Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 10/Solution 1

Let $L$ denote the line of intersection. the two planes have normal vectors $\langle 1,1,-2\rangle$ and $\langle 2,-5,7\rangle$ , respectively. A vector in the direction of the line is therefore

$\langle 1,1,-2\rangle \times \langle 2,-5,7\rangle =\langle -3,-11,-7\rangle .$

We only need to find a point $P$ on $L$.

We consider $P$ to be the point of $L$ on the plane $x=0$.

Thus substitute $x=0$ in the system in the question to get $(0,0,0)$ lies on the line. Thus

$x=-3t,y=-11t,z=-7t$

answer: $\color {blue}(x,y,z)=(-3,-11,-7)t$