find shortest distance between the line :
(x - 8)/ 3 = (y + 19)/-16 = (z - 10)/ 7
and
(x - 15)/ 3 = (y - 29)/8 = (z - 5)/ -5
To find the shortest distance between two lines, we can use the vector equation of each line and calculate the distance vector between them.
Given the two lines:
L1: (x - 8)/ 3 = (y + 19)/-16 = (z - 10)/ 7
L2: (x - 15)/ 3 = (y - 29)/8 = (z - 5)/ -5
We can rewrite these equations in vector form:
L1: r1 = (8, -19, 10) + t(3, -16, 7)
L2: r2 = (15, 29, 5) + s(3, 8, -5)
Now, let's calculate the distance vector (d) between the two lines. The distance vector can be defined as the vector connecting any point on L1 to the closest point on L2.
So, we need to find t0 and s0 such that the vector (d) between r1(t0) and r2(s0) is perpendicular to both lines L1 and L2.
To find t0 and s0, we can set up the following system of equations:
(r1(t0) - r2(s0)) · (L1 direction vector) = 0
(r1(t0) - r2(s0)) · (L2 direction vector) = 0
Let's calculate the direction vectors of L1 and L2:
L1 direction vector = (3, -16, 7)
L2 direction vector = (3, 8, -5)
Now, let's solve the system of equations to find t0 and s0. Using the dot product:
(8 + 3t0 - 15 - 3s0, -19 - 16t0 - 29 + 8s0, 10 + 7t0 - 5 + 5s0) · (3, -16, 7) = 0
(8 + 3t0 - 15 - 3s0, -19 - 16t0 - 29 + 8s0, 10 + 7t0 - 5 + 5s0) · (3, 8, -5) = 0
Expanding these equations, we get:
3(8 + 3t0 - 15 - 3s0) - 16(-19 - 16t0 - 29 + 8s0) + 7(10 + 7t0 - 5 + 5s0) = 0
3(8 + 3t0 - 15 - 3s0) + 8(-19 - 16t0 - 29 + 8s0) - 5(10 + 7t0 - 5 + 5s0) = 0
Simplifying and rearranging the equations, we get:
112t0 - 104s0 = -366
-264t0 + 116s0 = -13
Now, we can solve this system of equations to find the values of t0 and s0. Once we have these values, we can substitute them back into the vector equation of L1 and L2 to find the closest points p1 and p2 on L1 and L2, respectively.
Finally, we can calculate the distance vector (d) as the difference between p1 and p2:
d = p1 - p2
The length of the vector d give us the shortest distance between the two lines L1 and L2.