1. ## Vector equation of intersecting lines. Answer check

The question asks for the vector equation for the line of intersection by the planes x+2y-2z=5 and 6x-3y+2z=8

$\displaystyle \vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle + t\left\langle {2,14,15} \right\rangle$

But when I do the working

Labeling x+2y-2z=5 as equation 1
and 6x-3y+2z=8 as equation 2

Multiply equation 1 by 6 to get 6x+12y-12z=30 call this equation 3

equation 2 minus equation 3 to get

$\displaystyle \begin{array}{l} 15y = 22 - 14z \\ y = \frac{{22}}{{15}} - \frac{{14}}{{15}}z \\ \end{array}$

Let z=t

$\displaystyle y = \frac{{22}}{{15}} - \frac{{14}}{{15}}t$

Than from equation 1

$\displaystyle \begin{array}{l} x + 2y - 2z = 5 \\ x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) - 2t \\ x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t - 2t \\ x = \frac{{31}}{{15}} - \frac{2}{{15}}t \\ \end{array}$

and so

$\displaystyle \begin{array}{l} \vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle + t\left\langle {\frac{{ - 2}}{{15}},\frac{{ - 14}}{{15}},1} \right\rangle \\ \vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle + t\left\langle { - 2. - 14,15} \right\rangle \\ \end{array}$

Am I right with my negative signs in <-2,-14,15>

and the answer in the text of <2,14,15> is wrong?

2. Well, You've made two mistakes while 'rearranging' equations. There is:

$\displaystyle \begin{array}{l} 15y = 22 - 14z \\ y = \frac{{22}}{{15}} - \frac{{14}}{{15}}z \\ \end{array}$
and should be:
$\displaystyle \begin{array}{l} 15y = 22 + 14z \\ y = \frac{{22}}{{15}} + \frac{{14}}{{15}}z \\ \end{array}$
and the next minus is here:
$\displaystyle \begin{array}{l} x + 2y - 2z = 5 \\ x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) - 2t \\ x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t - 2t \\ x = \frac{{31}}{{15}} - \frac{2}{{15}}t \\ \end{array}$
whereas it should be
$\displaystyle \begin{array}{l} x + 2y - 2z = 5 \\ x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) + 2t \\ x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t + 2t \\ x = \frac{{31}}{{15}} + \frac{2}{{15}}t \\ \end{array}$
So, if You take it into consideration, it occurs that the answer in the text is correct.

best regards!