Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Justify your answer algebraically
The cross-product of the direction vectors on the given plane (1,1,-1) and (2,-1,1) is (0,3,3). ---- I assume you know a method to find that cross-product.
So the equation of the plane is 3y + 3z = k
with (3,0,2) on that plane, so
3(0) + 3(2) = k, k = 6 and the equation of the plane is
3y + 3z = 6 or
y + z = 2
let's see where
x = 5 + u
y = -4 + 4u
z = 6 - u
intersects that plane:
-4 + 4u + 6 - u = 2
3u = 0
u = 0
So the given line intersects the plane at the point (5,-4,6), so it cannot be part of the plane.
If the line is part of the plane, there would have to be an infinite number of solutions, that is, the above equation should have resulted in 0=0
another way:
To be on the plane, the direction vector of the line must be perpendicular to the normal of the plane.
We already found the normal to be (0,3,3) or (0,1,1) in reduced form
(0,1,1) dot (1,4,-1) = 9+4-1 ≠ 0
So the line intersects the plane at one point.
To determine if the line defined by the equation (x, y, z) = (5, -4, 6) + u(1, 4, -1) lies in the plane defined by (x, y, z) = (3, 0, 2) + s(1, 1, -1) + t(2, -1, 1), we can substitute the equation of the line into the equation of the plane and check whether a consistent solution exists.
Let's plug in the line equation into the plane equation:
(x, y, z) = (3, 0, 2) + s(1, 1, -1) + t(2, -1, 1)
=> (x, y, z) = (3, 0, 2) + s(1, 1, -1) + t(2, -1, 1) + u(1, 4, -1)
Expanding this equation, we have:
x = 3 + s + 2t + u
y = s - t + 4u
z = 2 - s + t - u
Now we can compare the coefficients of x, y, and z with the original equation of the line:
x = 5 + u
y = -4 + 4u
z = 6 - u
By comparing the coefficients, we can create a system of equations:
3 + s + 2t = 5 + u (1)
s - t + 4u = -4 (2)
2 - s + t - u = 6 (3)
To check if the line lies in the plane, we need to find a solution to this system of equations. If a consistent solution exists, then the line lies in the plane; otherwise, it does not.
By rearranging equation (1), we get:
u = s + 2t - 2 (4)
Substituting equation (4) into equations (2) and (3), we have:
s - t + 4(s + 2t - 2) = -4 (5)
2 - s + t - (s + 2t - 2) = 6 (6)
Expanding equations (5) and (6), simplifying, and combining like terms, we obtain:
5s + 7t = -6 (7)
-3s - 4t = -4 (8)
We can solve equations (7) and (8) simultaneously to find the values of s and t. If a solution exists, then the line lies in the plane; otherwise, it does not.
Using any suitable method (e.g., substitution or elimination), we solve equations (7) and (8):
Multiply equation (7) by 3 and equation (8) by 5 to eliminate s:
15s + 21t = -18 (9)
-15s - 20t = -20 (10)
Summing equations (9) and (10), we get:
t = -2
Substituting t = -2 into equation (7), we have:
5s + 7(-2) = -6
5s - 14 = -6
5s = 8
s = 8/5
Now, substituting the values of s = 8/5 and t = -2 into equation (4):
u = (8/5) + 2(-2) - 2
u = 8/5 - 4 - 2
u = 8/5 - 20/5 - 10/5
u = -22/5
We have found the values of s, t, and u, indicating a consistent solution. Therefore, the line defined by the equation (x, y, z) = (5, -4, 6) + u(1, 4, -1) lies in the plane defined by (x, y, z) = (3, 0, 2) + s(1, 1, -1) + t(2, -1, 1).