Let L1 be the line passing through the points Q1=(4, 2, −3) and Q2=(0, −2, 3). Find a value of k so the line L2 passing through the point P1 = P1(−11, 2, k) with direction vector →d=[3, −2, −3]T intersects with L1
don't reduce, the rest should be fine
To find the value of k such that the line L2 intersects with L1, we need to find the point of intersection.
First, let's find the equation of the line L1.
The direction vector of L1 is given by the difference of the two given points:
→v = Q2 - Q1 = [0, -2, 3] - [4, 2, -3] = [-4, -4, 6]
The parametric equation of the line L1 can be written as:
x = 4 - 4t
y = 2 - 4t
z = -3 + 6t
Next, let's find the equation of the line L2.
The direction vector of L2 is given as →d = [3, -2, -3]T.
The parametric equation of the line L2 can be written as:
x = -11 + 3s
y = 2 - 2s
z = k - 3s
To find the point of intersection, we need to equate the x, y, and z coordinates of L1 and L2.
Equating x:
4 - 4t = -11 + 3s (Equation 1)
Equating y:
2 - 4t = 2 - 2s (Equation 2)
Equating z:
-3 + 6t = k - 3s (Equation 3)
From Equation 2, we can see that 4t = 2s.
Substituting this value in Equation 1, we get:
4 - 2s = -11 + 3s
Rearranging:
5s = 15
Solving for s:
s = 3
Substituting this value of s in Equation 2:
2 - 4t = 2 - 6
Simplifying:
4t = 4
Solving for t:
t = 1
Now that we have the values of t and s, we can substitute them into Equation 3 to find the value of k:
-3 + 6(1) = k - 3(3)
Simplifying:
-3 + 6 = k - 9
9 = k - 9
k = 18
Therefore, the value of k such that the line L2 intersects with L1 is k = 18.
To find the value of k for which the line L2 intersects with L1, we need to see if there is a point that is on both lines.
First, let's find the parametric equations for L1. The direction vector of L1 can be found by subtracting the coordinates of Q1 from Q2:
→d1 = Q2 - Q1 = (0, -2, 3) - (4, 2, -3) = (-4, -4, 6).
Now, let's set up the parametric equations for L1:
x = 4 - 4t,
y = 2 - 4t,
z = -3 + 6t.
Next, we need to find the parametric equations for L2. We are given the point P1(-11, 2, k) and the direction vector →d2 = [3, −2, −3]T for L2. We can set up the parametric equations for L2:
x = -11 + 3t,
y = 2 - 2t,
z = k - 3t.
To find the intersection point, we need to find values of t that satisfy both sets of parametric equations. Equating corresponding components, we get:
4 - 4t = -11 + 3t,
2 - 4t = 2 - 2t,
-3 + 6t = k - 3t.
Simplifying these equations, we have:
7t = 15,
-2t = 0,
9t = k + 3.
From the second equation, we find t = 0. Substituting this value into the first equation, we get:
4 - 4(0) = -11 + 3(0),
4 = -11.
Since this equation is not true, there is no intersection when t = 0. Therefore, the line L2 does not intersect with L1 for any value of k.
Direction of L1 = <4,4,-6> or reduced to <2,2,-3>
direction of L2 = <3,-2,-3> , that was given
If L1 and L2 intersect they must lie on the same plane.
The normal of that plane is the cross product of <2,2,-3> with <3,-2,-3>
Using whatever method you learned, that normal is <12,3,10>
so the equation of the plane is 12x + 3y + 10z = k
but the point (4, 2, −3) lies on it, so 48 + 6 - 30 = k ----> k = 24
( I could have used the point (0, −2, 3) to get 0 -6 + 30 = k , k = 24)
So the equation of the plane containing our two lines is
12x + 3y + 10z = 24
That also contains any point on either line, and (-11,2,k) is supposed to be on it, so
-132 + 6 + 10k = 24
10k = 150
k = 15
check my arithmetic