Find the distance between the lines [x,y,z] = [-1,1,0] + s[3,4,-2] and [x,y,z] = [5,9,-4] + t[2,3,1]. Interpret the result.
As I recall, if you have two lines
L1: p1 + v1*t
L2: p2 + v2*t
the distance can be obtained by projecting the unit normal between the lines onto the line joining the two points:
v1 × v2 / |v1 × v2| • (p1 - p2)
For a brief example and discussion, see
physicsforums . com / showthread.php?t=276253
for this question, do i have to solve it? it just says interpret the results, does that mean only explain?
im trying to solve it right now. i have:
L1: [x,y,z]=[-1,1,0]+s[3,4,-2]
L2: [x,y,z]=[5,9,-4]+t[2,3,1]
First, cross product of the two directional vectors from both lines:
[3,4,-2] x [2,3,1] =
That's all I have so far :/
I don't know what to do next because in the example shown on the forum, I don't know how they got the answer [from the forum: (-2,3,-2) x (-2,-3,-1) = (-9,-2, 12)] I know 3 x (-3) is -9 but I don't understand where the -2 and +12 came from since (-2)(-2) = +4 and (-2)(-1)= +2.
Also on the forum there is:
And its unit vector is:
-9x-2y+12z
-----------
229^1/2
I don't know how they got this answer either. They mentioned at the end of it they may not be right so I don't know ....
To find the distance between two parallel lines in 3D space, you can use the formula:
Distance = |(A - B) · N| / ||N||
where:
- A and B are any two points on the two lines,
- · represents the dot product,
- N is the direction vector of one of the lines, and
- ||N|| represents the magnitude (or length) of the vector N.
Let's apply this formula to the given lines.
First, we need to find a point on each line. Since the lines are given in vector form, we can use any values for the parameters s and t to find the points.
For the first line:
Let's take s = 0. Plugging this into the vector equation, we get:
[x, y, z] = [-1, 1, 0] + 0[3, 4, -2]
So, a point on the first line is [-1, 1, 0].
For the second line:
Let's take t = 0. Plugging this into the vector equation, we get:
[x, y, z] = [5, 9, -4] + 0[2, 3, 1]
So, a point on the second line is [5, 9, -4].
Now, let's find the direction vector for one of the lines. We can use the coefficients of the parameters s and t from the vector equations.
For the first line, the direction vector is [3, 4, -2].
Now we can substitute the values into the distance formula:
Distance = |([-1, 1, 0] - [5, 9, -4]) · [3, 4, -2]| / ||[3, 4, -2]||
Simplifying this, we get:
Distance = |[-6, -8, 4] · [3, 4, -2]| / ||[3, 4, -2]||
Taking the dot product of [-6, -8, 4] and [3, 4, -2], we get:
Distance = |(-6)(3) + (-8)(4) + (4)(-2)| / ||[3, 4, -2]||
Distance = |-18 - 32 - 8| / ||[3, 4, -2]||
Distance = |-58| / ||[3, 4, -2]||
Finally, we need to compute the magnitude of the direction vector, ||[3, 4, -2]||:
||[3, 4, -2]|| = sqrt((3)^2 + (4)^2 + (-2)^2) = sqrt(9 + 16 + 4) = sqrt(29)
Substituting this into our equation, we have:
Distance = |-58| / sqrt(29)
Thus, the distance between the two given lines is |-58| / sqrt(29). The result may be interpreted as the shortest distance between any two points on the two lines.