Question

Equation of a plane passing through three non-collinear points is
Note Where $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ are coordinates of three non-collinear points.

• A. $( r - a ) \cdot[( b - a ) \times( c - a )]=0$, where $a , b , c$ are the position vectors of three non-collinear points
• B. $\begin{vmatrix}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$
• C. Both(a) and (b)
• D. None of the above

Answer 1

Answer: (C)

Let $R, S$ and $T$ be three non-collinear points on the plane with position vectors $a, b$ and $c$, respectively.

The vector $RS$ and $RT$ are in the given plane. Therefore, the vector $RS \times RT$ is perpendicular to the plane containing points $R, S$ and $T$. Let $r$ be the position vector of any point $P$ in the plane. Therefore, the equation of the plane passing through $R$ and perpendicular to the vector $RS \times RT$ is
$(r-a) \cdot(R S \times R T) =0 \quad \text { (if a } \perp b \text {, then } a \cdot b=0) $
$\Rightarrow \quad(r-a) \cdot[(b-a) \times(c-a)] =0$ ...(i)
This is the equation of the plane in vector form passing through three non-collinear points.
Note Why was it necessary to say that the three points had to be non-collinear? If the three points were on the same line, then there will be many planes that will contain them.

These planes will resemble the pages of a book where the line containing the points $R, S$ and $T$ are members in the binding of the book.
Cartesian Form
Let $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ be the coordinates of the points $R, S$ and $T$, respectively. Let $(x, y, z)$ be the coordinates of any point $P$ on the plane with position vector
r. Then,
$RP =\left(x-x_1\right) \hat{ i }+\left(y-y_1\right) \hat{ j }+\left(z-z_1\right) \hat{ k }$
$RS =\left(x_2-x_1\right) \hat{ i }+\left(y_2-y_1\right) \hat{ j }+\left(z_2-z_1\right) \hat{ k }$
$RT =\left(x_3-x_1\right) \hat{ i }+\left(y_3-y_1\right) \hat{ j }+\left(z_3-z_1\right) \hat{ k }$
Substituting these values in Eq. (i) of the vector form and expressing it in the form of a determinant, we have
$\begin{vmatrix}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$
which is the equation of the plane in cartesian form passing through three non-collinear points $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$