given is:
$ g: \vec x = \frac{1}{15} \begin{pmatrix} 14 \\ 13 \\ 7 \\ \end{pmatrix} + r* \begin{pmatrix}1\\0\\2 \end{pmatrix}$
and
$ E: ax_1 * 2x_2 + x_3 = 4 $
What is "a", when $ g \in E $ ?
I tried putting two points of g into E without getting the desired result. I also tried to argue that $ \vec n * \vec u = 0 $ (90° between the two vectors..
Maybe you can help :))
First let's try one point of $g$. You know the point $g(0) = (x_1, x_2,x_3) = \tfrac1{15}(14,13,7)$ lies on the line $g$, since it corresponds to the parameter value $r=0$. If this point also lies on the plane $E$, then $$ a\left(\frac{14}{15}\right) + 2\left(\frac{13}{15}\right) + \left(\frac{7}{15}\right) = 4 $$ We can solve this to find a value of $a$.
But if we do the same thing with the point $g(1)$ corresponding to $r=1$, then we get a different value for $a$.
This shows that there is no value of $a$ that will cause both $g(0)$ and $g(1)$ to lie on the plane. So, there is no value of $a$ that will cause the line to lie in the plane.