# Finding a Vector Perpendicular to Two Other Vectors w/o Cross Products

I need to find a vector that is perpendicular to the vectors $[2, 3, 2]$ and $[4, 9, 5]$. I have not been taught the method with cross-products using matrices so I cannot use that method while solving the problem.

I was thinking of using systems of equations to solve this problem and find the line that is the intersection of the two planes with the given normal vectors.

$$2x + 3y + 2z =0$$ $$4x+ 9y+5z=0$$

I used elimination method to get $3y + z = 0$. I know that if I choose that $y = 1$, then $z = -3$. I was thinking that since I eliminated $x$, then its value will be $0$. So I thought that the vector would be $[0, 1, -3]$, however, after computing the dot products I know that that vector is incorrect. Any help will be greatly appreciated.

## 1 Answer

Eliminating a variable does not mean that it is $0$, it means that you manipulate the original system of equations to another system of equations that say nothing at all about $x$. Hence in order to find the value of $x$, you have to resort to the original, uneliminated equations and plug in the values for $y$ and $z$ that you found.

• I have just used your method and have found a possible vector with components of $[1.5, 1, -3]$. Thank you. – geo_freak Apr 29 '18 at 21:11