00:01
In this problem, we are given with some points.
00:02
They are 1 .0 .1 and 2 .3 .2.
00:08
Now, we need to find a point that passes through these points and that is closest to the curve.
00:14
And the curve is given with the parameter equations.
00:18
That is, x of t is equals to t, y of t is equals to 2t, and z of t is equals to 2t.
00:26
Now, let us assume this point as p1, this point as p2.
00:30
Therefore, the line passing through these points would be l1 is equal to p1 plus lambda into p2 minus p1.
00:40
Therefore, l1 is equal to i -cap plus k -cap.
00:46
We are writing this in vector form plus lambda into p2 is 2i -cap plus 3j cap plus 2k -cap minus i -cap minus k -cap.
00:58
This implies l1 is equals to i cap plus k cap plus lambda into this is i cap plus 3j cap plus k cap.
01:10
Now this can be written as v1 bar vector v1 bar.
01:15
Now in this curve the parametric equations are x of t, y of t and z of t.
01:21
Now if we observe this curve closely, this curve is nothing but 2x is equals to y is equals to z.
01:29
We will write this in form of a 3d line equation.
01:33
Therefore, through x is equal to y is equals to z, which implies x minus 0 by 1 by 2 is equals to y minus 0 by 1 is equals to z minus 0 by 1.
01:45
This is a line that passes through the origin.
01:48
Therefore, l2 is equal to mu is some constant into 1 by 2 i cap plus j cap plus k cap.
01:59
Is vector 2.
02:01
Now, we have a situation that is there is two lines...