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A key question asked of a vector is, "Does it obey the usual rules of coordinate system transformations of vectors?" As expected, forces, accelerations, etc do. The number of people eating meals with you does not. Coordinate Transforms are discussed in detail here.

\[ |{\bf a}| = \sqrt{a^2_1 + a^2_2 + a^2_3} \]

\[ |{\bf a}| = \sqrt{3^2 + 7^2 + 2^2} = \sqrt{62} = 7.874 \]

\[ {\bf u} = {{\bf a} \over |{\bf a}|} = {(a_1, a_2, a_3) \over \sqrt{a^2_1 + a^2_2 + a^2_3}} \]

\[ {\bf u} = \left( {3 \over \sqrt{62}}, {7 \over \sqrt{62}}, {2 \over \sqrt{62}} \right) \]

\[ (1, 3, 2) + (4, 1, 7) = (1+4, 3+1, 2+7) = (5, 4, 9) \]

Vector addition can be written as

\[ {\bf c} = {\bf a} + {\bf b} \quad \quad \quad \text{or} \quad \quad \quad c_i = a_i + b_i \]

The first form is vector or matrix notation, where non-scalars are written in bold font. The second form has many names: index, indicial, tensor, and Einstein notation.

\[ {\bf a} \cdot {\bf b} = |{\bf a}| \, |{\bf b}| \cos \theta \]

where \(\theta\) is the angle between the two vectors. Applying this to the vector components gives

\[ \begin{eqnarray} {\bf a} \cdot {\bf b} & = & ( a_x {\bf i} + a_y {\bf j} + a_z {\bf k} ) \cdot( b_x {\bf i} + b_y {\bf j} + b_z {\bf k} ) \\ \\ & = & \matrix { a_x b_x ( {\bf i} \cdot {\bf i} ) & + & a_x b_y ( {\bf i} \cdot {\bf j} ) & + & a_x b_z ( {\bf i} \cdot {\bf k} ) & + \\ a_y b_x ( {\bf j} \cdot {\bf i} ) & + & a_y b_y ( {\bf j} \cdot {\bf j} ) & + & a_y b_z ( {\bf j} \cdot {\bf k} ) & + \\ a_z b_x ( {\bf k} \cdot {\bf i} ) & + & a_z b_y ( {\bf k} \cdot {\bf j} ) & + & a_z b_z ( {\bf k} \cdot {\bf k} ) } \\ \end{eqnarray} \]

but \( {\bf i} \cdot {\bf i} = {\bf j} \cdot {\bf j} = {\bf k} \cdot {\bf k} = 1 \) and \( {\bf i} \cdot {\bf j} = {\bf j} \cdot {\bf k} = {\bf k} \cdot {\bf i} = 0 \), leaving only

\[ {\bf a} \cdot {\bf b} = a_x b_x + a_y b_y + a_z b_z \]

Therefore, in summary, the dot product is

\[ {\bf a} \cdot {\bf b} = |{\bf a}| \, |{\bf b}| \cos \theta = a_x b_x + a_y b_y + a_z b_z \]

\[ {\bf a} \cdot {\bf b} = 3 * 1 + 7 * 2 + 2 * 3 = 23 \]

and since \( |{\bf a}| = 7.874 \), and \( |{\bf b}| = 3.742 \), then \(\theta\) can be solved for to find that the angle between the vectors is 38.7°.

This works because the length of \( {\bf b} \) along the direction of \( {\bf a} \) is given by \(|{\bf b}| \cos \theta\), where \(\theta\) is the angle between the two vectors. But this is the same as \(|{\bf u_a}| |{\bf b}| \cos \theta\), since \(|{\bf u_a}| = 1\). So it is the same as \({\bf u_a} \cdot {\bf b}\).

\[ a_i b_i \equiv a_1 b_1 + a_2 b_2 + a_3 b_3 \]

\[ W = \int {\bf F} \cdot d{\bf x} \]

And yes, W can be a negative quantity. If you are in a tug-of-war and your \( \int {\bf F} \cdot d{\bf x} \) is negative, then you are losing.

