00:01
In this question we have been given the vector y having the following entries as minus 4, 10 and 5.
00:08
We have vector u1 equals to minus 2, minus 4 and minus 1.
00:14
Then u2 which is equals to 0, 6, minus 24.
00:20
We have to find out the distance from y to the plane which is spanned by these both vectors.
00:27
So for that, so for that what we are going to do, we consider y cap to be the projection of y on the plane w which is given by the following formula y dot u1 over the magnitude of u1 whole square times u1 plus y dot u2 over the magnitude of u2 whole square times u2.
01:03
We know all these values so y dot u1 is going to be.
01:08
Just take the dot product of these two, so it will come out to be 8 minus 40 minus 5 over this magnitude mod of this u1 square.
01:22
So it comes out to be 4 plus 60 plus 1 times u1 minus 2 minus 4 minus 1 plus y dot u2, 0 plus 60 minus 120 over u2 square 0 plus 36 plus 576 times u2, 0, 6, minus 24.
01:51
This is equals to minus 37 over 21 times minus 2 minus 4 minus 1 plus minus 60 over 576 plus 36, 612 times of 0, 6, minus 24.
02:15
Calculate these values here first of all.
02:21
So, in minus 37 over 21, it comes out to be minus 1 .762 minus 1 .76 i will take times minus 2 minus 4 minus 1 plus 60 over 612, 0 .098 minus 0 .098 times 0, 6, minus 24.
02:52
Let us multiply this.
02:54
So, this will be 1 .76 multiplied with 2 which comes out to be 3 .52.
03:05
Next, 1 .76 multiplied with 4 minus 0 .098 multiplied with 6...