# Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 16/Solution 1

The first plane

${\displaystyle \displaystyle {}y-z=1}$

has normal n${\displaystyle _{1}}$=[0,1,-1] (just the coefficients on x,y,z). The second plane

${\displaystyle \displaystyle {}2x+y+z=2}$

has normal n${\displaystyle _{2}}$=[2,1,1]. Let's call the direction our line has v. Notice that since v is an intersection of the two planes then by definition it is on plane 1 and plane 2. Therefore

${\displaystyle \mathbf {n} _{1}\times \mathbf {v} =\mathbf {0} \quad {\text{and}}\quad \mathbf {n} _{2}\times \mathbf {v} =\mathbf {0} .}$

Therefore we seek v so that it is orthogonal to both normal vectors. The cross product, by definition, provides this direction. Therefore,

${\displaystyle \mathbf {v} =\mathbf {n} _{1}\times \mathbf {n} _{2}=[0,1,-1]\times [2,1,1]=[2,-2,-2].}$

We want this to have length 1 and so we see that

${\displaystyle |\mathbf {v} |={\sqrt {12}}=2{\sqrt {3}}.}$

Therefore we get that the unit norm vector, ${\displaystyle \mathbf {\hat {v}} }$ is

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