00:01
For the first subpart we have l is equal to x comma y comma z belongs to r3 such that x minus 2y plus z equal to 0 and y plus x equal to 0.
00:16
Now simplifying this equation and substituting the value of y as minus x in the first equation we get the set x y x y belongs to r3.
00:32
Such that y equal to minus x and z is equal to minus three x now we get this set of points x comma minus x comma minus 3x so clearly we have l as span of 1 comma minus 1 comma minus 3 so we have got the basis of l is equal to 1 comma minus 1 comma minus 3 so, ortho -normal basis will be given by 1 divided by under root of 11 comma minus 1 divided by under root of 11 and minus 3 divided by under root of 11 as the norm of 1 minus 1 and 3 is under root of 11 which is obtained as under root of 1 square plus minus 1 square plus minus 3 square.
01:28
So, now that we have got the orthonormal basis further, we will find in the second subpart equation for orthogonal complement of l.
01:43
So we have the set x, y, z where the points x, y, z, as we found above, 1 by root 11, minus 1 by root 11 and minus 3 by root 11 equated to 0.
02:02
So, solving further, we have x, comma, y, comma, z such that x minus y minus 3 z is equal to 0.
02:12
So, the equation of v is x minus y minus 3z equal to 0.
02:25
So writing further, we have y plus 3z comma y, comma z is equal to span of 1 .1 .0.
02:37
And the second one being 3 .0 .1...