Question

Let aand bbe two constant vectors. Show that

$\mathbf{\int }\left(a.\stackrel{⏜}{r}\right)\left(b.\stackrel{⏜}{r}\right)\mathbf{sin\theta d\theta d\varphi }\mathbf{=}\frac{\mathbf{4}\mathbf{\pi }}{\mathbf{3}}\left(a.b\right)$

(the integration is over the usual range:$\mathbf{0}\mathbf{<}\mathit{\theta }\mathbf{<}\mathbf{\pi }\mathbf{,}\mathbf{0}\mathbf{<}\mathbf{\varphi }\mathbf{<}\mathbf{2}\mathbf{\pi }$). Use this result to demonstrate that

$\left(\frac{3\left(S_p.\stackrel{⏜}{r}\right)\left(S_e.\stackrel{⏜}{r}\right)-S_p.S_e}{r^3}\right)\mathbf{=}\mathbf{0}$

For states with I=0. Hint:$\stackrel{\mathbf{⏜}}{\mathbf{r}}\mathbf{=}\mathbf{sin\theta cos\varphi }\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{sin\theta sin\varphi }\stackrel{\mathbf{⏜}}{\mathbf{\Phi }}\mathbf{+}\mathbf{cos\theta }\stackrel{\mathbf{⏜}}{\mathbf{k}}$.

Short answer

It is proved that$\int \left(\mathrm{a}.\stackrel{⏜}{\mathrm{r}}\right)\left(\mathrm{b}.\stackrel{⏜}{\mathrm{r}}\right)\mathrm{sin\vartheta d\vartheta d\phi }=\frac{4\mathrm{\pi }}{3}\left(\mathrm{a}.\mathrm{b}\right)$ .

Step-by-step solution

Definition of vectors.

Geometrical entities with magnitude and direction are known as vectors. A vector is represented as a line with an arrow pointing in the direction of the vector, and the length of the line denotes the vector's magnitude. As a result, vectors are represented by arrows and have two points: a beginning point and a terminal point.

Step2: Structural isomers of carboxylic acids.

To prove following relation:

$\int \left(a.\stackrel{⏜}{r}\right)\left(b.\stackrel{⏜}{r}\right)\sin \vartheta d\vartheta d\phi =\frac{4\mathrm{\pi }}{3}\left(a.b\right)$

It is known that:

$$\begin{gathered} a.\stackrel{⏜}{r}=a_x\sin \vartheta \cos \phi +a_y\sin \vartheta \sin \phi +a_z\cos \vartheta \\ I=\int \left(a_x\sin \vartheta \cos \phi +a_y\sin \vartheta \sin \phi +a_z\cos \vartheta \right)\left(b_x\sin \vartheta \cos \phi +b_y\sin \vartheta \sin \phi +b_z\cos \vartheta \right)\sin \vartheta d\vartheta d\phi {\int }_0^{2\mathrm{\pi }}\sin \phi d\phi \\ ={\int }_0^{2\mathrm{\pi }}\cos \phi d\phi \\ ={\int }_0^{2\mathrm{\pi }}\sin \phi \cos \phi d\phi \\ =0 \\ I=\int \left(a_x{b}_x{\sin }^2\vartheta {\cos }^2\phi +a_y{b}_y{\sin }^2\vartheta {\sin }^2\phi +a_z{b}_z{\cos }^2\vartheta \right)\sin \vartheta d\vartheta d\phi \end{gathered}$$

But

$$\begin{gathered} ={\int }_0^{2\mathrm{\pi }}{\cos }^2\phi d\phi \\ =\mathrm{\pi }, \end{gathered}$$

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}d\phi =2\mathrm{\pi } \\ \mathrm{I}={\int }_0^{\mathrm{\pi }}\left[\mathrm{\pi }\left({\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{y}}\right){\sin }^2\mathrm{\vartheta }+2{\mathrm{\pi a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{z}}{\cos }^2\mathrm{\vartheta }\right]\mathrm{sin\vartheta d\vartheta } \end{gathered}$$

But

$$\begin{gathered} {\int }_0^{\mathrm{\iota \iota }}{\sin }^3\vartheta d\vartheta =\frac{4}{3} \\ {\int }_0^{\mathrm{\iota \iota }}{\cos }^2\vartheta \sin \vartheta d\vartheta =\frac{2}{3} \end{gathered}$$

So,

$$\begin{gathered} I=\mathrm{\pi }\left({\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{y}}\right)\frac{4}{3}{\mathrm{\pi a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{z}}\frac{2}{3} \\ =\frac{4}{3}\mathrm{\pi }\left({\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{x}}+{\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{y}}+{\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{z}}\right) \\ =\frac{4}{3}\mathrm{\pi }\left(\mathrm{a}.\mathrm{b}\right) \end{gathered}$$