Which of the following planes is perpendicular to $(1+2t,4t,3-2t)$?

Question

$$(x,y,z)=(1+2t,4t,3-2t),t\in\Bbb R$$ A) $2x - y - z = 5$

B) $2x + 6z = 1$

C) $2x - y = -3$

D) $-3x -6y + 3z = 2$

Which one is it and why??

Answer 1

the trick is here to prove that the normal vector of the given plane and the direction vector of the straight line are linear depending. For A) $$2x-y-z=5$$ then we have $$\vec{n}=[2;-1;-1]$$ and $$\vec{a}=[2;4;-2]$$ then you will get $$\vec{n}\ne \lambda\vec{a}$$ and for D we get $$2=-3t$$ $$4=-6t$$ $$-2=-3t$$ thus $$t=-\frac{2}{3}$$ they are orthogonal