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I'm supposed to be making a study guide answer for this question, but I'm struggling with proof.


Show that the three planes

plane 1

plane 2

plane 3

intersect at the point

intersection point

provided that

condition


Note that the denominator of the intersection point contains a dot product, it's just poorly formatted, my apologies.

Now, each r vector is a point in 3D space, on the plane being described, the u hat vector is the normalize normal vector, and d seems to be the z offset of the plane. I was able to determine this by poking around online, and have learned that this is called Hessian form.

I could easily calculate intersection points given this info, but proofs are something I have always struggled with. I think an informal proof is completely sufficient. All I've been able to figure is that the denominator being zero happens in circumstances where planes are parallel to each other, which means there is no intersection, or intersections as lines rather than points.

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  • $\begingroup$ So, what is your formula for the intersection point? Presumably your calculation is enough to prove the formula at least. $\endgroup$ – Travis Jul 31 '15 at 17:45
  • $\begingroup$ This is just Cramer's Rule. See en.wikipedia.org/wiki/Cramer%27s_rule. $\endgroup$ – Batominovski Jul 31 '15 at 17:47
  • $\begingroup$ @Batominovski I'm struggling to understand what the system of equations I would be solving is. I'm assuming the formula for vector r would be the solution of the system, then? $\endgroup$ – Dan Jul 31 '15 at 17:57
  • $\begingroup$ The link I gave you contains a solution to a system of three linear equations. Then, when $r=(x,y,z)$ and $u_i=\left(a_i,b_i,c_i\right)$ for $i=1,2,3$, you can see that the denominators of $x$, $y$, and $z$ in that link are the same, which is exactly $u_1\cdot\left(u_2\times u_3\right)$. You only have to expand the coefficients of $d_1$, $d_2$, and $d_3$ to see that they coincide with the solution you got. $\endgroup$ – Batominovski Jul 31 '15 at 19:05
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Alternatively, if you don't want to fuss over expanding the determinants as I told you to try, you can use this argument. Since $V:=u_1\cdot\left(u_2\times u_3\right)=u_2\cdot\left(u_3\times u_1\right)=u_3\cdot\left(u_1\times u_2\right) \neq 0$, the vectors $u_1$, $u_2$, and $u_3$ are linearly independent. Therefore, the solution $r\in\mathbb{R}^3$ to the system of equations $r\cdot u_i-d_i=0$ for $i=1,2,3$ is unique. Now, if $$r=\frac{d_1\left(u_2\times u_3\right)+d_2\left(u_3\times u_1\right)+d_3\left(u_1\times u_2\right)}{V}\,,$$ then $$r\cdot u_1=\frac{d_1\left(u_2\times u_3\right)\cdot u_1}{V}=\frac{d_1 V}{V}=d_1$$ since $\left(u_3\times u_1\right)\perp u_1$ and $\left(u_1\times u_2\right)\perp u_1$. Likewise, $r\cdot u_2=d_2$ and $r\cdot u_3=d_3$. Hence, this formula for $r$ gives the unique solution to the system of equations.

There is still another solution. You can refer to this vector identity: $$ \left(\mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right)\right)\mathbf{D}=\left(\mathbf{A}\cdot\mathbf{D}\right)\left(\mathbf{B}\times\mathbf{C}\right)+\left(\mathbf{B}\cdot\mathbf{D}\right)\left(\mathbf{C}\times\mathbf{A}\right)+\left(\mathbf{C}\cdot\mathbf{D}\right)\left(\mathbf{A}\times\mathbf{B}\right) \,.$$ For reference, see https://en.wikipedia.org/wiki/Vector_calculus_identities.

In general, the solution $r\in\mathbb{R}^n$ to $r\cdot u_i-d_i=0$ for $i=1,2,\ldots,n$ is given by $$r=\frac{\sum_{i=1}^n\,(-1)^{i-1}d_i\left(u_1\wedge u_2\wedge \ldots \wedge u_{i-1} \wedge u_{i+1}\wedge \ldots \wedge u_{n-1}\wedge u_n\right)}{u_1\wedge u_2 \wedge \ldots \wedge u_{n}}\,,$$ where we identify $\bigwedge^{n-k}\,\mathbb{R}^n$ as $\bigwedge^{k}\,\mathbb{R}^n$.

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