finding a perpendicular vector to a tangent line

Question

I was given the function $$y=2\sin x$$ and was told to find a parallel vector and a perpendicular vector to the tangent line at the point $(\frac{\pi}{6},1)$ I found that $$x=t$$ and $$y=2\sin t$$ so that I can write a vector equation $$r(t)=it+2\sin tj$$ I found the vector that was parallel was $$r'(\frac{\pi}{6}) = i + \sqrt{3}$$

But where I got stuck was for with the perpendicular one, I know that the perpendicular vector would have an inverse reciprocal slope so then, $$(\frac{1}{-r'(\frac{\pi}{6})}) = \frac{1}{-i-\sqrt{3}j}$$

I'm not sure where to continue from here, I recently learned this method online and I kind of like it but my understanding is lacking at some parts but I get the general purpose of what I am suppose to do. But I do not know where to proceed with the perpendicular vector and how to get the actual correct values.

Answer 1

We have that

$$y=2\sin x\implies y'(\pi/6)=2\cos \dfrac{\pi}{6}=\sqrt{3}.$$ That is, the slope of the tangent line is $\sqrt{3}.$ So $$(1,\sqrt{3})$$ gives the direction of the line and $$(\sqrt{3},-1)$$ gives the direction of the perpendicular line.

(Note that $(b,-a)\perp (a,b)$ for any vector.)

Answer 2

note that the equation of your Tangent line is given by $$y=\sqrt{3}\left(x-\frac{\pi}{6}\right)+1$$ and two Points of this line are $$P_1\left(\frac{\pi}{6};1\right)$$ and $$P_2\left(0;1-\frac{\sqrt{3}\pi}{6}\right)$$ can you go on?