Bình phương 2 vế:

\(\Leftrightarrow\overrightarrow{a}^2+\overrightarrow{b^2}+2\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}^2+\overrightarrow{b}^2-2\overrightarrow{a}.\overrightarrow{b}\)

\(\Leftrightarrow\overrightarrow{a}.\overrightarrow{b}=0\)

\(\Leftrightarrow\overrightarrow{a}\perp\overrightarrow{b}\)

Vậy đẳng thức xảy ra khi 2 vecto có phương vuông góc nhau

a: \(\overrightarrow{x}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{BA}\)

=>\(\overrightarrow{x}+\overrightarrow{BC}=\overrightarrow{BC}\)

=>\(\overrightarrow{x}=\overrightarrow{0}\)

b: \(\overrightarrow{CA}-\overrightarrow{x}-\overrightarrow{CB}=\overrightarrow{AB}\)

=>\(\overrightarrow{BC}+\overrightarrow{CA}-\overrightarrow{x}=\overrightarrow{AB}\)

=>\(\overrightarrow{BA}-\overrightarrow{x}=\overrightarrow{AB}\)

=>\(\overrightarrow{x}=\overrightarrow{BA}-\overrightarrow{AB}=2\cdot\overrightarrow{BA}\)

Chọn C.

Ta có

Nên hai vecto đã cho là ngược hướng.

\(\overrightarrow{u}\) ⊥\(\overrightarrow{v}\)

=>\(\overrightarrow{u}\cdot\overrightarrow{v}=0\)

=>\(\left(\overrightarrow{a}-3\cdot\overrightarrow{b}\right)\left(2\cdot\overrightarrow{a}+\overrightarrow{b}\right)=0\)

=>\(2\cdot\left(\overrightarrow{a}\right)^2+\overrightarrow{a}\cdot\overrightarrow{b}-6\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot\left(\overrightarrow{b}\right)^2=0\)

=>\(2\cdot2-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot3=0\)

=>\(-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}=5\)

=>\(\overrightarrow{a}\cdot\overrightarrow{b}=-1\)

Chọn C

\(\cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\dfrac{\left(-1\right)\cdot2+1\cdot0}{\sqrt{\left(-1\right)^2+1^2}+\sqrt{2^2+0^2}}=-2+\sqrt{2}\)

\(\Leftrightarrow\left(\overrightarrow{a},\overrightarrow{b}\right)=125^0\)