Find the equations of the normal plane and osculating plane of the helix $r(t) = cost\mathbf{i} + sint\mathbf{i} + t\boldsymbol{k}$ at the point $P(0,1,2)$
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$\begingroup$You know that for describing a plane we need a point and a normal vector. This normal vector is perpendicular to the plane. Here, the normal desired vector is really the tangent vector $T$ (See here). But what is the point? The point you indicated is not on the curve, so I assumed another point like $A:r(\pi/2)=(0,1,\pi/2)$. We have $$\vec{T}(t)=r'(t)=(-\sin t, \cos t,1)\to \vec{T}(\pi/2)=(-1,0,1)=\vec{n}$$ and this $\vec{n}$ is normal vector. Therefore, we have the equation of normal plane at $A$ is: $$P: -1(x-0)+0(y-1)+1(z-\pi/2)=0$$. I made a nice plot for this curve and this resulted plane $P$:
For another part you need to follow the above link and try to find the vector $\vec{B}$. :-)
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