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Equation:

$ax_1 + bx_2 + cx_3 = d$ describes a plane in $R^3$. I need to examine all possible positions of the planes E and F to each other, which are given through:

$E: a_1x_1 + b_1x_2 + c_1x_3 = d_1 $

$F: a_2x_1 + b_2x_2 + c_2x_3 = d_2 $

The planes may intersect(cut) in a straight line, be parallel or equal. Share this with reference to the discussion of the rank of the matrix and the augmented matrix of the system of equations.

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  • $\begingroup$ You are told the three possible positions the planes may have in relation to each other. You are being asked to think about the rank of the matrix of coefficients, and the rank of the augmented matrix, and explain how those two numbers relate to the three ways the two planes can relate. Can you try to do that? $\endgroup$ – Gerry Myerson Apr 5 '16 at 12:30
  • $\begingroup$ I have tried but I this is totally false. $\endgroup$ – Alena Everdeen Apr 5 '16 at 13:10
  • $\begingroup$ If you will share with us the details of what you tried, it will be easier to see what assistance we can give. $\endgroup$ – Gerry Myerson Apr 5 '16 at 22:32
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Developing the ideas of @Gerry Myerson, let

$$M=\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\end{bmatrix}; \ \ N=\begin{bmatrix}a_1&b_1&c_1&d_1\\a_2&b_2&c_2&d_2\end{bmatrix}$$

(1) If rank(M)=2, the plane are intersecting along a line.

(2) otherwise, if rank(M)=1:

  • (2a) either rank(N)=2: in this case the planes are (strictly) parallel, or

  • (2b) rank(N)=1: in this case the planes are identical (what you call "equal").

Why that? let us enter into explanations.

A fundamental fact is that the normal vectors to the planes are $(a_k, b_k, c_k)^T$.

(1) If rank(M)=2, these normal vectors are not proportional, thus the planes are not parallel.

(2) If rank(M)=1: the normal vectors are proportional, therefore, the planes are parallel or identical; we thus have the system of equations:

$$\begin{cases}a_1 x_1 + b_1 x_2 + c_1 x_3& = &d_1\\ka_1 x_1 + kb_1 x_2 + kc_1 x_3& = &d_2 \end{cases} \ \ (1)$$

or, equivalently:

$$\begin{cases}a_1 x_1 + b_1 x_2 + c_1 x_3& = &d_1\\a_1 x_1 + b_1 x_2 + c_1 x_3& = &d_2/k \end{cases} \ \ (2)$$

(2b) either $d_1=d_2/k$ corresponding to the subcase rank(N)=1, we have the same equation, thus the 2 planes are identical;

(2a) or $d_1 \neq d_2/k$ (rank(N)=2). But, in this case, system (2) cannot have a solution (if there were one, by substraction, one would get $d_1-d_2/k=0$, impossible, thus it is impossible that there exist a common point: the planes are thus parallel.

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  • $\begingroup$ I don't understand what is k in this example? $\endgroup$ – Alena Everdeen Apr 5 '16 at 17:29
  • $\begingroup$ Rank(N)=1 is equivalent to the fact that the two (row) vectors are linearly dependent, which itself is equivalent to : one of the rows is a multiple of the other. Let $k$ be the coefficient of proportionality : $(a_2,b_2, c_2)=k(a_1, b_1,c_1)$, allowing you to replace $a_2$ by $ka_1$, $b_2$ by $kb_2$, $c_2$ by $kc_1$. $\endgroup$ – Jean Marie Apr 5 '16 at 17:36

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