Find a line through (xo, yo, zo) and normal to the plane {(x, y, z): ax + by + cz = d}
A line through (xo, yo, zo) can be represented as:
x = xo + at
y = yo + bt
z = zo + ct
where a, b, and c are direction ratios of the line, and t is a parameter.
Now, we need to find a line that is normal to the plane {(x, y, z): ax + by + cz = d}. A normal vector to the plane can be represented as (a, b, c).
So, a line through (xo, yo, zo) and normal to the plane is:
x = xo + at
y = yo + bt
z = zo + ct
This line will be perpendicular to the plane as the direction ratios (a, b, c) of the line are the same as the normal vector of the plane.
To find a line that passes through the point (xo, yo, zo) and is normal to the plane {(x, y, z): ax + by + cz = d}, we need to find the direction vector of the line.
The normal vector of the plane is (a, b, c). To find the direction vector of the line, we can take the cross product of the normal vector with any other vector.
Let's choose the vector (1, 0, 0) to find the direction vector. Take the cross product of (1, 0, 0) and (a, b, c):
Direction Vector = (1, 0, 0) × (a, b, c)
= (0, c, -b)
Now we can find the equation of the line passing through (xo, yo, zo) with the direction vector (0, c, -b):
(x - xo) / 0 = (y - yo) / c = (z - zo) / -b
This can also be written as:
x - xo = 0
y - yo = c * t
z - zo = -b * t
where t is a parameter.
Therefore, the equation of the line passing through (xo, yo, zo) and normal to the plane {(x, y, z): ax + by + cz = d} is:
x = xo
y = yo + c * t
z = zo - b * t