You are given the two lines L1: x= 4 + 2t, y = 4 + t, z = -3 - t, and
L2: x= -2 + 3s, y= -7 + 2s, z = 2 - 3s, If the point P1 lies on L1
and the point P2 lies on L2, determine the coordinates of these two points
if P1P2 is perpendicular to each of the two lines.
Don't really know what to do, I got the direction vector of each point and got their dot product (-1,3,1)
One of the classic problems in vector geometry.
In effect you are given two skew lines, (they don't intersect in 3D)
and you are finding the line of shortest distance between them.
We can label P1(4+2t, 4+t, -3-t) and P2(-2+3s, -7+2s, 2-3s)
then vector P2P1 = (-2+3s-4-2t, -7+2s-4-t, 2-3s+3+t)
= (3s-2t-6 , 2s-t-11 , -3s+t+5)
We then get two dot product equations, since this vector is perpendicular
with each of the lines:
(3s-2t-6 , 2s-t-11 , -3s+t+5) dot (2,1,-1) = 0
6s -4t -12 + 2s - t - 11 + 3s - t - 5 = 0
11s -6t = 28
(3s-2t-6 , 2s-t-11 , -3s+t+5) dot (3,2,-3) = 0
9s -6t -18 + 4s - 2t -22 + 9s - 3t - 15 = 0
22s -5t = 55
solve for s and t, sub into the corresponding definitions for P1 and P2
and you are done.
(I got s = 190/77 and t = -1/7
was expecting "nicer" results, so I might have made an arithmetic error.
Better check my calculations)
You are right with your steps, but you have indeed made an arithmetic error on the step when simplifying 9s -6t -18 + 4s - 2t -22 + 9s - 3t - 15 = 0. The correct answer should be s=3 and t=1
Should be s=2 and t=-1
cross product*
Well, it sounds like you're on the right track with finding the direction vectors and calculating the dot product! Now, let's use that information to determine the coordinates of the two points.
Since P1 lies on L1, we can substitute the coordinates of P1 into the equation of L1:
x = 4 + 2t
y = 4 + t
z = -3 - t
Now, let's say the coordinates of P1 are (x1, y1, z1). Substituting the values from L1, we have:
x1 = 4 + 2t
y1 = 4 + t
z1 = -3 - t
Similarly, since P2 lies on L2, we can substitute the coordinates of P2 into the equation of L2:
x = -2 + 3s
y = -7 + 2s
z = 2 - 3s
Let's say the coordinates of P2 are (x2, y2, z2). Substituting the values from L2, we have:
x2 = -2 + 3s
y2 = -7 + 2s
z2 = 2 - 3s
Now, we know that P1P2 is perpendicular to both L1 and L2. This means that the direction vector of P1P2 must be orthogonal, or perpendicular, to the direction vectors of L1 and L2.
You found the direction vectors of L1 and L2 to be (-1, 3, 1). Since P1P2 is perpendicular to both L1 and L2, the dot product between the direction vector of P1P2 and the direction vectors of L1 and L2 should be zero:
(-1, 3, 1) · (-1, 3, 1) = 0
This gives us the following equation:
(-1)(-1) + (3)(3) + (1)(1) = 0
1 + 9 + 1 = 0
11 = 0
Haha, looks like there's a little problem here! Since 11 does not equal 0, it seems we made a mistake somewhere in our calculations. I apologize for the confusion. Let me try to help you find the correct solution.
Unfortunately, I can't derive the solution myself, but I can recommend that you double-check your calculations and make sure you've correctly found the direction vectors and performed the dot product calculations. If you're still having trouble, it might be helpful to consult a math tutor or teacher for further assistance. Good luck!
To determine the coordinates of points P1 and P2, we need to find the values of t and s that satisfy the conditions given in the problem.
First, let's find the direction vectors of L1 and L2:
Direction vector of L1 = ⟨2, 1, -1⟩
Direction vector of L2 = ⟨3, 2, -3⟩
Next, we can find the dot product of the direction vectors with the vector P1P2 to check if they are perpendicular:
Vector P1P2 = ⟨x2 - x1, y2 - y1, z2 - z1⟩
= ⟨(-2 + 3s) - (4 + 2t), (-7 + 2s) - (4 + t), (2 - 3s) - (-3 - t)⟩
= ⟨-6 - 2t + 3s, -11 + t + 2s, 5 - 3t + 3s⟩
To find if P1P2 is perpendicular to L1, take the dot product of P1P2 and the direction vector of L1:
⟨-6 - 2t + 3s, -11 + t + 2s, 5 - 3t + 3s⟩ · ⟨2, 1, -1⟩ = 0
Solving this equation will give us the value of t.
Similarly, to find if P1P2 is perpendicular to L2, take the dot product of P1P2 and the direction vector of L2:
⟨-6 - 2t + 3s, -11 + t + 2s, 5 - 3t + 3s⟩ · ⟨3, 2, -3⟩ = 0
Solving this equation will give us the value of s.
Once we have the values of t and s, we can substitute them into the equations of L1 and L2 to find the coordinates of P1 and P2.