Cho hai vectơ a(1;1;-2) .b(1;0;m) Góc giữa chúng bằng 450 khi:

Góc giữa $\overrightarrow{u};\overrightarrow{v}$ bằng 45⁰ khi$\cos \left(\overrightarrow{u};\overrightarrow{v}\right)=\cos {45}^0$

Mà $\cos \left(\overrightarrow{u};\overrightarrow{v}\right)=\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|}$

$\Rightarrow cos{45}^0=\frac{1.1+1.0-2.m}{\sqrt{1^2+1^2+{(-2)}^2}.\sqrt{1^2+0^2+m^2}}$

$\iff \frac{1}{\sqrt{2}}=\frac{1-2m}{\sqrt{6}.\sqrt{1+m^2}}$

$\iff \sqrt{6(1+m^2)}=\sqrt{2}(1-2m)$

$\begin{array}{l}\iff \left\{\begin{array}{c}6(1+m^2)=2{(1-2m)}^2\\ 1-2m\ge 0\end{array}\right.\\ \iff \left\{\begin{array}{c}6+6m^2=2(1-4m+4m^2)\\ m\le \frac{1}{2}\end{array}\right.\end{array}$

$\iff \left\{\begin{array}{c}6+6m^2=2-8m+8m^2\\ m\le \frac{1}{2}\end{array}\right.$

$\iff \left\{\begin{array}{c}2m^2-8m-4=0\\ m\le \frac{1}{2}\end{array}\right.$

$\iff \left\{\begin{array}{c}m=2\pm \sqrt{6}\\ m\le \frac{1}{2}\end{array}\right.\iff m=2-\sqrt{6}$

Đáp án cần chọn là: C