Prove that if B.N = 0 and A is on the plane {p:(p – po).N = 0} then the entire line {p:p = A + tB} lies in the plane.
To prove that the entire line {p:p = A + tB} lies in the plane {p:(p - po).N = 0}, we need to show that for any value of t, the point A + tB satisfies the equation (p - po).N = 0.
Let A + tB be any point on the line. We can write this point as p = A + tB.
Now, let's substitute this value of p into the equation of the plane:
(p - po).N = 0
(A + tB - po).N = 0
Expanding the dot product, we have:
(A - po).N + t(B.N) = 0
Since B.N = 0 (given), the equation simplifies to:
(A - po).N = 0
But we also know that A is on the plane, so we can substitute A for p:
(A - po).N = 0
Since this equation holds for any value of t, we can conclude that the entire line {p:p = A + tB} lies in the plane {p:(p - po).N = 0}.
To prove that the entire line lies in the plane, we need to show that every point on the line satisfies the equation of the plane.
Let's start with the equation of the plane: (p - po) · N = 0, where po is a point on the plane and N is the normal vector to the plane.
Now, let's consider a point on the line, p = A + tB, where A is a point on the plane and B is a vector parallel to the line.
Substituting p = A + tB into the equation of the plane, we have:
(A + tB - po) · N = 0
Expand the dot product:
(A - po) · N + t(B · N) = 0
Since A is on the plane, (A - po) · N = 0. Therefore, the equation simplifies to:
t(B · N) = 0
Now, since B · N = 0 (given in the problem statement), we have:
t(0) = 0
This shows that for any value of t, the equation t(B · N) = 0 is satisfied.
Therefore, every point on the line {p:p = A + tB} satisfies the equation of the plane (p - po) · N = 0.
Hence, we have proved that if B · N = 0 and A is on the plane {p:(p - po) · N = 0}, then the entire line {p:p = A + tB} lies in the plane.