00:01
So here we have the two vectors a and b and we need to write these two vectors.
00:10
We need to express them in terms of icap and j cap because we know i -cap is representing the direction along x and j -cap is representing the direction along y.
00:22
So if we see vector a, so if we just break down its component, so we apply the normal trigonometry here because this hypertenuse of.
00:32
3 .6 meter is given.
00:34
This angle is given so we can use cosine.
00:37
So cost 70 would be equal to 3 point not 3 points so it would be equal to the base.
00:44
So let us say the base is a x and the hypotenuse which is 3 .6.
00:50
So this will give us the value of a x as 3 .6 times cost 70.
00:58
If you figure out the expression of cost 70 and multiply it with 3 .6, we're going to get 1 .23 meters.
01:08
So similarly, we can apply sign and get this a y.
01:12
So this sign 70, that would be equal to the opposite, which is a y over the hypotenuse, which is 3 .6.
01:20
So this will give us the value of a y here as 3 .83.
01:31
So approximately this would be 3 .38 meters.
01:37
So we got the components of vector a along x and y.
01:42
So we can express a vector as 1 .23 icap because this component is along x plus 3 .38 along j cap.
01:56
So now in the same manner we are going to express this vector b in terms of i cap and j cap.
02:02
So its component would be along the negative x direction so that would be 2 .4 cost 30 and along y direction it would be 2 .4 sine 30.
02:14
So exactly same approach we're gonna use here.
02:18
So b vector now directly we can express this as minus 2 .4 cost 30 minus 2 .4 cost 30 along icap minus 2 .4 sine 30 along j cap.
02:32
So this gives us the value of b vector.
02:35
So we simplify each of these terms here.
02:39
So cost 30 when multiplied by 2 .4, we get 2 .078.
02:49
So approximately minus 2 .1 icap.
02:53
And sine 30 when multiplied with 2 .4, we get 1 .2j cap...