If the dot product is negative, then the angle between the vectors is greater than 90°. If the two vectors happen to be forces, then a negative dot product implies that the forces are cancelling each other out to some degree because the angle between them is greater than 90°.

If the dot product is positive, then the angle between the vectors is less than 90° and the two are contributing constructively in a given direction.

\[ \begin{eqnarray} {\bf a} \times {\bf b} & = & ( a_x {\bf i} + a_y {\bf j} + a_z {\bf k} ) \times( b_x {\bf i} + b_y {\bf j} + b_z {\bf k} ) \\ \\ & = & \matrix { a_x b_x ( {\bf i} \times {\bf i} ) & + & a_x b_y ( {\bf i} \times {\bf j} ) & + & a_x b_z ( {\bf i} \times {\bf k} ) & + \\ a_y b_x ( {\bf j} \times {\bf i} ) & + & a_y b_y ( {\bf j} \times {\bf j} ) & + & a_y b_z ( {\bf j} \times {\bf k} ) & + \\ a_z b_x ( {\bf k} \times {\bf i} ) & + & a_z b_y ( {\bf k} \times {\bf j} ) & + & a_z b_z ( {\bf k} \times {\bf k} ) } \\ \end{eqnarray} \]

but \( {\bf i} \times {\bf j} = {\bf k} \) and \( {\bf j} \times {\bf k} = {\bf i} \) etc, while \( {\bf i} \times {\bf i} = {\bf j} \times {\bf j} = {\bf k} \times {\bf k} = 0 \), leaving

\[ {\bf a} \times {\bf b} = (a_y b_z - a_z b_y) {\bf i} + (a_z b_x - a_x b_z) {\bf j} + (a_x b_y - a_y b_x) {\bf k} \]

The result can be conveniently written as a determinant as follows

\[ {\bf a} \times {\bf b} = \left | \matrix { {\bf i\;} & {\bf j\;} & {\bf k\;} \\ a_x & a_y & a_z \\ b_x & b_y & b_z } \right | = (a_y b_z - a_z b_y) {\bf i} + (a_z b_x - a_x b_z) {\bf j} + (a_x b_y - a_y b_x) {\bf k} \]

The magnitude of a cross product is related to the sine of the angle between the two inputs.

\[ | {\bf a} \times {\bf b} | = |{\bf a}| \, |{\bf b}| \sin \theta \]

\[ c_i = \epsilon_{ijk} a_j b_k \]

where \( \epsilon_{123} = \epsilon_{231} = \epsilon_{312} = 1 \), while \( \epsilon_{321} = \epsilon_{213} = \epsilon_{132} = -1 \), and all other combinations equal zero. Summation of the \(j\) and \(k\) indices from 1 to 3 is implied because they are repeated as subscripts in the above equation. In other words, it is shorthand for

\[ \matrix { c_i \; = \; \epsilon_{ijk} a_j b_k & = & \epsilon_{i11} a_1 b_1 & + & \epsilon_{i12} a_1 b_2 & + & \epsilon_{i13} a_1 b_3 & + & \\ & & \epsilon_{i21} a_2 b_1 & + & \epsilon_{i22} a_2 b_2 & + & \epsilon_{i23} a_2 b_3 & + & \\ & & \epsilon_{i31} a_3 b_1 & + & \epsilon_{i32} a_3 b_2 & + & \epsilon_{i33} a_3 b_3 } \]

The equation is still general until a particular component is chosen for \(i\) to be evaluated.

\[ \matrix { c_3 \; = \; \epsilon_{3jk} a_j b_k & = & \epsilon_{311} a_1 b_1 & + & \epsilon_{312} a_1 b_2 & + & \epsilon_{313} a_1 b_3 & + & \\ & & \epsilon_{321} a_2 b_1 & + & \epsilon_{322} a_2 b_2 & + & \epsilon_{323} a_2 b_3 & + & \\ & & \epsilon_{331} a_3 b_1 & + & \epsilon_{332} a_3 b_2 & + & \epsilon_{333} a_3 b_3 } \]

