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Determine a so that the intersection line between the planes

$P_1: 2x+ay-z=3$

$P_2: x-2y+az=5$

are parallel to the plane $P_3: 2x+y+z=2$.

I want to solve this using determinants in some way.

Im thinking the intersection line between the planes $P_1$, $P_2$ is the solution the equation of system with $P_1$, $P_2$ and there are infinitely many solutions

$x=T$

$y=S$

$z=3-2T+aS$

and then this would be parallel to the plane $P_3$, can I use some determinant $=0$? Correct answer is $a=-3/2$ or $a=3$.

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2 Answers 2

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You can directly solve for $$ \begin{vmatrix} 2 & a & -1\\ 1 & -2 & a \\ 2 & 1 & 1 \end{vmatrix}=0$$

There are various justifications for this.

First, as $P_1$ and $P_2$ have linear independent normals, there is a single line of intersection. Because that line does not pass through $P_3$, there is no solution $(x, y, z)$ to the set of three equations, hence the determinant is $0$.

Secondly, the line is perpendicular to the normals of $P_1$ and $P_2$ so it's calculated via a cross product. The line is perpendicular to the third normal as well so their dot product is 0. Put it together we have the scalar triple product is $0$.

EDIT: One should verify that the line does not lie on the third plane, because both justifications also work if the planes meet at a single line (infinite solutions).

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Indirect approach: We first find the direction numbers of the line L resulting from intersection of two planes:

$l=\begin{vmatrix}a& -1\\-2& a\end{vmatrix}=a^2-1$

$m=\begin{vmatrix}-1& 2\\a& 1\end{vmatrix}=-1-2a$

$n=\begin{vmatrix}2& a\\1& -2\end{vmatrix}=-a-4$

The normal of plane are $N_p:(A=2, B=1, C=1)$ and we must have $L\bot N_p$ ;the condition is that:

$l\times A+m\times B+n\times C=0$

which gives this equation:

$2a^2-3a-9=0$

which it's roots are $a=3$ and $a=-\frac 32$

Using this method you can conclude the method mentioned in other answer.

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