Let L1 be the line passing through the point P1=(9, −1, 15) with direction vector →d1=[−2, −1, −3]T, and let L2 be the line passing through the point P2=(9, 4, 8) with direction vector →d2=[−2, −1, −1]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. Use the square root symbol '√' where needed to give an exact value for your answer.
I have found the distance by using the formula "distance = projection of the P1P2 vector onto N. N is the cross product of the direction vectors." "(PQ-> dotted N)/|N|" I tried to make that formula make as much sense as I could. The value I got for distance is 10/Sqrt(5). What I don't know how to do is find the Q1 and Q2 values that satisfy the equation above. Any help is appreciated.
To find the points Q1 and Q2 on L1 and L2 respectively, we can use the fact that the distance between two lines is achieved by connecting any point on one line with the point on the other line that lies on the line perpendicular to both lines.
1. The line equation for L1 can be expressed as:
L1: r = P1 + t * d1, where r is the position vector, P1 is a point on L1, and d1 is the direction vector of L1.
2. The line equation for L2 can be expressed as:
L2: r = P2 + s * d2, where r is the position vector, P2 is a point on L2, and d2 is the direction vector of L2.
3. Now, we need to find a point on L1 and a point on L2 that are closest to each other.
4. Let Q1 be a point on L1 given by Q1 = P1 + k * (P2 - P1).
5. To find the value of k, we can substitute the expression for Q1 into the equation for L1:
Q1 = P1 + k * (P2 - P1) = P1 + k * P2 - k * P1
Substitute r = Q1 and simplify the equation:
Q1 = (P1 - k * P1) + k * P2
Q1 = P1 - k * P1 + k * P2
Since r = P1 + t * d1, we can compare the coefficients:
P1 - k * P1 = P1
k * P2 = t * d1
6. Comparing the coefficients, we get:
-k * P1 = P1, which implies -k = 1
k * P2 = t * d1, which implies k = t * d1 / P2
7. Substituting -k = 1 into the expression for Q1:
Q1 = P1 - P1 + P2 = P2
8. Therefore, we have found that Q1 = P2.
9. Similarly, we can find Q2 by substituting the expression for Q1 into the equation for L2:
Q2 = P2 + t * d2
Since r = P2 + t * d2, we can compare the coefficients:
P2 + t * d2 = P2
t * d2 = 0
10. Therefore, we have found that Q2 = P2.
11. Finally, we can calculate the distance d using the formula you provided, which gives the squared distance:
d^2 = (P1P2 · N)^2 / |N|^2
d^2 = [(P2 - P1) · N]^2 / |N|^2
where N is the cross product of the direction vectors, N = d1 × d2.
12. Calculate N = d1 × d2:
N = [-2, -1, -3] × [-2, -1, -1]
N = [-1, 2, -1]
13. Substitute the values into the formula:
d^2 = [(P2 - P1) · N]^2 / |N|^2
d^2 = [(9, 4, 8) - (9, -1, 15)] · [-1, 2, -1]^2 / |-1, 2, -1|^2
d^2 = [0, 5, -7] · [-1, 2, -1]^2 / 6
d^2 = (0 + 10 + 7) / 6
d^2 = 17 / 6
14. Therefore, the shortest distance d is the square root of the above value, d = √(17/6).
15. The points Q1 and Q2 satisfying the distance equation are Q1 = P2 and Q2 = P2, which means Q1 = (9, 4, 8) and Q2 = (9, 4, 8).
So, the shortest distance d between the two lines is √(17/6), and the points Q1 and Q2 on L1 and L2 respectively are Q1 = (9, 4, 8) and Q2 = (9, 4, 8).
To find the points Q1 and Q2 on lines L1 and L2, respectively, such that the distance between them is d, we can follow these steps:
1. Determine the vector connecting the two points P1 and P2:
- P1P2 = P2 - P1 = (9, 4, 8) - (9, -1, 15) = (0, 5, -7).
2. Find the unit vector N that is perpendicular to both direction vectors →d1 and →d2 by taking their cross product:
- N = →d1 × →d2 = (-2, -1, -3) × (-2, -1, -1).
- Using the cross product formula for the components, we get:
N = (-2 * (-1) - (-3) * (-1), -3 * (-2) - (-3) * (-2), -2 * (-1) - (-1) * (-2))
= (-1 + 3, 6 - 6, 2 - 2) = (2, 0, 0).
3. Calculate the length of vector N:
- |N| = √(2^2 + 0^2 + 0^2) = √4 = 2.
4. Normalize vector N to obtain the unit vector:
- Ň = N / |N| = (2/2, 0/2, 0/2) = (1, 0, 0).
5. Project the vector P1P2 onto vector Ň to find the distance d:
- d = (P1P2 ∙ Ň) / |Ň|.
- The dot product P1P2 ∙ Ň is equal to the dot product of P1P2 and Ň since Ň is a unit vector.
- P1P2 ∙ Ň = (0, 5, -7) ∙ (1, 0, 0) = 0 * 1 + 5 * 0 + (-7) * 0 = 0.
- Since |Ň| = 1, the distance becomes: d = P1P2 ∙ Ň = 0.
6. Substitute the value of d back into the distance formula to determine the projections PQ1 and PQ2:
- PQ1 = d * N = 0 * (2, 0, 0) = (0, 0, 0).
- PQ2 = d * N = 0 * (2, 0, 0) = (0, 0, 0).
Therefore, the distance d between the two lines L1 and L2 is 0. The points Q1 and Q2 on L1 and L2, respectively, such that d(Q1, Q2) = d, are the same as P1 and P2, respectively.