All subscripts are now specified, and this permits evaluation of all alternating tensors. All of them will equal zero except two. This leaves

\[ c_3 \; = \; \epsilon_{3jk} a_j b_k \; = \; a_1 b_2 - a_2 b_1 \]

which is consistent with the determinant result (as it had better be). Results for the x

\[ M_i = \epsilon_{ijk} r_j F_k \]

Likewise, the velocity, \({\bf v}\), of a point due to an angular rotation rate, \({\bf \omega}\), is \({\bf \omega} \times {\bf r}\). In tensor notation, this is

\[ v_i = \epsilon_{ijk} \omega_j r_k \]

And finally, the area of a triangle bounded on two sides by vectors \({\bf a}\) and \({\bf b}\) is

\[ Area = {1 \over 2} | \; {\bf a} \times {\bf b}| \]

In tensor notation, this is written in two steps as

\[ c_i = \epsilon_{ijk} a_j b_k \quad \quad \quad \text{and} \quad \quad \quad Area = {1 \over 2} \sqrt{c_i c_i} \]

or in a single equation as

\[ Area = {1 \over 2} \sqrt{ \epsilon_{ijk} a_j b_k \epsilon_{imn} a_m b_n } \]

\[ \begin{eqnarray} {\bf a} \otimes {\bf b} & = & ( a_x {\bf i} + a_y {\bf j} + a_z {\bf k} ) \otimes( b_x {\bf i} + b_y {\bf j} + b_z {\bf k} ) \\ \\ \\ & = & \matrix { a_x b_x ( {\bf i} \otimes {\bf i} ) & + & a_x b_y ( {\bf i} \otimes {\bf j} ) & + & a_x b_z ( {\bf i} \otimes {\bf k} ) & + \\ a_y b_x ( {\bf j} \otimes {\bf i} ) & + & a_y b_y ( {\bf j} \otimes {\bf j} ) & + & a_y b_z ( {\bf j} \otimes {\bf k} ) & + \\ a_z b_x ( {\bf k} \otimes {\bf i} ) & + & a_z b_y ( {\bf k} \otimes {\bf j} ) & + & a_z b_z ( {\bf k} \otimes {\bf k} ) } \\ \\ \\ & = & \left[ \matrix { a_x b_x & a_x b_y & a_x b_z \\ a_y b_x & a_y b_y & a_y b_z \\ a_z b_x & a_z b_y & a_z b_z } \right] \\ \end{eqnarray} \]

In effect, the diadic products such as \( ( {\bf i} \otimes {\bf i} ) \) and \( ( {\bf i} \otimes {\bf j} ) \) simply dictate the location of the terms in the tensor. A diadic product is sometimes referred to as the

\[ {\bf C} \, = \, {\bf a} \otimes {\bf b} \, = \, \left\{ \! \matrix { a_x \\ a_y \\ a_z } \! \right\} \matrix { \{ b_x \quad b_y \quad b_z \} \\ \text{ } \\ \text{ } \\ } \, = \, \left[ \matrix { a_x b_x & a_x b_y & a_x b_z \\ a_y b_x & a_y b_y & a_y b_z \\ a_z b_x & a_z b_y & a_z b_z } \right] \]

\[ c_{ij} = a_i b_j \]

Diadic products will be used in the calculation of resolved shear stresses on the traction vector page.

\[ \begin{eqnarray} {\bf a} \otimes {\bf b} & = & \left[ \matrix { 3*1 & 3*2 & 3*3 \\ 7*1 & 7*2 & 7*3 \\ 2*1 & 2*2 & 2*3 } \right] \\ \\ \\ & = & \left[ \matrix { 3 & 6 & 9 \\ 7 & 14 & 21 \\ 2 & 4 & 6 } \right] \\ \end{eqnarray} \]

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