Planes \(\pi_1: x+y+z=3\) and \(\pi_2: x-y+z=1\). The line of intersection has direction \(\vec n_1\times\vec n_2 = (a,0,b)\). Find \(a\).
Solution:
\((1,1,1)\times(1,-1,1)=(1\cdot1-1\cdot(-1),\,1\cdot1-1\cdot1,\,1\cdot(-1)-1\cdot1)=(2,0,-2)\). \(a=2\).