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

The first plane

$\displaystyle {}y-z=1$

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

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

has normal n$_{2}$=[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

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

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

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

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

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

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

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