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I have 2 equation in terms of $a,b$ and $c$ .

$3a + 10b + 5c =0$ and $4a + 6b + 2c =0$

I need to find a:b:c and answer is $\dfrac a5 = \dfrac {b}{-7} = \dfrac {c}{11}$

I want to know how to get that?

My attempt:

Given equations can be written in form

$$\left ( \begin{matrix} 3 & 10 & 5 \\ 4 & 6 & 2 \\ \end{matrix} \right ) \left ( \begin{matrix} a\\ b\\ c\\ \end{matrix} \right ) = \left ( \begin{matrix} 0 \\ 0\\ \end{matrix} \right ) $$

But here I don't have any idea how to proceed. I can do for 3×3 matrix and I thought it can be done in same way.

Thanks.

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3 Answers 3

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$(a\,b\,c)$ is orthogonal to both $(3\,10\,5)$ and $(4\,6\,2)$. The obvious choice for a common orthogonal vector in 3-space is the cross product.

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  • $\begingroup$ Can you please elaborate? Cross product will be 0 but how to get relation between a,b and c ? $\endgroup$
    – Fawad
    Commented Feb 5, 2017 at 10:18
  • $\begingroup$ @Fawad Cross product of those two vectors is $(-10,14,-22)$, see WolframAlpha. Perhaps when you said it is zero, you meant dot product? $\endgroup$ Commented Feb 13, 2017 at 10:58
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HINT:

$$3a+10b=-5c$$

$$4a+6b=-2c$$

Solve the two simultaneous equations for $a,b$ in terms of $c$

See this.

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  • $\begingroup$ Thanks,but it is given that a:b:c is $$ \dfrac {a}{30-20} = \dfrac{b}{6-20} = \dfrac {c}{-18 + 14}$$ which is like denominators are determinant of cofactor matrices and I think it has something to do with matrices. Can you show how only using matrix? $\endgroup$
    – Fawad
    Commented Feb 5, 2017 at 10:10
  • $\begingroup$ @Ramanujan, See mathworld.wolfram.com/CramersRule.html $\endgroup$ Commented Feb 5, 2017 at 10:15
  • $\begingroup$ but we are aware that Cramer's rule will only work for square matrix $\endgroup$
    – Fawad
    Commented Feb 6, 2017 at 10:05
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    $\begingroup$ @F.Mirza, Isn't $$3a+10b=-5c$$ $$4a+6b=-2c$$ square? $\endgroup$ Commented Feb 6, 2017 at 10:08
  • $\begingroup$ If we want to write equation as AX=D where A is coefficient matrix , X is variable matrix and D is 0,0. $\endgroup$
    – Fawad
    Commented Feb 6, 2017 at 10:11
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Hint...first eliminate one of the letters, say $c$ so that you have one equation in $a$ and $ b$. Then sepatate these letters to either side of the equation and set each side equal to a parameter $\lambda$. You can then get each of $a,b,c$ in terms of $\lambda$ to get the set of ratios.

Note that this problem is equivalent to that of finding the line of intersection of two planes in three dimensional space.

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