Does the line [x,y,z] = [-2,6,5] + t[3,2,-1] lie in the plane 3x - 4y + z + 25 = 0 ? Please show work.
If the line
L: [x,y,z] = [-2,6,5] + t[3,2,-1]
lies in the plane
P: 3x - 4y + z + 25 = 0
Then both of the following must be true
1. [-2,6,5] must lie in the plane.
Check: 3(-2)-4(6)+5+25=0
so [-2,6,5] lies in the plane
2. the vector <3,2,-1> must be orthogonal to the vector <3,-4,1> of the plane.
Check: <3,-4,1>.<3,2,-1>=9-8-1=0
So the line is perpendicular to the normal of the plane.
Therefore the line lies in the plane.
Alternatively, locate any two points on the line and show that they both lie in the plane.
thank you!
You're welcome!
To check if the line lies in the given plane, we need to substitute the coordinates of the line [x, y, z] into the equation of the plane and see if the equation is satisfied for all values of t.
The equation of the plane is: 3x - 4y + z + 25 = 0
Let's substitute the coordinates of the line into the equation:
x = -2 + 3t
y = 6 + 2t
z = 5 - t
Substituting these values into the equation of the plane, we get:
3(-2 + 3t) - 4(6 + 2t) + (5 - t) + 25 = 0
Simplifying this equation:
-6 + 9t - 24 - 8t + 5 - t + 25 = 0
Combine like terms:
-6t = 10
Divide by -6:
t = -10/6
Simplifying further:
t = -5/3
Now, substitute this value of t back into the parametric equation for the line to find the coordinates at t = -5/3:
x = -2 + 3(-5/3) = -2 - 5 = -7
y = 6 + 2(-5/3) = 6 - 10/3 = 8/3
z = 5 - (-5/3) = 5 + 5/3 = 20/3
So, the coordinates of the line at t = -5/3 are (-7, 8/3, 20/3).
Now, let's substitute these coordinates into the equation of the plane to check if they satisfy it:
3(-7) - 4(8/3) + 20/3 + 25 = 0
Simplifying:
-21 - (32/3) + (20/3) + 25 = 0
Combining like terms:
-21 - (12/3) + (20/3) + 25 = 0
-21 - 4 + 20 + 25 = 0
0 = 0
Since the equation is satisfied, the line [x, y, z] = [-2, 6, 5] + t[3, 2, -1] lies on the plane 3x - 4y + z + 25 = 0.