00:01
Hi, today we are solving the question in which we are given with pt is equals to minus 3, 0, minus 5 plus t minus 2 minus 2 minus 1.
00:16
Similarly, qt is given as 2, 4 minus 3 plus t minus 5 minus 5 minus 2.
00:28
So here we have to find the distance between the skew lines for, 4, minus 3 plus t minus 5 minus 5 minus 2.
00:32
So here we have to find the distance between the skew lines for, vectors of p and q.
00:36
So here first of all we know that the formula of distance d is equals to a2 vector minus a1 vector whole into b1 vector cross product b2 vector divided by cross product mod of b1 vector into b2 vector.
01:05
So first of all we will find a1 a2 and b1 b2 so from here a 1 vector is nothing but this minus 3 0 and minus 5 from pt similarly a 2 vector is 2 4 minus 3 from this qt similarly b1 vector is minus 2 minus 2 minus 1 and b2 vector is equals to this minus 5 minus 5 and minus 2.
01:48
So from here now we'll find a2 vector minus a1 vector firstly.
01:56
So if we see that a2 vector minus a1 vector so 2 minus minus 3.
02:03
So we'll get 5.
02:05
Similarly 4 minus 0 will give 4 and minus 3.
02:10
Minus minus 5 will give us.
02:14
So this is a 2 vector minus a 1 vector.
02:18
Now if we find out the cross product of b1 vector and b2 vectors...