3 plus 120 minus 24 .1 i cap and 25 minus 20 .5 .27 .9 minus 31 .9 jcap.
04:56
So here we add 43 .3 with 120 and we subtract from this 24 .1...