Converting span{[1,1,1][2,-1,1][5,-7,1]} in R3 into a scalar equation. My answer different than the teacher's. Not sure where I went wrong.

Question

We have three vectors, v1 = (1,1,1), v2 = (2,-1,1), v3=(5,-7,1).

Together, these vectors form a plane in R3. For my (not-for-marks) homework question, I'm tasked with converting this to a scalar equation. However, my answer differs from that of the teachers. I'm not sure why.

Here is my answer: 2x+3y+6z = 0 (wrong)

Here is the teacher's: 2x+y-3z=0 (correct? I assume the teacher is right?)

To get my answer, I did this:

• Added the 3 spanning vectors to a 3x3 matrix
• Converted the matrix to RREF form
• Noticed 1 of the vectors is dependent
• Created the vector equation for the span: x=t[3,-4,1]
• Converted the vector equation to the parametric equation: x=3t, y=-4t, z=t
• Wrote each of the x,y,z in terms of t and substituted them together to get 2x+3y+6z = 0, which is wrong based on my teacher's answer.

Can anybody explain where I went wrong? Thanks so much!

Answer 1

Once you find that $v_3$ is a linear combination of $v_1$ and $v_2$, you know that $v_1$ and $v_2$ are the two independent vectors inside your plane. Then you can just take the cross product of $v_1$ and $v_2$, and its coordinates are the coefficients of the teacher's equation:

$$v_1 \times v_2=(2,1,-3)$$ $$\implies 2x+y-3z=0$$

Answer 2

We ought to be a little careful with this. I claim that the (column) null vector of $$ M = \left( \begin{array}{rrr} 1&1&1 \\ 2&-1&1 \\ 5&-7&1 \\ \end{array} \right) $$

gives the correct coefficients $A,B,C $ for your plane being $Ax+By+Cz=0$

I also claim that the same (nonzero) null vector is a null vector for the row-reduced matrix. It would be really nice to see your row reduced version of the $M.$ perhaps you could type in three rows in order, just nine numbers as a comment

The result should have two nonzero rows followed by a zero row. Each nonzero row has more than one nonzero entry

Here is the fully reduced version gotten by the most straightforward method

$$ M_R = \left( \begin{array}{rrr} 1&0&\frac{2}{3} \\ 0&1&\frac{1}{3} \\ 0&0&0 \\ \end{array} \right) $$

What is the (nonzero) null vector required to be? I tokk $C = -3$ for convenience