Find a line through (xo, yo, zo) and normal to the plane {(x, y, z): ax + by + cz = d}
The equation of a line passing through (xo, yo, zo) and normal to the plane {(x, y, z): ax + by + cz = d} can be found by taking the normal vector of the plane and using it as the direction vector of the line.
The normal vector of the plane is given by (a, b, c).
So, the equation of the line is:
x = xo + at
y = yo + bt
z = zo + ct
where t is a parameter.
This line is parallel to the plane and perpendicular to its normal vector.
To find a line through the point (xo, yo, zo) that is normal to the plane {(x, y, z): ax + by + cz = d}, follow these steps:
Step 1: Determine the direction vector of the line
Let the direction vector of the line be (a, b, c), which is perpendicular to the plane. Since the plane is given by ax + by + cz = d, the coefficients of x, y, and z in the plane equation give the components of the direction vector. Therefore, the direction vector is (a, b, c).
Step 2: Find a second point on the line
To find a second point on the line, you can either choose any arbitrary point or use the equation of the plane to find one. Let's use the equation of the plane to find the second point.
Substitute xo, yo, and zo into the equation of the plane:
axo + byo + czo = d
Now solve for one variable, let's say x:
axo = d - byo - czo
x = (d - byo - czo) / a
So, the second point on the line is ((d - byo - czo) / a, yo, zo).
Therefore, the line through (xo, yo, zo) and normal to the plane {(x, y, z): ax + by + cz = d} can be described by the parametric equations:
x = xo + at
y = yo + bt
z = zo + ct
where (a, b, c) is the direction vector of the line.