MHB Vectors in $R^3$: Is $\begin{bmatrix}4\\3\\0\end{bmatrix}$ Valid?

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Is $\begin{bmatrix} 4 \\ 3 \\ 0 \end{bmatrix}$ in $R^3$?
 
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bwpbruce said:
Is $\begin{bmatrix} 4 \\ 3 \\ 0 \end{bmatrix}$ in $R^3$?

Yes, it is!
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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