Plane $x-2y-z=4$ which parametric line is normal?
A) $x=1$, $y=-2t$, $z=-t$
B) $x=2t$, $y=5-4t$, $z=-7-2t$
C) $x=-t$, $y=5+2t$, $z=-7-t$
D) $x=\cos(t)$, $y=-2\sin(t)$, $z=-2\sin(t)\cos(t)$
E) NONE
Its forbidden to use options to find answer,options are only for that we check if we find answer correct
And this is my first post here forgive me if a made something wrong,and my english is not quite good for math.And I tried so much to solve this but with out a dot ı couldnt please dont scold and help me to solve this
If you found that $N=(1,-2,-1)$ is a normal vector, then a normal line is given by multiples of this vector, hence $r(t)=(x(t),y(t),z(t))=(t,-2t,-t)$. Now if you translate the line in the space you still get an orthogonal line to the plane. What matters are the coefficients of $t$, which must be proportional to $(1,-2,-1)$. So the line $s(t) = 2r(t) = (2t,-4t,-2t)$ will be orthogonal to your plane, and by translating it you obtain that all lines of the form $(2t+a,-4t+b,-2t+c)$ are orthogonal. Choose $a=0,b=5,c=-7$.