Prove that if B.N = 0 & A is on the plane {P(P-p1).N = 0} then the entire line {p:p = A+tB} lies in the plane
We want to prove that every point on the line {p:p = A+tB} satisfies the equation {P(P-p1).N = 0}, given that B.N = 0 and A is on the plane {P(P-p1).N = 0}.
Let p = A+tB be any point on the line. Then we have:
P(P-p1) = P(A+tB-p1) = P(A-p1) + tP(B) - P(p1)
Since A is on the plane, we know that P(A-p1).N = 0. Also, we are given that B.N = 0, which means that P(B).N = 0. Therefore, the above equation simplifies to:
P(P-p1).N = (P(A-p1) + tP(B) - P(p1)).N = P(A-p1).N - P(p1).N
Since P(A-p1).N = 0 and B.N = 0, we have:
P(P-p1).N = - P(p1).N
But we know that p1 is on the plane, so P(p1).N = 0. Therefore, we have:
P(P-p1).N = 0
This shows that every point on the line {p:p = A+tB} satisfies the equation {P(P-p1).N = 0}, which means that the entire line lies in the plane {P(P-p1).N = 0}.