Let L1 be the line passing through the points Q1=(4, 1, 2) and Q2=(0, −1, 4) and let L2 be the line passing through the point P1=(20, 8, −7) with direction vector →d=[−3, −1, 2]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.
umhhh, why not look back at your earlier posts before reposting the same question ?
http://www.jiskha.com/display.cgi?id=1445204064
Ah, lines intersecting...a classic case of "Love at first Coordinate"! Let's see if these two lines have a meeting point or if they're destined to go their separate ways.
First, let's find the equations of the two lines:
For L1, the vector form of the line equation is given by:
⎡ x ⎤ ⎡ 4 ⎤ ⎡ 4 − 0 ⎤
⎢ y ⎥ = ⎢ 1 ⎥ + t ⎢ 1 − (-1) ⎥
⎣ z ⎦ ⎣ 2 ⎦ ⎣ 2 − 4 ⎦
Simplifying this, we get:
x = 4 + 4t
y = 1 + t
z = 2 - 2t
Now, for L2, the vector form of the line equation is given by:
⎡ x ⎤ ⎡ 20 ⎤ ⎡ -3t ⎤
⎢ y ⎥ = ⎢ 8 ⎥ + t ⎢ -t ⎥
⎣ z ⎦ ⎣ -7 ⎦ ⎣ 2t ⎦
Simplifying this, we get:
x = 20 - 3t
y = 8 + t
z = -7 + 2t
Now, let's see if there's any common ground for these two lines! We'll set the x, y, and z coordinates equal to each other and solve for t:
4 + 4t = 20 - 3t
1 + t = 8 + t
2 - 2t = -7 + 2t
Simplifying these equations, we get:
7t = 16
0 = 7
4t = 9
Oops! That last equation seems to be missing a solution. So, it looks like these two lines never intersect. They're just not meant to be.
Keep looking for that perfect point of intersection! Don't worry, it's out there somewhere.
To determine whether L1 and L2 intersect, we can set up a system of equations using the parameterization of each line.
The parameterization of line L1 is given by:
x = 4 - t
y = 1 + t
z = 2 + 2t
The parameterization of line L2 is given by:
x = 20 - 3s
y = 8 - s
z = -7 + 2s
To find the point of intersection Q, we need to find values of t and s that satisfy the equations for both lines simultaneously.
Setting the x, y, and z coordinates of L1 and L2 equal to each other, we have:
For x:
4 - t = 20 - 3s
For y:
1 + t = 8 - s
For z:
2 + 2t = -7 + 2s
We can solve this system of equations to find the values of t and s that satisfy all three equations.
1. From the first equation, we get:
t = 16 - 3s
2. Substituting t = 16 - 3s into the second equation, we get:
1 + 16 - 3s = 8 - s
17 - 3s = 8 - s
-2s = -9
s = 4.5
3. Substituting s = 4.5 into t = 16 - 3s, we get:
t = 16 - 3(4.5)
t = 16 - 13.5
t = 2.5
Now that we have the values of t and s, we can plug them back into the parameterization equations for L1 and L2 to find the point of intersection Q.
For L1:
x = 4 - t = 4 - 2.5 = 1.5
y = 1 + t = 1 + 2.5 = 3.5
z = 2 + 2t = 2 + 2(2.5) = 7
Therefore, the point Q of intersection between L1 and L2 is Q(1.5, 3.5, 7).
To determine whether L1 and L2 intersect, we need to check if there is a common point that lies on both lines. We can find this point by solving the equations of the lines.
The equation of a line in three-dimensional space can be expressed in parametric form as:
L: P = P0 + t * →d
Where:
- P is a point on the line L.
- P0 is a known point on the line L.
- →d is the direction vector of the line L.
- t is a parameter that determines the position of the point P on the line L.
For L1, we have:
P1 = (4, 1, 2) + s * →v1 (Equation 1)
Where:
- P1 is a point on line L1.
- s is a parameter that determines the position of the point P1 on the line L1.
- →v1 is the direction vector of line L1, which can be calculated as →v1 = Q2 - Q1.
Substituting the known values, we have:
P1 = (4, 1, 2) + s * (0, -2, 2)
P1 = (4, 1, 2) + s * (0, -2, 2)
For L2, we have:
P2 = (20, 8, -7) + t * →d (Equation 2)
Where:
- P2 is a point on line L2.
- t is a parameter that determines the position of the point P2 on line L2.
- →d is the given direction vector of line L2.
Substituting the known values, we have:
P2 = (20, 8, -7) + t * (-3, -1, 2)
P2 = (20, 8, -7) + t * (-3, -1, 2)
To find if L1 and L2 intersect, we can equate the points P1 and P2 and solve for s and t:
(4, 1, 2) + s * (0, -2, 2) = (20, 8, -7) + t * (-3, -1, 2)
By comparing the corresponding components, we get the following system of equations:
0s + 4 = -3t + 20 (Equation 3)
-2s + 1 = -t + 8 (Equation 4)
2s + 2 = 2t - 7 (Equation 5)
By solving this system of equations, we can find the values of s and t. If there is a solution, then L1 and L2 intersect at a point Q, and we can find the coordinates of Q by substituting the values of s or t back into one of the line equations.
Please note that I have provided the steps to solve for the intersection point. You can now use any algebraic solving method such as substitution, elimination, or matrix operations to find the solution for s and t.