[SOLVED] a linear equation in 3 variables

[karush]

write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$

 

[chisigma]

write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$

You have the linear system...

$\displaystyle a + b + c = d$

$\displaystyle 2 b = d$

$\displaystyle a = d\ (1)$

... the solution of which is $a=d,\ b= \frac{d}{2},\ c=- \frac{d}{2}$, so that You can write...

$\displaystyle a x + \frac{d}{2} y - \frac{d}{2} z = d \implies x + \frac{y}{2} - \frac{z}{2} = 1\ (2)$

Kind regards

$\chi$ $\sigma$

 

[HallsofIvy]

write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$

I would have done this just a little differently from chisigma. First, note that there is no "unique" solution. for any ax+ by+ cz= d describing the line, you could multiply through by any number and get a new equation for the same line. You could. for example, divide both sides by d to get (a/d)x+ (b/d)y+ (c/d)z= 1, then let a'= a/d, b'= b/d, and c'= c/d so that the equation is a'x+ b'y+ c'z= 1, with only three unknown values.

I suspect that the answer you are asked for is the one that has all integer coefficients with no common factor.

Now, from ax+ by+ cz= d, (1, 1, 1) on the line means we must have a+ b+ c= d. (0, 2, 0) on the line means 0a+ 2b+ 0c= 2b= d. (1, 0, 0) on the line means 1a+ 0b+ 0c= a= d. Since both 2b and a are equal to d, a= 2b and we can replace both a and d with 2b in the first equation: 2b+ b+ c= 2b so that b+ c= 0 or b= -c.

Now we can write everything in terms of b: a= 2b, c= -b, d= 2b, and so the equation is
2bx+ by- bz= 2b. Here, b can be any (non-zero) number and still give an equation describing the line but dividing both sides of the equation by b gives
2x+ y- z= 2, the simplest form for the equation as all numbers are integers and they do not all have a common factor so cannot be reduced.

 

[Prove It]

write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$

There will be a unique plane which satisfies these three points simultaneously. The plane will have the same coefficients as its normal vector, and the normal vector will also be normal to any vectors in the plane. So if we can get two vectors that lie in the plane, taking their cross product will give the normal vector. Call two of these vectors [tex]\displaystyle \begin{align*} \mathbf{a} \end{align*}[/tex] and [tex]\displaystyle \begin{align*} \mathbf{b} \end{align*}[/tex]. We could have

[tex]\displaystyle \begin{align*} \mathbf{a} &= ( 1 - 0, 1 - 2, 1 - 0) \\ &= ( 1 , -1, 1 ) \\ \\ \mathbf{b} &= (1 - 0, 0 - 2, 0 - 0 ) \\ &= ( 1 , -2, 0) \end{align*}[/tex]

[tex]\displaystyle \begin{align*} \mathbf{n} &= \mathbf{a} \times \mathbf{b} \\ &= \left| \begin{matrix} \mathbf{i} & \phantom{-}\mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -2 & 0 \end{matrix} \right| \\ &= \mathbf{i} \, \left| \begin{matrix} -1 & 1 \\ -2 & 0 \end{matrix} \right| - \mathbf{j} \, \left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| + \mathbf{k} \, \left| \begin{matrix} 1 & -1 \\ 1 & -2 \end{matrix} \right| \\ &= \mathbf{i} \, \left[ -1 \cdot 0 - 1 \cdot (-2) \right] - \mathbf{j} \, \left( 1 \cdot 0 - 1 \cdot 1 \right) + \mathbf{k} \, \left[ 1 \cdot (-2) - (-1) \cdot 1 \right] \\ &= 2\,\mathbf{i} + \mathbf{j} - \mathbf{k} \end{align*}[/tex]

The plane will have the same coefficients as the normal vector, so that gives [tex]\displaystyle \begin{align*} 2x + y - z = d \end{align*}[/tex]. Since we have three points that lie on the plane, any of them can be substituted to find [tex]\displaystyle \begin{align*} d \end{align*}[/tex]. If we substitute [tex]\displaystyle \begin{align*} (1, 0, 0) \end{align*}[/tex] that gives

[tex]\displaystyle \begin{align*} 2 \cdot 1 + 0 - 0 &= d \\ 2 &= d \end{align*}[/tex]

and thus the plane is [tex]\displaystyle \begin{align*} 2x + y - z = 2 \end{align*}[/tex].