0:00
Hello everyone.
00:02
We need to compute the electrical power for that first we need to compute the vector n which is given to be ab vector cross product ad vector so we are given with the point a is 1329 b is given to be 5 5 9 and the point d is given to be 8 to 11.
00:38
So using this first let us compute ab vector.
00:43
This is nothing but the distance between the points a and b.
00:48
So this will give us 5 minus 13i plus 5 minus 2 j plus 9m minus 9 k.
01:06
So this is minus 8i plus 3j.
01:11
Similarly, we can compute ad vector that is equal to 8 minus 13 i plus 2 minus 2 j plus 11 minus 9 k clearly this is equal to 5i plus 2k using this, now we can compute our vector n, that is equal to determinant i .j .k.
01:51
Minus 8, 3, 0, 5 ,02.
01:59
So if we simplify this determinant, we'll get the n vector.
02:06
So that is, n vector is equal to 6i plus 16 j.
02:15
Minus 15 k.
02:21
So now it is easy for us to compute the electrical power p.
02:26
So that is given by the dot product of f into n vector.
02:36
Here f is equal to 900 s so this can be written as 900 by 900 by root 3 into i plus j plus k.
02:54
So now let us take the drought product of f and n to compute p.
03:01
So p is equal to 900 by root 3 into icap plus j cap plus k cap into 6i plus 16 j j minus 15 k k so further simplifying this we would get 900 by root 3 into 6 plus 16 minus 15.
03:37
Clearly the electrical power p is equal to 3 ,637 .31 watts...