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

${\mathbf{𝐧}}_1\times \mathbf{𝐯}=\mathbf{𝟎}\text{and}{\mathbf{𝐧}}_2\times \mathbf{𝐯}=\mathbf{𝟎}.$

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, $\hat{\mathbf{v}}$ is

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