00:01
In this problem we are provided with two vectors.
00:05
The vector u equals to 1 -1 -1 and the vector v equals to 4 -6 -3.
00:15
Further it is given that the vectors unv span s.
00:21
We are asked to find out the point p that exists in s such that p is closest to close to the vector b which equals to 6, 7, 4.
00:40
So this can be obtained by finding out the projection of b on s.
00:48
And this is given by the formula the dot product of b and u over the dot product of u and you times you plus the dot product of b and v over the dot product of b and v times u plus the dot product of b and v over the dot product of b and v.
01:06
Of v with itself times v.
01:09
So let us begin by finding out the dot product of v and u.
01:14
So we have b dot u to be equal to the vector b which is 6 -7 -4 times the vector u which is 1 -1 -1 and this equals to 6 times 1 which is 6 plus 7 times 1 which is 7 plus 4 times 1 which is 4 and adding these up we obtain 17.
01:37
Next we evaluate the dot product of u with itself.
01:42
So we have u dot u to be equal to 1 .11 times 1 1 1 and this equals to 1 plus 1 plus 1.
01:53
Adding them up we obtain 3.
01:55
Next we evaluate the dot product of b with v so we have b dot v to be equal to the vex b which is 6 -7 -4 times the vector v which is 4, 6, 3...
