Tangent, Normal and Binormal Vectors — Question 10

Question 10 – Solution

Strategy. First audit unit length and orthogonality. Then use vector projection to remove the tangential component from the proposed bending direction.

Step 1: Diagnose the supplied normal The tangent is unit because $(4+1+4)/9=1$. But $$\mathbf T\cdot\mathbf N =\frac{2(-4)+1(7)+2(1/2)}{9\sqrt 5}=0,$$ while $$\|\mathbf N\|^2=\frac{16+49+1/4}{45}=\frac{29}{20}\ne 1.$$ Thus it is perpendicular to $\mathbf T$ but is not a unit vector, so the data are not a Frenet frame.

Step 2: Correct the bending direction Let $\mathbf w=\left\langle -4,7,1\right\rangle$. Its tangential scalar component is $$\mathbf w\cdot\mathbf T=\frac{-8+7+2}{3}=\frac 13.$$ Therefore $$\mathbf w_\perp=\mathbf w-(\mathbf w\cdot\mathbf T)\mathbf T =\left\langle -4,7,1\right\rangle-\frac 19\left\langle 2,1,2\right\rangle =\frac 19\left\langle -38,62,7\right\rangle.$$ Since $\sqrt{(-38)^2+62^2+7^2}=\sqrt{5337}=3\sqrt{593}$, $$\boxed{\widehat{\mathbf N}=\frac 1{3\sqrt{593}}\left\langle -38,62,7\right\rangle}.$$

Step 3: Binormal and osculating plane $$\mathbf B=\mathbf T\times\widehat{\mathbf N} =\boxed{\frac 1{\sqrt{593}}\left\langle -13,-10,18\right\rangle}.$$ The numerator has length $\sqrt{169+100+324}=\sqrt{593}$. Since $\mathbf B$ is normal to the osculating plane through $P$, $$-13(x-1)-10(y+1)+18(z-2)=0,$$ or $$\boxed{13x+10y-18z+33=0}.$$ Substitution of $P$ gives zero, and the construction guarantees a right-handed orthonormal frame.