Determine the Scalar equation of the plane that passes through the point (2, -4, -3), and is parallel to the plane 3x - 2y -4z - 9 = 0
Well , that's easy, it will differ only in the constant. The new equation will be
3x - 2y - 4z = c
Plug in the given point (2,-4,-3) to find c and you are done
would I sub 2 into x, -4 into y, and -3 into z??
To determine the scalar equation of a plane, we need to find the coefficients of its variables (x, y, and z) along with the constant term.
Given that the plane is parallel to the equation 3x - 2y - 4z - 9 = 0, we know that the normal vector (a, b, c) of the desired plane will be the same as the coefficients of x, y, and z in the parallel plane equation.
From the equation 3x - 2y - 4z - 9 = 0, we can see that a = 3, b = -2, and c = -4.
Next, we need to substitute the coordinates of the given point (2, -4, -3) into the scalar equation of the plane.
The scalar equation of a plane is given by: ax + by + cz + d = 0
Using the values obtained above, we have:
3x - 2y - 4z + d = 0
Substituting the x, y, and z values from the given point (2, -4, -3) yields:
3(2) - 2(-4) - 4(-3) + d = 0
Simplifying, we get:
6 + 8 + 12 + d = 0
26 + d = 0
Solving for d, we have:
d = -26
Therefore, the scalar equation of the plane parallel to 3x - 2y - 4z - 9 = 0 and passing through the point (2, -4, -3) is:
3x - 2y - 4z - 26 = 0