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If the intersection of planes $a^{k}x+2^{k}y+3^{k}z+d^{k}=0(0\leq k\leq 4)$ is a line, then $a=?$ and $d=?$.

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    $\begingroup$ Why do you want to solve this problem ? & what attempts have you made ? (Please add some context to your question). $\endgroup$ – Donald Splutterwit Oct 2 '17 at 10:32
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    $\begingroup$ @DonaldSplutterwit Why does he want to solve it? It's part of his homework ... What attempts has he made? None, because he wants other people to do his homework for him. $\endgroup$ – terrace Oct 2 '17 at 10:34
  • $\begingroup$ @ Donald Splutterwit It is part of my homework. I only obtained $a=2$ or $a=3$ because the intersections of arbitrary two planes, if they exist, share with the same direction vector. However, I can not find a point to calculate $d=?$ $\endgroup$ – stuxuf Oct 3 '17 at 8:05
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The matrix of coefficients is $$A=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & 2 & 3 & d \\ a^2 & 4 & 9 & d^2 \\ a^3 & 8 & 27 & d^3 \\ a^4 & 16 & 81 & d^4 \\ \end{array} \right)$$ The system has One and only one solution, that is one point, if $\text{rank }A=4$

We know that the planes intersect along a line, and this means that $\text{rank }A=3$ which happens if determinants of order $4$ are all zero.

$$\det \left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & 2 & 3 & d \\ a^2 & 4 & 9 & d^2 \\ a^3 & 8 & 27 & d^3 \\ \end{array} \right|=\\=-a^3d^2+5 a^3 d-6 a^3+a^2 d^3-19 a^2 d+30 a^2-5 a d^3+19 a d^2-36 a+6 d^3-30 d^2+36 d$$ can be easily factored as $$(a-3) (a-2) (d-3) (d-2) (d-a)=0$$ $(a=2,\;d=2);\;(a=2,\;d=3);\;(a=3,\;d=2);\;(a=3,\;d=3)$

The solution $a=d$ leads to planes intersecting in one point so it's been eliminated

Hope this helps

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  • $\begingroup$ You didn't need to calculate the determinant! Just see if you substitute $a = 2,3,d$, determinant will be zero. So $(a-2), (a-3) ...$ are factors. Similarly for $d$. In all 5 factors as its a quintic. Nice answer nevertheless! $\endgroup$ – samjoe Oct 2 '17 at 16:33
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If you rewrite it as $a^kx + 2^ky + 3^k z + d^kw=0$ then it's the intersection of the null-space of the matrix $\begin{pmatrix}a^k & 2^k & 3^k & d^k\end{pmatrix}$ and the hyper-plane $w=1$. This being a line would require the null-space to be a plane - which means that the rank of the matrix is $2$.

Since the second and third columns are independent this means that the first and last must be a linear combination of those which is only possible if $a$ and $d$ are $2$ or $3$.

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  • $\begingroup$ The idea of "hyper-plane" is very useful, thank you! $\endgroup$ – stuxuf Oct 3 '17 at 10:14

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