Tangent, Normal and Binormal Vectors — Question 7

Question 7 – Solution

Strategy. The osculating plane is spanned by $\mathbf T$ and $\mathbf N$, so its normal is $\mathbf B$. The normal plane is perpendicular to $\mathbf T$.

Step 1: Point and tangent direction $$P=\mathbf r(1)=(1,1,1),\qquad \mathbf r'(1)=\left\langle 1,2,3\right\rangle.$$ We may use $\left\langle 1,2,3\right\rangle$ as the normal vector of the normal plane.

Step 2: Binormal direction $$\mathbf r''(t)=\left\langle 0,2,6t\right\rangle,\qquad \mathbf r''(1)=\left\langle 0,2,6\right\rangle.$$ Then $$\mathbf r'(1)\times\mathbf r''(1) =\left\langle 1,2,3\right\rangle\times\left\langle 0,2,6\right\rangle=\left\langle 6,-6,2\right\rangle=2\left\langle 3,-3,1\right\rangle.$$ Hence an osculating-plane normal is $\left\langle 3,-3,1\right\rangle$.

Step 3: Plane equations Through $P=(1,1,1)$, the osculating plane is $$3(x-1)-3(y-1)+(z-1)=0 \quad\Longrightarrow\quad \boxed{3x-3y+z=1}.$$ The normal plane is $$(x-1)+2(y-1)+3(z-1)=0 \quad\Longrightarrow\quad \boxed{x+2y+3z=6}.$$ Substitution of $(1,1,1)$ verifies both equations, and their normals have the required Frenet directions.