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A plane can be taken to be a quadruple $(a, b, c, d)$. Then a point is taken to be a triple of planes $(P, Q, R)$. The point $(p, q, r)$ is valid if and only if the determinant of the matrix $A$ below is non-zero.

$$A= \left[ \begin{array}{ccc} Pa & Pb & Pc\\ Qa & Qb & Qc\\ Ra & Rb & Rc\\ \end{array} \right] $$

We can get it by simple linear algebra: The point $(P, Q, R)$ can be presented as the equation root of $Ax = b$. If this is valid then $Ax = b$ has an unique solution which implies $\det(A) \neq 0$.

Given that the point $(P, Q, R)$ is valid, it lies behind, on, or in-front of the plane $S$ if and only if the following expression is negative, zero or positve.

$$ \left\vert \begin{array}{cccc} Pa & Pb & Pc & Pd\\ Qa & Qb & Qc & Qd\\ Ra & Rb & Rc & Rd\\ Sa & Sb & Sc & Sd \end{array} \right\vert \left\vert \begin{array}{ccc} Pa & Pb & Pc\\ Qa & Qb & Qc\\ Ra & Rb & Rc \end{array} \right\vert $$

My question is: What is the mechanism behind this?

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I suggest taking a course on linear algebra. –  Mehrdad Dec 29 '11 at 4:50
2  
Can you specify what your first sentence means? Do you associate $(a, b, c, d)$ with the plane $ax + by + cz = d$? What do you mean by "valid"? –  Qiaochu Yuan Dec 29 '11 at 6:06
    
Do you associate (a,b,c,d) with the plane ax+by+cz=d? YES What do you mean by "valid"? That means (P, Q, R) is the only point shared by 3 planes. –  CChen Dec 29 '11 at 6:40
    
The Question has a type error: det(A) = 0 should be det(A) != 0 –  CChen Dec 29 '11 at 6:58
    
But for your matrix $A$ we always have $\det A=0$? Where did $d$ disappear? –  Jyrki Lahtonen Dec 29 '11 at 7:40
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1 Answer 1

The only way the question makes sense to me is that $Pa$ actually means $a_P$ instead of the product of the two numbers $P$ and $a$ as I would normally expect. IOW, the coefficient of $x$ in the equation of the plane $P$. Similarly for others. So we are given 4 planes determined by the four equations $$ a_ix+b_iy+c_iz=d_i, $$ where the subscript $i$ takes values from the set $\{P,Q,R,S\}$. From the problem description we know that the first three planes are aligned in such a way that they have a unique point of intersection $(x_0,y_0,z_0)$. Let us look at the augmented matrix $$ M=\left(\begin{array}{cccc}a_P&b_P&c_P&d_P\\a_Q&b_Q&c_Q&d_Q\\a_R&b_R&c_R&d_R\end{array}\right). $$ Let $M_j, j\in\{a,b,c,d\}$ be the determinant of the 3x3 matrix gotten from $M$ by removing the column corresponding to $j$. So $\det A$ from the OP is now equal to $M_d$, and we know that $M_d\neq0$.

Cramer's rule tells us that the point of intersection has coordinates $$ x_0=\frac{M_a}{M_d},\quad y_0=-\frac{M_b}{M_d},\quad z_0=\frac{M_c}{M_d}.$$

Let us then compute the determinant of the 4x4 matrix $$ N=\left(\begin{array}{cccc}a_P&b_P&c_P&d_P\\a_Q&b_Q&c_Q&d_Q\\a_R&b_R&c_R&d_R\\ a_S&b_S&c_S&d_S\end{array}\right). $$ We expand along the bottom row and get $$ \det N=-a_SM_a+b_SM_b-c_SM_c+d_SM_d=-a_S(x_0M_d)-b_S(y_0M_d)-c_S(z_0M_d)+d_SM_d, $$ so $$ \det N=-M_d(a_Sx_0+b_Sy_0+c_Sz_0-d_S). $$ Let us write $\vec{n}=(a_S,b_S,c_S)$ for this normal of the plane $S$. Let $\vec{r}=(x_S,y_S,z_S)$ be the vector from the origin to some point $(x_S,y_S,z_S)$ on the plane $S$, so we know that $\vec{n}\cdot \vec{r}=d_S.$ Let $\vec{r}_0=(x_0,y_0,z_0)$ be the vector from the origin to the point of intersection $(x_0,y_0,z_0)$. Then $a_Sx_0+b_Sy_0+c_Sz_0=\vec{n}\cdot\vec{r}_0$, so our earlier equation reads $$ \det N=-\det A (\vec{n}\cdot \vec{r_0}-\vec{n}\cdot\vec{r})=(\det A)\ \vec{n}\cdot(\vec{r}-\vec{r}_0). $$

This tells us that the sign of the product $\det N \ \det A$ is the same as the sign of the dot product $\vec{n}\cdot(\vec{r}-\vec{r}_0)$. A geometric interpretation of this last dot produt is that it is positive/zero/negative, if the projection of the difference vector $\vec{r}-\vec{r}_0$ onto the chosen normal vector $\vec{n}$ is a positive/zero/negative multiple of $\vec{n}$. IOW, we get a plus sign, if the point of intersection $(x_0,y_0,z_0)$ is on the opposite side of the plane $S$ from the one pointed at by $\vec{n}$, and vice versa. Of course, the inner product equals zero, if and only if the point of intersection is on the plane $S$.

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The answer is accepted. Clear presentation and excellent deduction! Thanks, Jyrki:) –  CChen Dec 30 '11 at 3:09
    
@CChen No, the answer is not accepted, as you forgot to actually accept it. –  Christian Rau Feb 27 '12 at 16:53
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