Trong không gian \(Oxyz\), cho các vectơ \(\vec{a}=(m;1;0)\), \(\vec{b}=(2;m-1;1)\), \(\vec{c}=(1;m+1;1)\). Tìm \(m\) để ba vectơ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) đồng phẳng.
![]() | \(m=\dfrac{3}{2}\) |
![]() | \(m=-2\) |
![]() | \(m=-\dfrac{1}{2}\) |
![]() | \(m=-1\) |
Chọn phương án C.
Ta có: \(\left[\vec{a},\vec{b}\right]=\left(1;-m;m^2-m-2\right)\).
\(\begin{aligned}\Rightarrow\left[\vec{a},\vec{b}\right]\cdot\vec{c}&=1\cdot1-m(m+1)+1\left(m^2-m-2\right)\\
&=-2m-1.\end{aligned}\)
Để \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) đồng phẳng thì $$\left[\vec{a},\vec{b}\right]\cdot\vec{c}=0\Leftrightarrow-2m-1=0\Leftrightarrow m=-\dfrac{1}{2}.$$