Plane through a Point with a given Normal Vector:
If you find this page helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor
Find the plane, P, through A(2, -3, -4) with normal vector n = ai + bj + ck = 4i - j + 3k . See the diagram.
Here we can use the scalar equation of the plane:
a*[x - xA] + b*[y - yA] + c*[z - zA] = 0, where A(xA, yA, zA) is the required point and a, b, c are the components of the normal vector.
Upon substitution, we get
(4)*[x - 2] + (-1)*[y - (-3)] + (3)*[z - (-4)] = 0.
Simplifying, we get
4x - y + 3z = -1.
If you find this page helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor