Let L1 be the line passing through the points Q1=(4, −1, −2) and Q2=(1, 0, −1) and let L2 be the line passing through the point P1=(−6, 21, −8) with direction vector →d=[−3, 9, −3]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.
screw lyryx
hi wahts up
To determine whether L1 and L2 intersect, we need to solve the system of equations formed by the parametric equations for the lines L1 and L2. The parametric equation of a line in 3D is given by:
L1: x = 4 + t(1-4)
y = -1 + t(0+1)
z = -2 + t(-1+2)
L2: x = -6 + s(-3)
y = 21 + s(9)
z = -8 + s(-3)
To find the point of intersection Q, we need to solve the system of equations formed by equating the x, y, and z components of L1 and L2 and solving for the values of t and s.
Equating the x-components:
4 + t(1-4) = -6 + s(-3)
Equating the y-components:
-1 + t(0+1) = 21 + s(9)
Equating the z-components:
-2 + t(-1+2) = -8 + s(-3)
Simplifying these equations, we get:
-3t + 3s = -10 ...(1)
t + s = 22 ...(2)
t = 6s - 6 ...(3)
Now, we can substitute equation (3) into equation (2) and solve for s:
6s - 6 + s = 22
7s - 6 = 22
7s = 28
s = 4
Substituting the value of s back into equation (3), we can solve for t:
t = 6(4) - 6
t = 18
Therefore, the values of s and t are 4 and 18, respectively.
Now, we can substitute these values of s and t into the parametric equation for L2 to find the point of intersection Q:
x = -6 + 4(-3)
= -6 - 12
= -18
y = 21 + 4(9)
= 21 + 36
= 57
z = -8 + 4(-3)
= -8 - 12
= -20
Therefore, the point of intersection Q is Q=(-18, 57, -20).
To determine whether the lines L1 and L2 intersect, we need to check if there is a common point that lies on both lines. If such a point exists, it will be the point of intersection Q.
Line L1 is defined by two points Q1 and Q2, while line L2 is defined by a point P1 and a directional vector →d.
To find the intersection point Q, we can set up a system of equations. We want the coordinates of Q to satisfy both the line equation for L1 and the line equation for L2 simultaneously.
1. Equation for L1:
Since L1 passes through Q1=(4, −1, −2) and Q2=(1, 0, −1), we can define L1 with the parametric equation:
x = 4 - t (1)
y = -1 + t (2)
z = -2 + t (3)
where t is a parameter. These equations describe a line in 3D space.
2. Equation for L2:
Since L2 passes through P1=(−6, 21, −8) with the direction vector →d=[−3, 9, −3]T, we can define L2 with the parametric equation:
x = -6 - 3t (4)
y = 21 + 9t (5)
z = -8 - 3t (6)
where t is also a parameter. These equations describe another line in 3D space.
To find the point of intersection Q, we can set the corresponding coordinates of L1 and L2 equal to each other:
4 - t = -6 - 3t (7)
-1 + t = 21 + 9t (8)
-2 + t = -8 - 3t (9)
Now, we can solve this system of equations to find the value of t.
From equation (7):
4 - t = -6 - 3t
Add t to both sides:
4 = -6 - 2t
Add 6 to both sides:
10 = -2t
Divide by -2:
t = -5
Using this value of t, we can substitute it into equations (1), (2), and (3) to find the coordinates of point Q:
x = 4 - (-5) = 9
y = -1 + (-5) = -6
z = -2 + (-5) = -7
Therefore, the point of intersection Q is Q=(9, -6, -7).
In conclusion, the lines L1 and L2 do intersect at the point Q=(9, -6, -7).