To find an intersection point between two planes with only the direction vector

Question

Find the intersection between two planes $x−3y−2z = 2$ and $2x+y+3z = 1$

Solution:

$(1)$$\quad n_1 \times n_2 =\langle −7,7,7\rangle =7 \langle −1,1,1\rangle$.

$(2)$ To find one intersection point, we let $z = 0$. Then $x = \frac{5}{7},y = - \frac{3}{7}$.

where did they get:

$z = 0$

$x = \frac{5}{7}$

$y = - \frac{3}{7}$

For the point on the line?

Answer 1

Finding a point in common requires us to solve the system \begin{align*} x−3y−2z &= 2\\ 2x+y+3z &= 1 \end{align*} Depending on your level of sophistication, we can either use linear algebra to solve this system, or maybe multiply the first equation by $-2$ and add them, etc...