The plane $\pi$ has equation r = $3$i + $2$j + $s$(i + k) + $t$(j - k) and the line $\ell$ has equation r = $2$i + j -$3$k + $\lambda$ ($2$i - j).
(a) Find a vector which is normal to $\pi$.
My working so far:
r = $3$i + $2$j + $s$(i + k) + $r$(j - k)
If n is a vector perpendicular to the plane,
then r$\cdot$n =(a+$s$b+$t$c)$\cdot$n
= a$\cdot$n+$s$b$\cdot$n+$t$c$\cdot$n
r$\cdot$n = ($3$i + $2$j)$\cdot$n + $s$(i + k)$\cdot$n+ $t$(j - k)$\cdot$n
Let n=$\left(\begin{matrix}x\\y\\z\\\end {matrix}\right)$
b$\cdot$n = c$\cdot$n = $0$ since b and c are perpendicular to n.
Taking b$\cdot$n $=0$,
$\left(\begin{matrix}s\\0\\s\\\end {matrix}\right)\cdot\left(\begin{matrix}x\\y\\z\\\end {matrix}\right)=0$
$s$x+ $s$z $=0$
x = -z
Taking c$\cdot$n $=0$,
$\left(\begin {matrix}0\\t\\-t\\\end {matrix}\right)\cdot\left(\begin{matrix}x\\y\\z\\\end {matrix}\right) =0$
$t$y - $t$z $=0$
y = z
If what I have been doing is correct, the vector normal to $\pi$ would be (in terms of z), n= $\left(\begin{matrix} -z\\z\\z\\\end {matrix}\right)$
Is this the right way to solve this question?
What is a better way to solve the question?
The answer given is -i + j + k.
Yes your answer is correct although an easier way is to use cross-product of the vectors parallel to the plane (the ones in brackets).
So the normal is $$\left(\begin{matrix}1\\0\\1\end{matrix}\right)\times\left(\begin{matrix}0\\1\\-1\end{matrix}\right)=\left(\begin{matrix}-1\\1\\1\end{matrix}\right)$$