00:01
In this problem in sub part a we are provided with the vector v which equals to 2 2 2 and we are asked to find out the magnitude of this vector v and also a unit vector which we denote by u in the direction that is opposite to the vector v.
00:28
So first let us find out the magnitude.
00:30
The magnitude is given by the square root.
00:33
Of the sum of the squares of the components.
00:35
So that is square root of 4 plus 4 plus 4 which equals to square root of 4 times 3 which simplifies to 2 root 3.
00:45
So this is the required magnitude of v.
00:48
Next we need to find out a unit vector.
00:53
So here so this unit vector u is given by the negative of the vector v divided by the magnitude of the vector v.
01:06
The negative sign denotes that the vector is in the opposite direction.
01:11
So here we have negative 1 over 2 root 3 times the vector 2 to 2 .2.
01:18
So simplifying this, we get you to be equal to negative 1 over square root of 3, negative 1 over square root of 3, negative 1 over square root of 3...