Let L1 be the line passing through the point P1=(16, −1, −12) with direction vector →d1=[3, −1, −2]T, and let L2 be the line passing through the point P2=(4, 7, −17) with direction vector →d2=[3, 1, −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. Use the square root symbol '√' where needed to give an exact value for your answer.
I have wrecked my brain several times, but cannot solve this question whatsoever. Please help!
To find the shortest distance between two lines, we can use the formula:
d = ∥(P1 - P2) - ((P1 - P2) · d1) d1 - ((P1 - P2) · d2) d2∥ / ∥d1 × d2∥
Where ∥v∥ represents the magnitude (length) of vector v, · denotes the dot product of two vectors, and × denotes the cross product of two vectors.
Let's calculate step-by-step:
1. Calculate P1 - P2:
P1 - P2 = (16, -1, -12) - (4, 7, -17) = (12, -8, 5)
2. Calculate (P1 - P2) · d1:
(P1 - P2) · d1 = (12, -8, 5) · (3, -1, -2) = 12(3) + (-8)(-1) + 5(-2) = 36 + 8 - 10 = 34
3. Calculate (P1 - P2) · d2:
(P1 - P2) · d2 = (12, -8, 5) · (3, 1, -2) = 12(3) + (-8)(1) + 5(-2) = 36 - 8 - 10 = 18
4. Calculate d1 × d2:
d1 × d2 = (3, -1, -2) × (3, 1, -2)
To find the cross product, we can use the formula:
(| i j k |
| x1 y1 z1 |
| x2 y2 z2 |)
= (y1z2 - y2z1, x2z1 - x1z2, x1y2 - x2y1)
Plugging in the values, we get:
(3, -1, -2) × (3, 1, -2) = ((-1)(-2) - (1)(-2), (3)(-2) - (3)(-2), (3)(1) - (3)(-1)) = (-2, -6, 6)
5. Calculate ∥(P1 - P2) - ((P1 - P2) · d1) d1 - ((P1 - P2) · d2) d2∥:
∥(P1 - P2) - ((P1 - P2) · d1) d1 - ((P1 - P2) · d2) d2∥ = ∥(12, -8, 5) - 34(3, -1, -2) - 18(3, 1, -2)∥
= ∥(12, -8, 5) - (102, -34, -68) - (54, 18, -36)∥ = ∥(-144, -12, 109)∥ = √((-144)^2 + (-12)^2 + 109^2)
Calculating the square root, we get:
√((-144)^2 + (-12)^2 + 109^2) ≈ √(20736 + 144 + 11881) ≈ √32661 ≈ 180.65
Therefore, the shortest distance, d, between these two lines is approximately 180.65.
6. To find a point Q1 on L1 and a point Q2 on L2 such that d(Q1, Q2) = d, we can choose the same t parameter for both lines:
Q1 = P1 + t * d1
Q2 = P2 + t * d2
Replacing the values, we get:
Q1 = (16, -1, -12) + t * (3, -1, -2)
Q1 = (16 + 3t, -1 - t, -12 - 2t)
Q2 = (4, 7, -17) + t * (3, 1, -2)
Q2 = (4 + 3t, 7 + t, -17 - 2t)
By setting d(Q1, Q2) = d, we can equate the distances:
∥Q1 - Q2∥ = d
Using the distance formula, we have:
∥(16 + 3t - 4, -1 - t - 7, -12 - 2t - (-17))∥ = 180.65
Simplifying this expression, we get:
√((12 + 3t)^2 + (-8 - t)^2 + (5 + 2t + 17)^2) = 180.65
Now, we solve this equation to find the values of t, Q1, and Q2.
To find the shortest distance between two lines, we need to find a point on each line and calculate the vector between them. Then, we project this vector onto the vector that is perpendicular to both lines in order to find the shortest distance.
First, let's find a point Q1 on line L1. Since L1 passes through point P1 = (16, -1, -12), we can take Q1 = P1.
Next, let's find a point Q2 on line L2. Since L2 passes through point P2 = (4, 7, -17), we can take Q2 = P2.
Now, let's calculate the vector between Q1 and Q2. We subtract the coordinates of Q2 from those of Q1 to get the vector.
→Q1Q2 = (16 - 4, -1 - 7, -12 + 17) = (12, -8, 5)
Next, we need to find the vector that is perpendicular to both →d1 and →d2. The cross product of two vectors is perpendicular to both, so we can calculate the cross product of →d1 and →d2.
→n = →d1 × →d2 = [3, -1, -2] × [3, 1, -2]
To find the cross product, we can use the determinant method:
→n = [(−1)(−2) − (−2)(1)]i - [(3)(−2) − (−2)(3)]j + [(3)(1) − (3)(−1)]k
= (-2 - (-2))i - (-6 - (-6))j + (3 - (-3))k
= 0i + 0j + 6k
= 6k
Therefore, →n = 6k.
Now, we can project →Q1Q2 onto →n to find the shortest distance d between the two lines. The dot product of →Q1Q2 and →n will give us the magnitude of the projection.
→Q1Q2 · →n = (12, -8, 5) · (0, 0, 6)
= 0 + 0 + 30
= 30
To obtain the magnitude of →n, we take the square root of the dot product.
|→n| = √(6^2)
= √36
= 6
Therefore, the shortest distance between the two lines is d = 30/6 = 5.
To find the points Q1 and Q2 that achieve this distance, we can use the equations of the lines.
For Q1, we already determined that Q1 = P1 = (16, -1, -12).
For Q2, we use the parameterization of the line L2:
x = 4 + 3t
y = 7 + t
z = -17 - 2t.
Since the distance between Q1 and Q2 is 5, we can equate the magnitudes of the vectors Q1Q2 and →d2.
|→Q1Q2| = |→d2|
|Q2 - Q1| = |(3, 1, -2)|
Using the distance formula:
√((4 + 3t - 16)^2 + (7 + t + 1)^2 + (-17 - 2t + 12)^2) = √(3^2 + 1^2 + (-2)^2)
Simplifying:
√((3t - 12)^2 + (t + 8)^2 + (-2t - 5)^2) = √14
Squaring both sides:
(3t - 12)^2 + (t + 8)^2 + (-2t - 5)^2 = 14
Expanding and collecting like terms:
14t^2 - 138t + 285 = 0
This quadratic equation can be solved for t using factoring, completing the square, or the quadratic formula.
The final step is to substitute the value of t into the equations of line L2 to find the coordinates of point Q2.
there are several ways to solve this type of problem.
The easiest method in my opinion goes like this:
A normal to both lines would be the cross product of
<3, -1, -2> with <3, 1, -2> , the direction of the 2 given lines
which is <2, 0, 3> , (I assume you know how to find the cross-product)
Now consider 2 planes with that normal, the first one containing
the point (16, -1, -12) and the second plane containing the point (4, 7, -17)
The distance between these two parallel planes will of course be always
the same , this is the key point using this method
equation of first plane is 2x + 0y + 3z = c
but (16, -1, -12) lies on it, so
32 + 0 - 36 = c
c = -4
2x + 3y + 4 = 0
All we need now is the distance between the two planes using the distance between a point and a plane
We also have a point on the 2nd plane: namely (4, 7, -17)
distance = | 2(4) + 0(7) + 3(-17) + 4| / √(2^2 + 0^2 + 3^2)
= 39/√13 or 39√13/13 = 3√13