00:01
In this problem, we're dealing with a pipeline system in a 3d space, consisting of two straight line segments, segment ab, and bc, that are both connected at point b.
00:14
The two segments have their endpoints fixed at coordinates a is equal to 0 minus 40, 0, coordinate b equal to 40, 0, and point c equal to a, b, 0, where a and b are unknown and we'll grab itself.
00:32
If segment b is in a direction of vector 3i plus 4j plus k, we want to find, firstly, the distance of segment ab, the angle between the sections ab and bc, and the vector line equation describing segment bc.
00:51
First, let's start with an illustration of what's going on.
00:55
So here we have our two segments, ab and bc, in a 3d coordinate system, where we have indicated their endpoint.
01:04
A, b, and c.
01:09
And the first question, we want to find the distance of the segment a -b.
01:24
First, let's describe the vector that goes from b to a, essentially correspond to the difference of points b minus a.
01:32
So the vector starting at point b going to point a is equal to the difference of points, 40 minus 0, and the x coordinate, 0 minus 0, 0 ,000, minus 40 in the y coordinate, and minus 20 minus 0 in the z coordinate.
02:01
Thus, simplifying to 40, 40 minus 20.
02:11
And to simplify things a little bit, we can factor out a factor of 4.
02:22
So this is the vector that goes from point b to point a, and the magnitude of this vector is going to correspond to our distance of ab.
02:35
And to find the magnitude, essentially what we're going to apply is the p.
02:39
The gorin theorem.
02:40
So our distance of av is going to correspond to the square root of 40 squared plus 40 squared minus 20 squared.
02:51
But here i affected out factor 10 to simplify things a little bit, so take out our 10 is going to correspond to 4 squared plus 4 squared plus 2 squared.
03:04
It's going to give us 10 times the square root of 16 plus 16 plus 16 plus for in the square root we're going to have a square root of 36 times 10 which sensually suffice to 10 times 6 so 60.
03:25
Our distance ab is equal to 60.
03:30
Great.
03:31
Now for the next question we want to find the angle between segment a .v and b c.
03:53
To do so we note that we have just found in the previous question the vector starting from point b going to point a.
04:02
So we have a vector here, i'm going to call a vector v.
04:05
And in our problem statement, we are given a vector u that goes from b to c.
04:16
And we have an identity stating that the dot product of two vectors, say v .u, is equal to the product of the magnitude of v times magnitude of u times cost theta or theta is the angle between our vectors b and u and this is the angle that we precisely want to find we want to find this angle theta the angle between vectors b and u that describe our both lines are two line segments so vector re we just found it previously this equals to the vector four for minus two and the vector u is given in our problem statements those the vector 3, 4, 1.
05:17
And the problem statement is written in igk form.
05:20
I written it here in victorial form.
05:24
So now i want to calculate is the dot product and the two magnitudes.
05:29
So let's do that.
05:33
V .u is going to be equal to the sums of 4 times 13, so 12 plus 4 times 4, 16, plus minus 2 times 1, so minus 2.
05:47
Giving us 26.
05:50
Now let's calculate the magnitudes of v and you.
05:54
The magnitude of v corresponds to the square root of 4 squared plus 4 squared plus 2 squared.
06:00
The square root of 16 plus 16 plus 2, giving us the square root of 36, which is 6...