00:01
Hello student from this figure we can write the position vectors.
00:05
So these are the position vectors of the object.
00:10
So now we have to find out the vector ab that is is equal to 1 .532i plus 2 .036j minus 3k.
00:30
So this is the ab vector and the magnitude of this vector is equal to 3 .936.
00:39
And now we have to find out the univector of ab that is equal to x divided by abi plus y divided by ab plus j divided by ab into k unit vector.
01:04
And this univector which is equal to 0 .389i i plus 0 .517j minus 0 .762k.
01:21
Now we can find the force of ab vector that is fab vector is equal to the magnitude of force ab to the univector of k.
01:39
So here the magnitude of force is given in the figure that is 260 into 2 by 2b vector.
01:48
So on performing the multiplication we can get the force ab vector that is equal to 97 .3 plus 129 .3j minus 190 .7k.
02:10
So similarly we can find for force ab which is equal to fac univector of ab which is equal to 400 into the univector is 0 .353i minus 0 .69j minus 0 .83k.
02:38
So the ab force vector is equal to 221 .45 minus 27 .1j minus 332.
02:58
Now we can find the resultant vector for that that is the resultant force vector is equal to fab plus fac.
03:21
So on substitution we get 3187i plus 101 .4j minus 527 .7 and the magnitude of this force is equal to root of 318 .7 square plus 101 .4 square plus minus 527 .7 square which is equal to 620 .5k.
04:03
Now we can find the angle between them.
04:06
So here the magnitude of resultant force is equal to 620 .53 newton.
04:18
So now we can find the answer for the angle between the force and axis that is the alpha is equal to cos inverse of x divided by resultant vector and that is equal to 59 .1 degree and beta is equal to cos inverse of y by fr that is equal to 80 .6 degree and gamma which is equal to cos z by fr and that is equal to 17 .3.
05:03
Rate is 173 .4 degree...