Let the line L1 passes through the points (5,1,7) and (6,0,8) and the line L2 passes through the points (3,1,3) and . Find the value of for which the two lines intersect.
To find the intersection point of two lines, we need to find the values of x, y, and z that satisfy the equations of both lines.
Let's start with the equation for line L1. The direction vector of L1 can be found by subtracting the coordinates of the two points on the line:
dL1 = (6-5, 0-1, 8-7) = (1, -1, 1)
Using the point-slope form of a line, the equation for L1 can be written as:
L1: (x, y, z) = (5, 1, 7) + t(1, -1, 1) ... (1)
where t is a parameter.
Similarly, we can find the equation for L2. The direction vector of L2 is:
dL2 = (3- ,1,3) = (- ,0,- )
Using the point-slope form, the equation for L2 can be written as:
L2: (x, y, z) = (3, 1, 3) + s(- , 0, - ) ... (2)
where s is a parameter.
Now we can set the x, y, and z values of L1 and L2 equal to each other and solve for t and s:
5 + t = 3 -
1 - t = 1 + 0
7 + t = 3 -
From the second equation, we get t = 0. Substituting this into the first and third equations, we have:
5 = 3 -
7 = 3 -
Solving these equations, we find that = 0.5.
Therefore, the two lines intersect when = 0.5.
To find the value of λ for which the two lines L1 and L2 intersect, we need to set the parametric equations of the two lines equal to each other.
Let's start by finding the parametric equations for the two lines:
Line L1: P1 = (5, 1, 7), P2 = (6, 0, 8)
Let the parametric equation for L1 be:
x = 5 + λ(d1x) (1)
y = 1 + λ(d1y) (2)
z = 7 + λ(d1z) (3)
Line L2: P3 = (3, 1, 3), P4 = (a, b, c)
Let the parametric equation for L2 be:
x = 3 + μ(d2x) (4)
y = 1 + μ(d2y) (5)
z = 3 + μ(d2z) (6)
Since the two lines intersect, the point of intersection will satisfy the parametric equations of both lines.
Now, we can set the x, y, and z values of equation (1), (2), and (3) equal to the x, y, and z values of equation (4), (5), and (6) respectively, and solve for λ and μ.
Setting equation (1) and equation (4) equal:
5 + λ(d1x) = 3 + μ(d2x)
Similarly, setting equation (2) and equation (5) equal, and equation (3) and equation (6) equal, we have:
1 + λ(d1y) = 1 + μ(d2y)
7 + λ(d1z) = 3 + μ(d2z)
From equation (1) and equation (4), we get:
λ(d1x) - μ(d2x) = -2 (7)
From equation (2) and equation (5), we get:
λ(d1y) - μ(d2y) = 0 (8)
From equation (3) and equation (6), we get:
λ(d1z) - μ(d2z) = -4 (9)
Now, substitute the values of the points P1, P2, P3, and P4 into equations (7), (8), and (9):
(5-6)λ - aμ = -2 (10)
(1-0)λ - bμ = 0 (11)
(7-8)λ - cμ = -4 (12)
Now we have three equations with two unknowns (λ and μ).
To solve for the unknowns, we can solve the system of equations using matrix methods or substitution, which I will do using substitution:
From equation (11), we have:
λ - bμ = 0
Solving for λ, we get:
λ = bμ (13)
Now substitute the value of λ from equation (13) into equation (10):
(5-6)(bμ) - aμ = -2
-bμ - aμ = -2
(-b-a)μ = -2
Simplifying, we have:
μ = 2/(a+b) (14)
Now substitute the value of μ from equation (14) into equation (13):
λ = b(2/(a+b))
λ = 2b/(a+b) (15)
Therefore, the value of λ for which the two lines intersect is 2b/(a+b).