Let L1 be the line passing through the points Q1=(−2, −3, −4) and Q2=(−5, −2, −3) and let L2 be the line passing through the point P1=(−27, 31, −25) with direction vector →d=[−6, 9, −6]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.
Having a hard time with this one..
direction of L1 is [-3, 1, 1]
equation of L1
L1 = (-2,-3,-4) + s(-3,1,1)
equation for L2
L2 = (−27, 31, −25) + t(−6, 9, −6)
so : -2 - 3s = -27 - 6t ---> 6t - 3s = -25
and -3 + s = 31 + 9t ---> s = 9t + 34
using substitution :
6t - 3(9t+34) = -25
6t - 27t - 102 = -25
-21t = 77
t = -11/3, then s = 9(-11/3) + 34 = 1
check in third equation:
-4 + s = -25 - 6t
LS = -4 + 1 = -3
RS = -25 - 6(1) = -31
LS ≠ RS
They do not intersect.
Check my arithmetic, I did not write it down first.
The method is correct.
thank you so much!
To determine whether the lines L1 and L2 intersect, we need to find the point of intersection if it exists.
Step 1: Find the parametric equations of lines L1 and L2.
- Line L1: We have two points, Q1 and Q2. The parametric equations for L1 can be written as:
x = -2 + t(-5 + 2)
y = -3 + t(-2 + 3)
z = -4 + t(-3 + 4)
Simplifying the equations:
x = -2 - 3t
y = -3 + t
z = -4 + t
- Line L2: We have the point P1 and the direction vector →d. The parametric equations for L2 can be written as:
x = -27 + s(-6)
y = 31 + s(9)
z = -25 + s(-6)
Simplifying the equations:
x = -27 - 6s
y = 31 + 9s
z = -25 - 6s
Step 2: Equate the x, y, and z equations for lines L1 and L2:
Equating x equations:
-2 - 3t = -27 - 6s
Equating y equations:
-3 + t = 31 + 9s
Equating z equations:
-4 + t = -25 - 6s
Step 3: Solve the system of equations to find values for t and s:
From the x equation:
3t - 6s = -25
From the y equation:
t - 9s = 34
From the z equation:
t + 6s = -21
We can solve this system of equations using any method, such as substitution or elimination. In this case, let's use the elimination method.
Multiply the first equation by 2:
6t - 12s = -50
Subtract the second equation from the first:
8t + 3s = -16 ---> (Equation A)
Multiply the third equation by 3:
3t + 18s = -63
Subtract the second equation from the third:
2t - 27s = -55 ---> (Equation B)
Multiply Equation A by 2:
16t + 6s = -32 ---> (Equation C)
Add Equation B and Equation C:
18t - 21s = -87
Solving the above equation for t:
t = (-87 + 21s) / 18
Step 4: Substitute the value of t into any of the original parametric equations (we'll use the x equation):
x = -2 - 3t
x = -2 - 3[(-87 + 21s) / 18]
x = -2 + (87 - 21s) / 6
x = (85 - 21s) / 6
Thus, the x-coordinate of the point of intersection is (85 - 21s) / 6.
Step 5: Substitute the value of t into the y and z equations (we'll use the y equation):
y = -3 + t
y = -3 + (-87 + 21s) / 18
y = (-69 + 21s) / 18
Thus, the y-coordinate of the point of intersection is (-69 + 21s) / 18.
Similarly, substituting the value of t into the z equation:
z = -4 + t
z = -4 + (-87 + 21s) / 18
z = (-67 + 21s) / 18
Thus, the z-coordinate of the point of intersection is (-67 + 21s) / 18.
So, the point of intersection Q is given by Q((85 - 21s) / 6, (-69 + 21s) / 18, (-67 + 21s) / 18) if the values of t and s satisfy the system of equations.