Question

Verify equation (10.7). Hint: Use equations (2.4) to (2.6) and (2.10). For example,$\mathbf{\partial }{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{/}\mathbf{\partial }\mathit{z}\mathbf{=}\mathbf{\partial }\mathit{z}\mathbf{/}\mathbf{\partial }{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}\mathbf{=}{\mathit{a}}_{\mathbf{23}}$.

Short answer

$\frac{\partial {x^{\text{'}}}_i}{\partial x_j}=a_{ij}=\frac{\partial x_j}{\partial {x^{\text{'}}}_i}$

Thus,the cosines angles are equal.

Step-by-step solution

Step 1:Determine the cartesian vectors.

Let the rectangular axes be$(x_1,x_2,x_3)$ and the other set by rotating another set is $({x^{\text{'}}}_1,{x^{\text{'}}}_2,{x^{\text{'}}}_3)$.

Now we transform equation from the coordinate system$(x_1,x_2,x_3)$and $({x^{\text{'}}}_1,{x^{\text{'}}}_2,{x^{\text{'}}}_3)$is,

$\begin{array}{l}{x^{\text{'}}}_1=a_{11}{x}_1+a_{12}{x}_2+a_{13}{x}_3\\ {x^{\text{'}}}_2=a_{21}{x}_1+a_{22}{x}_2+a_{23}{x}_3\\ {x^{\text{'}}}_3=a_{31}{x}_1+a_{32}{x}_2+a_{33}{x}_3\end{array}$

Now we transform equation from the coordinate system$({x^{\text{'}}}_1,{x^{\text{'}}}_2,{x^{\text{'}}}_3)$and $(x_1,x_2,x_3)$is,

$\begin{array}{l}{x}_1=a_{11}{x^{\text{'}}}_1+a_{21}{x^{\text{'}}}_2+a_{31}{x^{\text{'}}}_3\\ x_2=a_{12}{x^{\text{'}}}_1+a_{22}{x^{\text{'}}}_2+a_{32}{x^{\text{'}}}_3\\ x_3=a_{13}{x^{\text{'}}}_1+a_{23}{x^{\text{'}}}_2+a_{33}{x^{\text{'}}}_3\end{array}$

Determine the co-ordinates.

So, the coordinated in terms of x is,

$\begin{array}{c}{x}^{\text{'}}=\sum _{j=1}^3{a}_{ij}{x}_{j}i=1,2,3\\ x^{\text{'}}=a_{ij}{x}_{j}i,j=1,2,3\end{array}$

So now we differentiate ${x^{\text{'}}}_i$with respect$x_j$. So, we get,

$\frac{\partial {x^{\text{'}}}_i}{\partial x_j}=a_{ij}$

So, the x coordinated in terms of $x^{\text{'}}$is,

$\begin{array}{c}x=\sum _{j=1}^3{a}_{ij}{x^{\text{'}}}_{j}i=1,2,3\\ x=a_{ij}{x^{\text{'}}}_{j}i,j=1,2,3\end{array}$

So now we differentiate $x_j$with respect ${x^{\text{'}}}_i$. So, we get,

$\frac{\partial x_j}{\partial {x^{\text{'}}}_i}=a_{ij}$

So from the above expressions we get,

$\frac{\partial {x^{\text{'}}}_i}{\partial x_j}=a_{ij}=\frac{\partial x_j}{\partial {x^{\text{'}}}_j}$

Thus, the cosine of the angle between ${x^{\text{'}}}_i$and $x_j$are equal between the axes.