I am getting the wrong answer.
Real values x and y such that a vector is orthogonal to 2 other vectors
-3
$\begingroup$
linear-algebra
vectors
2 Answers
1
As an alternative strategy, compute the cross product $(-2,3,-1)\times(3,3,-3)$ and scale it as necessary to make the first component $12$.
0
I did this quickly on a post-it in the bus, so I hope it's right. You should try to include your $=0$ portions of the equations. Your two dot products are correct, but your arithmetic appears wrong (note eq4)
$$-24+3y-x=0 \implies x=-24+3y$$ $$36+3y-3x=0 \implies x=12+y$$
As 0=0, you should be able to set x=x. This will give you your y value. From there you can solve for x using either of the relations.