Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 16/Solution 2

Since a vector on the line has to be on each plane, it has to be a solution of a linear system. We have the following equations

$\left\{{\begin{aligned}y-z&=1\\2x+y+z&=2\end{aligned}}\right.$

which is a linear system with 2 equations in 3 unknowns. We can write it in an augmented matrix as

$\left[{\begin{array}{ccc|c}0&1&-1&1\\2&1&1&2\end{array}}\right]$

We will swap the first two rows and then row reduce to get

$\left[{\begin{array}{ccc|c}1&0&1&1/2\\0&1&-1&1\end{array}}\right].$

We notice there is a free variable and we let x = t. We then get

${\begin{aligned}z&={\frac {1}{2}}-t\\y&=1+z={\frac {3}{2}}-t\end{aligned}}.$

Therefore we get that the solution, s, is

$\mathbf {s} =\mathbf {s} _{0}+t\mathbf {v} =\left[0,{\frac {3}{2}},{\frac {1}{2}}\right]+t[1,-1,-1].$

The direction part of the line, v is [1,-1,-1]. However to make it length 1 we must divide by its magnitude, $|\mathbf {v} |={\sqrt {3}}$ to get,

$\mathbf {\hat {v}} ={\frac {1}{\sqrt {3}}}[1,-1,-1]$

just like in solution 1.