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Prove that three planes with independent normal vectors intersect in exactly one point

Prove that three planes with normal vectors which are linearly independent intersect in exactly one point.


Proof. Let the normals of the three planes be given by

    \begin{align*}  N_1 &= (a_1, b_1, c_1) \\   N_2 &= (a_2, b_2, c_2) \\  N_3 &= (a_3, b_3, c_3). \end{align*}

Then, the Cartesian equations of the three planes are given by

    \begin{align*}  a_1 x + b_1 y + c_1 z &= d_1 \\  a_2 x + b_2 y + c_2 z &= d_2 \\  a_3 x + b_3 y + c_3 z &= d_3. \end{align*}

Since the normals are independent we know that they span the zero vector uniquely (by definition of independence). BY Theorem 12.7 (page 463 of Apostol) this implies that they span every vector in \mathbb{R}^3. Hence, the vector equation x N_1  + y N_2 + z N_3 = (d_1, d_2, d_3) has a unique solution (x,y,z). Therefore, the system of equations has a unique solution. Hence, there is exactly one point (x,y,z) on the intersection of three planes with linearly independent normal vectors. \qquad \blacksquare

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