Does the point of intersection between L1:(4,9,2)+s(1,-1,-1), seR and Ls: (2,11,4)+t(5,-3,-4), teR lie on the plane 3x-2y+3z+4=0?
To determine whether the point of intersection between the two lines lies on the given plane, we need to find the coordinates of the point of intersection and then substitute them into the equation of the plane to check if it satisfies the equation.
First, let's find the point of intersection between the two lines. We need to find values for s and t that make the parametric equations of the lines equal to each other.
For L1:
x1 = 4 + s
y1 = 9 - s
z1 = 2 - s
For Ls:
x2 = 2 + 5t
y2 = 11 - 3t
z2 = 4 - 4t
To find the point of intersection, we need to set the x, y, and z values equal to each other and solve the resulting equations:
4 + s = 2 + 5t
9 - s = 11 - 3t
2 - s = 4 - 4t
From the first equation, we can isolate s:
s = 2 + 5t - 4
Substituting this value of s into the second equation:
9 - (2 + 5t - 4) = 11 - 3t
9 - 2 - 5t + 4 = 11 - 3t
11 - 5t = 11 - 3t
Simplifying the equation, we get:
-5t + 4 = -3t
Adding 5t to both sides:
4 = 2t
Dividing by 2:
t = 2
Now that we have the value of t, we can substitute it back into the equation for s that we found earlier:
s = 2 + 5t - 4
s = 2 + 5(2) - 4
s = 2 + 10 - 4
s = 8
So the values of s and t that make the parametric equations of the lines equal to each other are s = 8 and t = 2.
Now, let's find the coordinates of the point of intersection by substituting these values into the parametric equations of either line. Let's use L1:
x = 4 + s = 4 + 8 = 12
y = 9 - s = 9 - 8 = 1
z = 2 - s = 2 - 8 = -6
Therefore, the point of intersection between the two lines is (12, 1, -6).
Next, we need to check if this point satisfies the equation of the given plane, 3x - 2y + 3z + 4 = 0. We substitute the coordinates of the point into the equation:
3(12) - 2(1) + 3(-6) + 4 = 0
36 - 2 + (-18) + 4 = 0
36 - 2 - 18 + 4 = 0
20 - 20 = 0
Since the equation is true, the point of intersection (12, 1, -6) lies on the plane 3x - 2y + 3z + 4 = 0.