Find a line through (x, y, z) & normal to the plane {(x1, y1, z1): ax + by + cz = d}
The normal vector of the plane is given by (a, b, c).
To find a line through (x, y, z) that is normal to the plane, we can use the cross product of the normal vector and a vector that starts at (x, y, z) and ends at any point on the plane. This will give us a vector that is perpendicular to both the normal vector and the vector from (x, y, z) to the plane, which means it lies on the line we are looking for.
Let's choose a point on the plane that is not (x, y, z) and call it (x1, y1, z1). The vector from (x, y, z) to (x1, y1, z1) is (x1 - x, y1 - y, z1 - z).
To find the cross product, we can use the following formula:
(u x v) = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)
where u = (a, b, c) and v = (x1 - x, y1 - y, z1 - z).
So the cross product is:
(a x (x1 - x), b x (y1 - y), c x (z1 - z))
This is the direction vector of the line we are looking for. We can then write the equation of the line in parametric form as:
x = x + t(a)
y = y + t(b)
z = z + t(c)
where t is any real number.