Two tangent vectors at a point on a surface are $T_1 = 4i + 2j + 3k$ and $T_2 = -2i - 3j + 1k$
Using the property of the dot product of two normal vectors, determine the unit vector normal to the surface at the point
Two tangent vectors at a point on a surface are $T_1 = 4i + 2j + 3k$ and $T_2 = -2i - 3j + 1k$
Using the property of the dot product of two normal vectors, determine the unit vector normal to the surface at the point
First, to find the direction of the normal vector, you want to take the cross product of the vectors for $T_1$ and $T_2$. You get the direction of the normal vector $n$ as such: $ n=T_1\times T_2=11i-10j-8k. $
You can verify that this vector is orthogonal to both $T_1$ and $T_2$, by seeing that their dot product with $n$ is $0$ in both cases, and thus normal to the surface. All that remains is to make the vector a unit vector, which you can do by dividing each coordinate by the length of the vector, and then you are done.
You can use a null-space method. Note that if we are trying to find $\mathbf u$ then we require that $\mathbf u \cdot T1 = 0$ and $\mathbf u \cdot T2 = 0$. We solve this by finding the null space of the matrix $[ T1 \hspace{3pt} T2]^T$.