Trong không gian Oxyz cho hai điểm A(0;−2;3),B(1;0;−1). Tính sin góc hợp

Ta có:

$\begin{array}{l}\overrightarrow{OA}=\left(0;-2;3\right)\Rightarrow \left|\overrightarrow{OA}\right|=\sqrt{0^2+{\left(-2\right)}^2+3^2}=\sqrt{13}\\ \overrightarrow{OB}=\left(1;0;-1\right)\Rightarrow \left|\overrightarrow{OB}\right|=\sqrt{1^2+0^2+{\left(-1\right)}^2}=\sqrt{2}\end{array}$

Suy ra

$\left[\overrightarrow{OA},\overrightarrow{OB}\right]=\left(\left|\begin{array}{cc}\begin{array}{l}-2\\ 0\end{array}& \begin{array}{l}3\\ -1\end{array}\end{array}\right|;\left|\begin{array}{cc}\begin{array}{l}3\\ -1\end{array}& \begin{array}{l}0\\ 1\end{array}\end{array}\right|;\left|\begin{array}{cc}\begin{array}{l}0\\ 1\end{array}& \begin{array}{l}-2\\ 0\end{array}\end{array}\right|\right)=\left(2;3;2\right)$

$\Rightarrow \left|\left[\overrightarrow{OA},\overrightarrow{OB}\right]\right|=\sqrt{2^2+3^2+2^2}=\sqrt{17}$

Do đó $\sin \left(\overrightarrow{OA},\overrightarrow{OB}\right)=\frac{\left|\left[\overrightarrow{OA},\overrightarrow{OB}\right]\right|}{\left|\overrightarrow{OA}\right|.\left|\overrightarrow{OB}\right|}=\frac{\sqrt{17}}{\sqrt{13}.\sqrt{2}}=\sqrt{\frac{17}{26}}$

Đáp án cần chọn là: D