For $\mathbf { C } , \mathbf { D } , \mathbf { E } , \mathbf { F } , \mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below. For the other $\square$, enter the correct number. Consider a regular tetrahedron OABC with sides of length 1. Let $x$ be a number satisfying $0 < x < 1$, and let P be the point that divides side AB by the ratio $x : ( 1 - x )$ and Q be the point that divides side BC by the ratio $x : ( 1 - x )$. Also, let $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$. We are to find the range of values of $\cos \angle \mathrm { POQ }$. The vectors $\vec { a } , \vec { b }$ and $\vec { c }$ satisfy $$\vec { a } \cdot \vec { b } = \vec { b } \cdot \vec { c } = \vec { c } \cdot \vec { a } = \frac { \mathbf { A } } { \mathbf { B } }$$ Next, since we can express $\overrightarrow { \mathrm { OP } }$ and $\overrightarrow { \mathrm { OQ } }$ as $\overrightarrow { \mathrm { OP } } = \square \mathbf { C }$ and $\overrightarrow { \mathrm { OQ } } = \mathbf { D }$, we have $$| \overrightarrow { \mathrm { OP } } | = | \overrightarrow { \mathrm { OQ } } | = \sqrt { \vec { E } } , \quad \overrightarrow { \mathrm { OP } } \cdot \overrightarrow { \mathrm { OQ } } = \square \mathbf { F } .$$ Hence we obtain $$\cos \angle \mathrm { POQ } = \frac { 1 } { \mathbf { G } } - \frac { \mathbf { H } } { \mathbf { I } } .$$ From this we finally obtain $$\frac { \square \mathbf { J } } { \mathbf { K } } < \cos \angle \mathrm { POQ } \leqq \frac { \mathbf { L } } { \mathbf { M } } .$$ (0) $( 1 - x ) \vec { a } + x \vec { b }$ (1) $x \vec { a } + ( 1 - x ) \vec { b }$ (2) $( 1 - x ) \vec { b } + x \vec { c }$ (3) $x \vec { b } + ( 1 - x ) \vec { c }$ (4) $x ^ { 2 } + x + 1$ (5) $x ^ { 2 } - x + 1$ (6) $x ^ { 2 } - x - 1$ (7) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x + 1 \right)$ (8) $\frac { 1 } { 2 } \left( - x ^ { 2 } - x + 1 \right)$ (9) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x - 1 \right)$
For $\mathbf { C } , \mathbf { D } , \mathbf { E } , \mathbf { F } , \mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below. For the other $\square$, enter the correct number.
Consider a regular tetrahedron OABC with sides of length 1. Let $x$ be a number satisfying $0 < x < 1$, and let P be the point that divides side AB by the ratio $x : ( 1 - x )$ and Q be the point that divides side BC by the ratio $x : ( 1 - x )$. Also, let $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$. We are to find the range of values of $\cos \angle \mathrm { POQ }$.
The vectors $\vec { a } , \vec { b }$ and $\vec { c }$ satisfy
$$\vec { a } \cdot \vec { b } = \vec { b } \cdot \vec { c } = \vec { c } \cdot \vec { a } = \frac { \mathbf { A } } { \mathbf { B } }$$
Next, since we can express $\overrightarrow { \mathrm { OP } }$ and $\overrightarrow { \mathrm { OQ } }$ as $\overrightarrow { \mathrm { OP } } = \square \mathbf { C }$ and $\overrightarrow { \mathrm { OQ } } = \mathbf { D }$, we have
$$| \overrightarrow { \mathrm { OP } } | = | \overrightarrow { \mathrm { OQ } } | = \sqrt { \vec { E } } , \quad \overrightarrow { \mathrm { OP } } \cdot \overrightarrow { \mathrm { OQ } } = \square \mathbf { F } .$$
Hence we obtain
$$\cos \angle \mathrm { POQ } = \frac { 1 } { \mathbf { G } } - \frac { \mathbf { H } } { \mathbf { I } } .$$
From this we finally obtain
$$\frac { \square \mathbf { J } } { \mathbf { K } } < \cos \angle \mathrm { POQ } \leqq \frac { \mathbf { L } } { \mathbf { M } } .$$
(0) $( 1 - x ) \vec { a } + x \vec { b }$\\
(1) $x \vec { a } + ( 1 - x ) \vec { b }$\\
(2) $( 1 - x ) \vec { b } + x \vec { c }$\\
(3) $x \vec { b } + ( 1 - x ) \vec { c }$\\
(4) $x ^ { 2 } + x + 1$\\
(5) $x ^ { 2 } - x + 1$\\
(6) $x ^ { 2 } - x - 1$\\
(7) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x + 1 \right)$\\
(8) $\frac { 1 } { 2 } \left( - x ^ { 2 } - x + 1 \right)$\\
(9) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x - 1 \right)$