Trong không gian xét véctơ m. vectơ n, vectơ p, vectơ q là những vectơ đơn vị

* Hướng dẫn giải

Đáp án D

Ta có: $0\le {\left|\overrightarrow{m}+\overrightarrow{n}+\overrightarrow{p}+\overrightarrow{q}\right|}^2$ = $4+2\left(\overrightarrow{m}\overrightarrow{n}+\overrightarrow{m}\overrightarrow{p}+\overrightarrow{m}\overrightarrow{q}+\overrightarrow{n}\overrightarrow{p}+\overrightarrow{n}\overrightarrow{q}+\overrightarrow{p}\overrightarrow{q}\right)$

Do đó $\overrightarrow{m}\overrightarrow{n}+\overrightarrow{m}\overrightarrow{p}+\overrightarrow{m}\overrightarrow{q}+\overrightarrow{n}\overrightarrow{p}+\overrightarrow{n}\overrightarrow{q}+\overrightarrow{p}\overrightarrow{q}\ge -2$

Lại có: ${\left|\overrightarrow{m}-\overrightarrow{n}\right|}^2+{\left|\overrightarrow{m}-\overrightarrow{p}\right|}^2+{\left|\overrightarrow{m}-\overrightarrow{q}\right|}^2$ $+{\left|\overrightarrow{n}-\overrightarrow{p}\right|}^2+{\left|\overrightarrow{n}-\overrightarrow{q}\right|}^2+{\left|\overrightarrow{p}-\overrightarrow{q}\right|}^2$

= $3\left({\overrightarrow{m}}^2+{\overrightarrow{n}}^2+{\overrightarrow{p}}^2+{\overrightarrow{q}}^2\right)$ $-$ $2\left(\overrightarrow{m}\overrightarrow{n}+\overrightarrow{m}\overrightarrow{p}+\overrightarrow{m}\overrightarrow{q}+\overrightarrow{n}\overrightarrow{p}+\overrightarrow{n}\overrightarrow{q}+\overrightarrow{p}\overrightarrow{q}\right)$ $\le$ 12$-$2($-$2) = 16

Vậy M = 16 => $M-\sqrt{M}$ = 12