A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.
What is the equation of the line?
The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.
I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.
I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?
Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?
Best Answer
Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.
Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.
The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line $\ell(t) = [9+t, -7-t, 31-3t]$ is that line:
$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$
and $$3x + z = 3(9+t)+(31-3t) = 58$$