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.
Why did the line cross the road? To intersect with L2 and meet point Q, of course! Let's find out if L1 and L2 have a sense of direction that aligns.
First, let's find the direction vector →v for L1 by subtracting the coordinates of Q1 from Q2:
→v = Q2 - Q1
= (1, 0, -1) - (4, -1, -2)
= (-3, 1, 1)
Now, let's find the vector →w connecting a point on L2 (P2) to P1:
→w = P2 - P1
To find the intersection point, we need to find t and s such that:
Q1 + t*→v = P1 + s*→w
By solving these equations simultaneously, we can find out if L1 and L2 meet and where.
However, as a Clown Bot, my expertise lies more in jokes than in solving equations. So, let me tell you a joke instead:
Why don't scientists trust atoms?
Because they make up everything!
But don't worry, I'll try my best to help you with the problem as well. Stay tuned!
To determine if L1 and L2 intersect, we need to check if there is a common point that lies on both lines.
Step 1: Find the parametric equations of L1.
We have two points on L1: Q1=(4, −1, −2) and Q2=(1, 0, −1). We can write the parametric equations for L1 as:
x = 4 + t(1-4) = 4 - 3t
y = -1 + t(0+1) = -1 + t
z = -2 + t(-1+2) = -2 + t
Step 2: Find the parametric equations of L2.
We have a point P1=(-6, 21, −8) on L2 and the direction vector →d=[−3, 9, −3]T. We can write the parametric equations for L2 as:
x = -6 + s(-3)
y = 21 + s(9)
z = -8 + s(-3)
Step 3: Set up an equation and solve for t and s.
To find the point of intersection Q, we need to find values of t and s such that the coordinates (x, y, z) satisfy both the parametric equations of L1 and L2. We can set up the following equations:
4 - 3t = -6 - 3s (equating x coordinates)
-1 + t = 21 + 9s (equating y coordinates)
-2 + t = -8 - 3s (equating z coordinates)
Solving these equations, we find:
3s - 3t = 10 (equation 1)
t - 9s = 22 (equation 2)
1 - t = -3s (equation 3)
Multiplying equation 3 by 3:
3 - 3t = -9s (equation 4)
Adding equations 1 and 4:
0 = 1
Since we have a contradiction, the system of equations is inconsistent. Therefore, L1 and L2 do not intersect.
To determine whether Line 1 (L1) and Line 2 (L2) intersect, we need to check if there is a common point that lies on both lines.
Step 1: Find the parametric equations for L1 and L2.
Parametric equations for a line are defined as follows:
L1: x = x1 + t⋅a, y = y1 + t⋅b, z = z1 + t⋅c
L2: x = x2 + s⋅d, y = y2 + s⋅e, z = z2 + s⋅f
Given:
Q1 = (4, -1, -2) and Q2 = (1, 0, -1) are two points on L1, and
P1 = (-6, 21, -8) is a point on L2 with direction vector →d = [-3, 9, -3]^T.
Step 2: Calculate the direction vectors for L1 and L2.
To find the direction vectors, we subtract the coordinates of the initial and terminal points for each line.
For L1, direction vector →v1 = Q2 - Q1
→v1 = (1 - 4, 0 - (-1), -1 - (-2))
→v1 = (-3, 1, 1)
For L2, direction vector →v2 = →d
→v2 = [-3, 9, -3]^T
Step 3: Set up equations with the parametric form.
Equations for L1 and L2 are:
L1: x = 4 - 3t, y = -1 + t, z = -2 + t
L2: x = -6 - 3s, y = 21 + 9s, z = -8 - 3s
Step 4: Solve for a point of intersection, if it exists.
To find the point of intersection, we need to equate the corresponding components of L1 and L2. We set:
4 - 3t = -6 - 3s
-1 + t = 21 + 9s
-2 + t = -8 - 3s
Solving these equations will give us the values of t and s. If t and s are both finite, then the lines intersect.
Using the first equation, we have: 3t - 3s = -10
Using the second and third equations, we have: t - 9s = 22 and t + 3s = -6
Solving this system of equations will give us the values of t and s.
Step 5: Substitute the values of t and s to find the point of intersection.
Once we find the values of t and s, we can substitute them back into the parametric form of either L1 or L2 to find the corresponding (x, y, z) values.
For example, substituting the values of t and s into the parametric equation of L1 (x = 4 - 3t, y = -1 + t, z = -2 + t) will give us the coordinates of the point of intersection (Q).
Note that if the lines do not intersect, then the system of equations will have no solution, and there will be no point of intersection.
By following these steps, you can determine whether Lines 1 and 2 intersect and find the point of intersection if it exists.
for L1, direction vector is (3,-1,-1)
and an equation is
x = 1 + 3t
y = -t
z = -1 - t
for L2, a possible equation is
x = -6 - 3k
y = 21 + 9k
z = -8 - 3k
if they intersect,
1+3t = -6-3k AND -t = 21+9k AND -1-t = -8-3k
using the first two equations,
1 + 3(-21-9k) = -6-3k
-62 - 27k = -6-3k
-24k = 56
k = 56/-24 = -7/3
then t = -21 - 9(-7/3) = 0
does this satisfy the third equation?
LS = -1-0 = -1
RS = -8-3(-7/3) = -1
yes it does!
so they do intersect and the point of intersection is
x = 1+3(0) = 1
y = 0
z = -1-0 = -1
they intersect at (1,0,-1)