Question

Given 3 points given by position vectors $\overline{a},\overline{b}$ and $\overline{c}$. The plane which passes through these 3 points can be given by

• A. $(\overline{r}-\overline{a}).\left(\right(\overline{a}+\overline{b})\times (\overline{b}+\overline{c}\left)\right)=0$
• B. $(\overline{r}-\overline{a}).\left(\right(\overline{a}+\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0$
• C. $(\overline{r}-\overline{b}).\left(\right(\overline{a}-\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0$
• D. $(\overline{r}-\overline{a}).\left(\right(\overline{a}-\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0$

Answer 1

Answer: (C), (D)

$(\overline{r}-\overline{a}).\left(\right(\overline{a}-\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0$

For constructing a plane, one of the ways is to have a point through which the plane passes and perpendicular vector to the plane. Here we already have 3 points. We just have to calculate the direction vector. For that if we can get 2 vectors parallel to the plane we can get the perpendicular vector by cross product. If you see closely $(\overline{a}-\overline{b}),(\overline{b}-\overline{c})$ and $(\overline{c}-\overline{a})$ are parallel to the plane we want. So $(\overline{a}-\overline{b})\times (\overline{b}-\overline{c})$ serves as a direction vector. Now we have a point (either $\overline{a}or\overline{b}or\overline{c})$and a direction vector perpendicular to the plane , therefore equation of the required plane can be given by $(\overline{r}-\overline{a}).\left(\right(\overline{a}-\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0or(\overline{r}-\overline{b}).\left(\right(\overline{a}-\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0or(\overline{r}-\overline{c}).\left(\right(\overline{a}-\overline{b})\times (\overline{b}-\overline{c}\left)\right)=0$. Therefore options c and d are correct.