Determine the Scalar equation of the plane that passes through the point (2, -4, -3), and contains the vectors a = [-3, 5, -1] and b = [2, -4, 6]
We need the normal to both a and b which would be the cross-product a X b
I assume you know how to find the cross-product and that you got <13,8,1> in reduced form
So the equation of the plane is 13x + 8y + z = c
but (2,-4,-3) lies on it, so
26 - 32 - 3 = c = -9
Plane: 13x + 8y + z = -9
better check my arithmetic
To determine the scalar equation of the plane that passes through the point (2, -4, -3) and contains the vectors a = [-3, 5, -1] and b = [2, -4, 6], we can use the following steps:
1. Find the normal vector to the plane:
To find the normal vector, we need to take the cross product of vector a and vector b.
Let's denote the normal vector as n, then we have:
n = a x b = [-3, 5, -1] x [2, -4, 6]
Using the cross product formula, we have:
nx = (5 * 6) - (-1 * -4) = 14
ny = (-3 * 6) - (-1 * 2) = -16
nz = (-3 * -4) - (5 * 2) = 2
Therefore, the normal vector to the plane is n = [14, -16, 2].
2. Use the point-normal form of the equation of a plane:
The scalar equation of a plane can be written as:
A(x - x0) + B(y - y0) + C(z - z0) = 0
Where A, B, and C are the coefficients of the equation, and (x0, y0, z0) is a point on the plane (in this case, the point (2, -4, -3)).
Substituting the values, we have:
14(x - 2) - 16(y - (-4)) + 2(z - (-3)) = 0
3. Simplify the equation:
Expand and simplify the equation:
14x - 28 - 16y + 64 + 2z + 6 = 0
Simplify further:
14x - 16y + 2z + 42 = 0
Hence, the scalar equation of the plane that passes through the point (2, -4, -3) and contains the vectors a = [-3, 5, -1] and b = [2, -4, 6] is:
14x - 16y + 2z + 42 = 0.
To determine the scalar equation of the plane passing through the point (2, -4, -3) and containing the vectors a = [-3, 5, -1] and b = [2, -4, 6], we can use the following steps:
1. Find the normal vector of the plane:
- Since the plane contains vectors a and b, the normal vector of the plane can be found by taking the cross product of vectors a and b:
n = a x b
- Let's calculate the cross product:
n = [-3, 5, -1] x [2, -4, 6]
Using the formula for the cross product:
n = [(5 * 6) - (-1 * -4), (-3 * 6) - (-1 * 2), (-3 * -4) - (5 * 2)]
= [34, -16, -14]
- Therefore, the normal vector of the plane is n = [34, -16, -14].
2. Write the scalar equation of the plane using the general form:
- The scalar equation of a plane is given by:
Ax + By + Cz = D
- Substitute the coordinates of the point (2, -4, -3) into the equation:
34x - 16y - 14z = D
- Multiply each coefficient by a constant to make the equation simpler:
2x - y - z = D
- Therefore, the scalar equation of the plane passing through the point (2, -4, -3) and containing the vectors a and b is:
2x - y - z = D.