Which one of the following is a normal vector of the plane $\mathbf x=s(1,2,3)+t(2,3,4)$ where $s,t\in\mathbb R$?
$\begin{align}&(a) \ (3,5,7) \phantom{XXXXXXXXXX} & (c) \ \ \ \ (1,2,1) \\
&(b) \ 20 \phantom{XXXXXXXXXX} & (d) \ (1,-2,1)
\end{align}$
To find the normal of the plane I need to find the coefficient of the general equation, right?
However I am not sure how to do that in this case. This is what I have
$$x= 1s + 2t $$
$$y= 2s + 3t$$
$$z=3s + 4t$$
so by cancelling each one out I get
$$y-2x = -t$$
and
$$z+4(y-2x)=3s$$
… But how can I simplify this to get the general equation??
Best Answer
The answer is (d). Parametric equations are $$x=s+2t,\quad y=2s+3t,\quad z=3s+4t.$$ From the first two equations we have $t=2x-y$ and $s=2y-3x$. Substituting these into the third equation we get the equation of the plane $x-2y+z=0$ and hence the normal vector is $(1, -2, 1)$.