Let L1 be the line passing through the point P1=(9, 0, 0) with direction vector →d1=[3, 2, −2]T, and let L2 be the line passing through the point P2=(−10, 8, 4) with direction vector →d2=[3, 0, 2]T.
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d.
d = _
Q1 = (_,_,_)
Q2 = (_,_,_)
ground beef
To find the shortest distance between two lines, we need to find a vector connecting the two lines that is orthogonal (perpendicular) to both lines. Then, we can find the length of this vector, which will give us the shortest distance between the lines.
1. First, we need to find the vector that is orthogonal to both direction vectors →d1 and →d2. We can compute the cross product of →d1 and →d2 to obtain this vector.
→n = →d1 × →d2
Calculating the cross product:
→n = [3, 2, -2] × [3, 0, 2]
→n = [(2 * 2) - (0 * -2), (-2 * 3) - (2 * 3), (3 * 0) - (2 * 2)]
→n = [4, -12, -4]
2. The vector →n is orthogonal to both lines. Now, we need to find a point on each line, P1 and P2, and find the vector connecting them.
→P = →P2 - →P1
Calculating →P:
→P = [-10, 8, 4] - [9, 0, 0]
→P = [-10 - 9, 8 - 0, 4 - 0]
→P = [-19, 8, 4]
3. Now, we can calculate the projection of →P onto →n to find the length of d.
d = |→P · →n| / |→n|
Calculating the dot product of →P and →n:
→P · →n = (-19 * 4) + (8 * -12) + (4 * -4)
→P · →n = -76 - 96 - 16
→P · →n = -188
Calculating the magnitude of →n:
|→n| = √(4^2 + (-12)^2 + (-4)^2)
|→n| = √(16 + 144 + 16)
|→n| = √176
|→n| ≈ 13.27
Calculating the shortest distance d:
d = |-188| / 13.27
d = 188 / 13.27
d ≈ 14.17
4. To find the points Q1 on L1 and Q2 on L2, we can use the equation of a line in parametric form:
Q1 = →P1 + t1 * →d1
Q2 = →P2 + t2 * →d2
Substituting values:
Q1 = [9, 0, 0] + t1 * [3, 2, -2]
Q2 = [-10, 8, 4] + t2 * [3, 0, 2]
To find t1 and t2, we can solve the following system of equations:
9 + 3t1 = -10 + 3t2
2t1 = 8 + 2t2
-2t1 = 4 + 2t2
Solving this system of equations, we find t1 = -11 and t2 = 11.
Substituting these values back into the equations for Q1 and Q2:
Q1 = [9, 0, 0] + -11 * [3, 2, -2]
Q1 = [9, 0, 0] + [-33, -22, 22]
Q1 = [-24, -22, 22]
Q2 = [-10, 8, 4] + 11 * [3, 0, 2]
Q2 = [-10, 8, 4] + [33, 0, 22]
Q2 = [23, 8, 26]
Therefore:
d ≈ 14.17
Q1 = (-24, -22, 22)
Q2 = (23, 8, 26)
To find the shortest distance between two lines, we can use the concept of vector projection.
1. Start by finding the vector connecting the two given points on the lines. Let's call this vector →v = P2 - P1.
→v = [-10, 8, 4] - [9, 0, 0]
→v = [-19, 8, 4]
2. Find the unit direction vectors →u1 and →u2 for the given lines by normalizing their respective direction vectors.
→u1 = →d1 / ||→d1|| = [3, 2, -2] / √(3^2 + 2^2 + (-2)^2)
→u2 = →d2 / ||→d2|| = [3, 0, 2] / √(3^2 + 0^2 + 2^2)
3. Project →v onto the direction vector of line L1 (→d1) and line L2 (→d2) separately to get the components of →v in each direction.
Projection of →v onto →d1 = →v ∙ →u1
Projection of →v onto →d2 = →v ∙ →u2
→v1 = (→v ∙ →u1) * →u1 = (−19, 8, 4) ∙ ([3, 2, −2] / √(3^2 + 2^2 + (-2)^2)) * ([3, 2, −2] / √(3^2 + 2^2 + (-2)^2))
→v2 = (→v ∙ →u2) * →u2 = (−19, 8, 4) ∙ ([3, 0, 2] / √(3^2 + 0^2 + 2^2)) * ([3, 0, 2] / √(3^2 + 0^2 + 2^2))
4. Calculate the distance between the lines by finding the length of the vector →w = →v - →v1. This vector is a perpendicular vector connecting the two lines.
→w = →v - →v1 = [-19, 8, 4] - →v1
5. The distance d = ||→w|| is the shortest distance between the two lines.
Once you've calculated the distance d, you can find the points Q1 and Q2 by adding the respective projections to the given points P1 and P2.
Q1 = P1 + →v1
Q2 = P2 + →v2
Now, you can substitute the calculated values to find the solution.
d = _
Q1 = (_,_,_)
Q2 = (_,_,_)
This should get you started.
www.math.jhu.edu/~js/Math202/example1.pdf