7. For any vectors $\vec { a } , \vec { b }$, which of the following relations always holds?
A. $| \vec { a } \bullet \vec { b } | \leq | \vec { a } | | \vec { b } |$
B. $| \vec { a } - \vec { b } | \leq | | \vec { a } | - | \vec { b } | |$
C. $( \vec { a } + \vec { b } ) ^ { 2 } = | \vec { a } + \vec { b } | ^ { 2 }$
D. $( \vec { a } + \vec { b } ) ( \vec { a } - \vec { b } ) = \vec { a } ^ { 2 } - \vec { b } ^ { 2 }$

Let $n$ be an integer such that $n \geqslant 2$. We denote by $E' = \operatorname{Vect}(e_{1}, \ldots, e_{n-1})$ and by $\pi$ the orthogonal projection onto $E'$. We set $X' = \pi \circ X = \sum_{i=1}^{n-1} \varepsilon_{i} e_{i}$. For $t$ in $\{-1, 1\}$ we denote $C_{t} = \pi(C \cap H_{t})$ where $H_t = E' + te_n$. For $t$ in $\{-1, 1\}$, we denote by $Y_{t}$ the projection of $X'$ onto the non-empty closed convex set $C_{t}$. Let $\lambda$ be a real such that $0 \leqslant \lambda \leqslant 1$.
Show that
$$d(X, C) \leqslant \left\|(1 - \lambda)(Y_{\varepsilon_{n}} + \varepsilon_{n} e_{n}) + \lambda(Y_{-\varepsilon_{n}} - \varepsilon_{n} e_{n}) - X\right\|$$

Let $n$ be an integer such that $n \geqslant 2$. We denote by $E' = \operatorname{Vect}(e_{1}, \ldots, e_{n-1})$ and by $\pi$ the orthogonal projection onto $E'$. We set $X' = \pi \circ X = \sum_{i=1}^{n-1} \varepsilon_{i} e_{i}$. For $t$ in $\{-1, 1\}$ we denote $C_{t} = \pi(C \cap H_{t})$ where $H_t = E' + te_n$. For $t$ in $\{-1, 1\}$, we denote by $Y_{t}$ the projection of $X'$ onto the non-empty closed convex set $C_{t}$. Let $\lambda$ be a real such that $0 \leqslant \lambda \leqslant 1$.
Deduce that
$$d(X, C)^{2} \leqslant 4\lambda^{2} + \left\|(1 - \lambda)(Y_{\varepsilon_{n}} - X') + \lambda(Y_{-\varepsilon_{n}} - X')\right\|^{2}$$
then that
$$d(X, C)^{2} \leqslant 4\lambda^{2} + (1 - \lambda) \|Y_{\varepsilon_{n}} - X'\|^{2} + \lambda \|Y_{-\varepsilon_{n}} - X'\|^{2}$$
Thus, we have shown the inequality
$$d(X, C)^{2} \leqslant 4\lambda^{2} + (1 - \lambda) d(X', C_{\varepsilon_{n}})^{2} + \lambda d(X', C_{-\varepsilon_{n}})^{2}$$

26. If If $\vec { a } , \vec { b }$, and $\vec { c }$ are unit vectors, then
$$| \vec { a } - \vec { b } | ^ { 2 } + | \vec { b } - \vec { c } | ^ { 2 } + | \vec { c } - \vec { a } | ^ { 2 }$$
does not exceed :
(A) 4
(B) 9
(C) 8
(D) 6

8. Let V be the volume of the parallelepiped formed by the vectors
$$\begin{aligned} & \vec { a } = a _ { 1 } \hat { \imath } + a _ { 2 } \hat { \jmath } + a _ { 3 } \hat { k } \\ & \vec { b } = b _ { 1 } \hat { \imath } + b _ { 2 } \hat { \jmath } + b _ { 3 } \hat { k } \\ & \vec { c } = c _ { 1 } \hat { \imath } + c _ { 2 } \hat { \jmath } + c _ { 3 } \hat { k } \end{aligned}$$
$r$, bsi,fą, where $\mathrm { r } = 1,2,3$ are non-negative real numbers and $\sum \mathrm { r } = 13 ( \mathrm { ar } + \mathrm { br } + \mathrm { cr } ) = 3 \mathrm {~L}$. Show that $\mathrm { V } < \mathrm { L } ^ { 3 }$.

2. $\vec { a } , \vec { b } , \vec { c } , \vec { d }$ are four distinct vectors satisfying the conditions $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$, then prove that $\vec { a } \cdot \vec { b } + \vec { c } \cdot \vec { d } \neq \vec { a } \cdot \vec { c } + \vec { b } \cdot \vec { d }$.
Sol. Given that $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$ $\Rightarrow \overrightarrow { \mathrm { a } } \times ( \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { c } } ) = ( \overrightarrow { \mathrm { c } } - \overrightarrow { \mathrm { b } } ) \times \overrightarrow { \mathrm { d } } = \overrightarrow { \mathrm { d } } \times ( \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { c } } ) \Rightarrow \overrightarrow { \mathrm { a } } - \overrightarrow { \mathrm { d } } \| \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { c } }$ $\Rightarrow ( \vec { a } - \vec { d } ) \cdot ( \vec { b } - \vec { c } ) \neq 0 \Rightarrow \vec { a } \cdot \vec { b } + \vec { d } \cdot \vec { c } \neq \vec { d } \cdot \vec { b } + \vec { a } \cdot \vec { c }$.

2. $\vec { a } , \vec { b } , \vec { c } , \vec { d }$ are four distinct vectors satisfying the conditions $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$, then prove that $\vec { a } \cdot \vec { b } + \vec { c } \cdot \vec { d } \neq \vec { a } \cdot \vec { c } + \vec { b } \cdot \vec { d }$.
Sol. Given that $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$ $\Rightarrow \vec { a } \times ( \vec { b } - \vec { c } ) = ( \vec { c } - \vec { b } ) \times \vec { d } = \vec { d } \times ( \vec { b } - \vec { c } ) \Rightarrow \vec { a } - \vec { d } \| \vec { b } - \vec { c }$ $\Rightarrow ( \vec { a } - \vec { d } ) \cdot ( \vec { b } - \vec { c } ) \neq 0 \Rightarrow \vec { a } \cdot \vec { b } + \vec { d } \cdot \vec { c } \neq \vec { d } \cdot \vec { b } + \vec { a } \cdot \vec { c }$.