I have attempted a question from the OCR paper, AS level Further Maths B Core Pure. It's "A vector v$=a$i+$b$j+$c$k has magnitude $1$ unit. The angle between v and the positive $z$-axis is $60°$, and v is parallel to the plane $x-2y=0$. Given that $a,b,c$ are all positive real numbers, find v. "
So my attempt goes:
Since the angle to positive $z$-axis is $60°$, $(a\textbf{i}+b\textbf{j}+c\textbf{k})\bullet{\textbf{k}}=|\textbf{v}||\textbf{k}|\cos{60}$ which gives $c=\frac{1}{2}$.
Then, since the magnitude is $1$, $a^2+b^2+c^2=1$, which is $a^2+b^2=\frac{3}{4}$. The next part I'm unsure about, I don't know exactly how to use the fact that it is parallel to the $x-2y=0$ plane. So I just assumed that that means $a-2b=0$.
So that gives $5b^2=\frac{3}{4}$, or $b=\sqrt{\frac{3}{20}}$ and hence $a=\sqrt{\frac{3}{5}}$.
So could someone please explain how to use the fact that v is parallel to the $x-2y=0$ plane, and to further help my understanding, how would I do it if they instead gave that it was perpendicular to the $x-2y=0$ plane?
Since $v$ is parallel to the plane $x - 2y = 0$, it is perpendicular to the normal vector of this plane which is $(1, -2, 0)$, hence,
$ a - 2 b = 0 $
and you also have
$ a^2 + b^2 = \dfrac{3}{4} $
Substituting for $ a $ from the first equation into the second equation
$ b^2 + 4 b^2 = \dfrac{3}{4} $
Hence,
$ b = \pm \dfrac{1}{2} \sqrt{ \dfrac{3}{5} } $
and then
$ a = \pm \sqrt{ \dfrac{3}{5} } $
So the vector $ v = ( \pm \sqrt{ \dfrac{3}{5} } ,\pm \dfrac{1}{2} \sqrt{ \dfrac{3}{5} }, \dfrac{1}{2} ) $