3D geometry

**alexmahone** · January 23rd 2009, 11:41 PM

Find the value of $a$ such that the planes $ax+y+z=0$ , $x+3z=0$ , and $5y+6z=0$ have a line in common.

**earboth** · January 24th 2009, 01:02 AM

Originally Posted by alexmahone

Find the value of $a$ such that the planes $ax+y+z=0$ , $x+3z=0$ , and $5y+6z=0$ have a line in common.

Let
$p_1:x+3z=0$

$p_2:5y+6z=0$ and

$p_3:ax+y+z=0$

denote the three planes.

1. Calculate

$p_1 \cap p_2 = l_{p_1,p_2} : \left\{\begin{array}{l}x=-3t \\y=-\frac65t \\z = t\end{array}\right.$

2. Calculate

$p_2 \cap p_3 = l_{p_2,p_3} : \left\{\begin{array}{l}x=\frac1{5a} t \\y=-\frac65t \\z = t\end{array}\right.$

3. Compare the coefficients:

$l_{p_1,p_2} = l_{p_2,p_3}\ if\ -3=\frac1{5a}~\implies~ a=\dfrac1{15}$

4. Thus the third plane has the equation:

$p_3:\dfrac1{15} x+y+z=0$

5. I've attached a sketch of the three planes.

**alexmahone** · January 24th 2009, 01:43 AM

Originally Posted by earboth

$-3=\frac1{5a}~\implies~ a=\dfrac1{15}$

Thus the third plane has the equation:

$p_3:\dfrac1{15} x+y+z=0$

Thanks for your solution, but shouldn't $a=-\frac{1}{15}$ ?

**mr fantastic** · January 24th 2009, 03:39 AM

Originally Posted by alexmahone

Thanks for your solution, but shouldn't $a=-\frac{1}{15}$ ?

Yes. Earboth made a simple and obvious typo.

**earboth** · January 24th 2009, 04:12 AM

Originally Posted by mr fantastic

Yes. Earboth made a simple and obvious typo.

Thanks!

Originally Posted by alexmahone

Thanks for your solution, but shouldn't $a=-\frac{1}{15}$ ?

Of course, you are right. The drawing is correct. The negative sign somehow slipped away on the way from my notes to the MHF editor. Sorry.

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