00:01
Here, consider a plane which passes through point 6, minus 4, 3 and it contains a line whose equation is x by 2 it is equal to y minus 4 upon minus 1 it is equal to z.
00:16
Suppose this equation it is equal to t, then from here we can find x, y and z.
00:22
So, x will be equal to twice 3, y will be y minus 4 will be equal to minus 1.
00:29
That means y is equal to minus t here, y will be equal to 4 minus t and z is equal to t.
00:39
So, x is twice t, y is 4 minus t and z is equal to t.
00:44
Therefore, the parametric equation of line will x, y, z will equals to twice t, 4 minus t and t.
00:54
This is the required parametric equation of line.
00:58
Now, observe that the equation this equation can be written as 0 plus twice t, 4 minus t and 0 minus not 0 plus t.
01:13
So, here 0, 4, 0 are the coordinates through which the line passes and let this 0, 4, 0 point is denoted by b, it is equivalent to b.
01:28
So, the direction ratios of the line will become b bar is equal to 2, minus 1, 1.
01:38
This coefficients of t, 2, minus 1 and 1 this will be the direction ratios here.
01:44
So, then we will consider this b point and that t point which we have taken in the starting, these two points.
01:53
So, b point was 0, 4, 0 and the p point was 6, minus 4 and 3.
02:03
Then we will consider the vector p, b bar.
02:06
It is nothing but p bar minus b bar minus t bar means o, b bar minus o, p bar.
02:14
This will be equal to 0, minus 6, then second coordinate will be 4, minus of minus 4, third coordinate will be 0, minus 3...