The normal vector of the plane is (1, -1, 0).
To find a line through the point po and normal to the plane, we can use the point-normal form of the equation of a line:
r = po + t * n
where r is a point on the line, po is the given point, t is a scalar parameter, and n is the normal vector.
Substituting the values, we have:
r = (-5, 2, 1) + t * (1, -1, 0)
Simplifying, we get:
r = (-5 + t, 2 - t, 1)
Therefore, the line through the point po = (-5, 2, 1) and normal to the plane {(x, y, z): x = y} is given by the equation:
x = -5 + t
y = 2 - t
z = 1
where t is a scalar parameter.
Find a line through the point po = (-5, 2, 1) and normal to the plane
{(x, y, z): x = y}
1 answer