Write a scalar equation of each plane, given its parametric equations.
x = 3 - 2s + 2t
y = 1 + 3s + t
z = 5 - s - 2t
The parametric equations show us two direction vectors in the plane, namely
[-2,3,-1] and [2,1,-2]
using those two we can find a normal, I will assume you know how to do that, I got
[5,6,8)
To the scalar equation would be
5x + 6y + 8z + c = 0
but (3,1,5) is a point on it, so
15 + 6 + 40 + c = 0
c = -61
equation: 5x + 6y + 8z - 61 = 0
To write a scalar equation of a plane given its parametric equations, we can follow these steps:
Step 1: Identify two direction vectors in terms of s and t from the given parametric equations. These direction vectors will lie in the plane.
In this case, the direction vectors can be formed by taking the coefficients of s and t in the parametric equations:
Direction Vector 1: < -2, 3, -1 >
Direction Vector 2: < 2, 1, -2 >
Step 2: Find the cross product of the two direction vectors to obtain the normal vector to the plane.
The cross product of Direction Vector 1 and Direction Vector 2 can be found by computing:
N = Direction Vector 1 × Direction Vector 2 = (-2i + 3j - k) × (2i + j - 2k) = 5i + 6j +7k
So, the normal vector to the plane is N = < 5, 6, 7 >.
Step 3: Substitute the coordinates of a point on the plane (x, y, z) with the normal vector (A, B, C) in the scalar equation of the plane: Ax + By + Cz = D.
Using the coordinates of the point (3, 1, 5) on the plane, we can substitute the values into the equation:
5x + 6y + 7z = D
Plugging in (3, 1, 5) we have:
5(3) + 6(1) + 7(5) = D
15 + 6 + 35 = D
D = 56
Therefore, the scalar equation of the plane is:
5x + 6y + 7z = 56