00:01
Hi, in the question it is given that, given that we have charge q is equal to 2 .40 multiplied by 10 to the power minus 18 coulomb and we have v vector is equal to 5 i cap plus 4 j cap minus k cap meter upon second and we have b vector is equal to 5 i cap plus 5 j cap plus k cap tesla and we have e vector is equal to 2 i cap minus j cap minus 3 k cap volt upon meter.
00:57
So, since we know that f vector can be written as f e vector plus f b vector and f e vector can be written as q multiplied by e vector.
01:16
So, we will have f e vector will be equal to 2 .40 multiplied by 10 to the power minus 18 multiplied by 2 i vector minus j vector minus 3 k vector and f b vector is equal to q vector multiplied by b vector multiplied by v vector multiplied by b vector.
01:44
So, this will be further equal to 2 .40 multiplied by 10 to the power minus 18 determinant with elements i cap j cap k cap 5 4 minus 1 5 5 1.
02:11
So, this will be further equal to 2 .40 multiplied by 10 to the power minus 18 multiplied by i cap multiplied by 4 plus 5 minus j cap multiplied by minus 5 minus 5 plus k cap multiplied by 25 minus 20.
02:34
So, this will be further equal to 2 .40 multiplied by 10 to the power minus 18 9 i cap plus 10 j cap plus 5 k cap.
02:50
So, therefore f vector will be equal to 2 .40 multiplied by 10 to the power minus 18 multiplied by 2 i cap minus j cap minus 3 k cap plus 9 i cap plus 10 j cap plus 5 k cap.
03:13
So, this is further equal to f vector is equal to 26 .4 i cap plus 21 .6 j cap plus 4 .8 k cap multiplied by 10 to the power minus 18.
03:38
So, therefore from here we will have f x will be equal to 26 .4 multiplied by 10 to the power minus 18 newton.
03:49
We have f y will be equal to 21 .6 multiplied by 10 to the power minus 18 newton and we have f z is equal to 4 .8 multiplied by 10 to the power minus 18 newton...