Question

The vector equation of the plane passing through $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\mathrm{is}\overrightarrow{r}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}+\gamma \overrightarrow{c},$ provided that
(a) α + β + γ = 0
(b) α + β + γ =1
(c) α + β = γ
(d) α² + β² + γ² = 1

Answer 1

(b) α + β + γ =1
Given: A plane passing through $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$.

⇒ Lines $\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{c}-\overrightarrow{a}$ lie on the plane.

The parmetric equation of the plane can be written as:

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+{\lambda }_1(\overrightarrow{a}-\overrightarrow{b})+{\lambda }_2(\overrightarrow{c}-\overrightarrow{a})\\ \overrightarrow{r}=\overrightarrow{a}(1+{\lambda }_1-{\lambda }_2)-{\lambda }_1\overrightarrow{b}+{\lambda }_2\hspace{0.17em}\overrightarrow{c}\\ \mathrm{Given}\hspace{0.17em}\hspace{0.17em}\mathrm{that}\overrightarrow{r}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}+\gamma \overrightarrow{c}\\ \therefore \hspace{0.17em}\hspace{0.17em}\alpha +\beta +\gamma =1+{\lambda }_1-{\lambda }_2-{\lambda }_1+{\lambda }_2\\ \alpha +\beta +\gamma =1\end{array}$