ข้อสอบคณิต 9 สามัญ - ปี 2560

$$\begin{gathered} \mathrm{สูตร} \overline{u}\times \overline{v}=-\left(\overline{v}\times \overline{u}\right) \overline{u}·\overline{u}=1 \\ \left( \overline{u}\times \overline{v}\right)·\overline{w}=\left(\overline{v}\times \overline{w}\right)·\overline{u} \overline{u}\times \overline{u}=0 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\overline{a}\times \overline{c}\right)·\left(\overline{a}+\overline{b}+\overline{c}\right) \mathrm{เท่ากับข้อใด} \\ =\left[\left(\overline{a}\times \overline{c}\right)·\overline{a}\right]+\left[\left(\overline{a}\times \overline{c}\right)·\overline{b}\right]+\left[\left(\overline{a}\times \overline{c}\right)·\overline{c}\right] \\ =\left[-\left(\overline{c}\times \overline{a}\right)·\overline{a}\right]+\left[\left(\overline{a}\times \overline{c}\right)·\overline{b}\right]+\left[\left(\overline{c}\times \overline{c}\right)·\overline{a}\right] \\ =-\left[\left(\overline{c}\times \overline{a}\right)·\overline{a}\right]+\left[\left(\overline{c}\times \overline{b}\right)·\overline{a}\right]+\left[0·\overline{a}\right] \\ =-\left[\left(\overline{a}\times \overline{a}\right)·\overline{c}\right]+\left[-\left(\overline{b}\times \overline{c}\right)·\overline{a}\right]+0 \end{gathered}$$
$$\begin{gathered} =-\left[0·\overline{c}\right]-\left[\left(\overline{b}\times \overline{c}\right)·\overline{a}\right] \\ =-\left[\left(\overline{b}\times \overline{c}\right)·\overline{a}\right] \\ =-\left[\left(3i+2j-k\right)·\left(2i-j+k\right)\right] \\ =-\left[3\left(2\right)+2\left(-1\right)-1\left(1\right)\right] \\ =-3 \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\overline{a}\times \overline{c}\right)·\left(\overline{a}+\overline{b}+\overline{c}\right)=-3 \end{gathered}$$