00:01
In this portion we have given the point as 2 ,1 ,1 and x of t is given as minus 1 plus 4t here.
00:12
Y of t is given as 1 plus 2t and z of t is given as 3 minus t.
00:21
So the plane contains the line x is equal to minus 1 plus 4t, y is equal to 1 plus 2t and z is equal to 3 minus t.
00:38
This is not 3t, this is our 3 minus t.
00:42
Ok, this is 3 minus t.
00:45
So therefore the vector in plane is as 4 ,2 ,minus 1.
00:57
Here given 2 ,1 ,1 is point on the plane.
01:09
This is given in the quotient and now put t is equal to 0 give us another point on plane.
01:22
Another point on plane.
01:26
Ok, so this is coming out to minus 1 ,1 ,3.
01:29
Here the vector form is equal to 1 ,4 ,3...