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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

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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.

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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 $[ T1 \hspace{3pt} T2]^T$.

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