Let L1 be the line passing through the points Q1=(3, −3, 4) and Q2=(−3, 6, 7). Find a value of k so the line L2 passing through the point P1 = P1(−6, 11, k) with direction vector →d=[1, −1, −2]T intersects with L1.
Not sure how to go on..
It is the different type of question though. The point P1 includes k
I did this for you here, you even thanked me for it
??
https://www.jiskha.com/questions/1768758/Let-L1-be-the-line-passing-through-the-points-Q1-2-3-4-and-Q2-5
I also answered your other question from yesterday,
you answer was incorrect.
https://www.jiskha.com/questions/1768663/The-parametric-equations-for-a-line-L1-are-as-follows-x-1-2t-y-4-4t-z
To find the value of k such that the line L2 intersects with L1, we need to find the coordinates of the intersection point of the two lines.
Step 1: Write the parametric equations of L1 and L2.
L1: x = 3 + t(-3 - 3), y = -3 + t(6 + 3), z = 4 + t(7 - 4)
x = 3 - 6t, y = -3 + 9t, z = 4 + 3t
L2: x = -6 + s(1), y = 11 + s(-1), z = k + s(-2)
x = -6 + s, y = 11 - s, z = k - 2s
Step 2: Set up the equations for the coordinates of the intersection point (x, y, z).
For the x-coordinates:
3 - 6t = -6 + s
6t + s = 9
For the y-coordinates:
-3 + 9t = 11 - s
-9t - s = 14
For the z-coordinates:
4 + 3t = k - 2s
2s + 3t - k = -4
Step 3: Solve the system of equations for t, s, and k.
To solve the system of equations, we can use any method such as substitution or elimination. Here, we will use the method of elimination.
From the first two equations:
6t + s = 9 (Equation 1)
-9t - s = 14 (Equation 2)
Adding Equation 1 and Equation 2, we get:
-3t = 23
Dividing both sides by -3, we find:
t = -23/3
Substituting t into Equation 1:
6(-23/3) + s = 9
-38 + s = 9
s = 47
Substituting t = -23/3 and s = 47 into Equation 3:
2(47) + 3(-23/3) - k = -4
94 - 23 - k = -4
k = 94 - 19 + 4
k = 79
Therefore, the value of k such that the line L2 intersects with L1 is k = 79.