Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors - Practice Questions & MCQ

Vector Form

Let a plane passes through a point $A$ with position vector $\overrightarrow{\mathbf{a}}$ and parallel to two vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$

Let $\overrightarrow{\mathbf{r}}$ be the position vector of any point P on the plane.
$
\overrightarrow{A P}=\overrightarrow{O P}-\overrightarrow{O A}=\vec{r}-\vec{a}
$

Since, $\overrightarrow{A P}$ lies in the plane, hence, $\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are coplanar.
We have,
$
\begin{aligned}
(\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c}) & =0 \\
(\overrightarrow{\mathbf{r}}) \cdot(\vec{b} \times \vec{c}) & =(\vec{a}) \cdot(\vec{b} \times \vec{c}) \\
{\left[\begin{array}{lll}
\vec{r} & \vec{b} & \vec{c}
\end{array}\right] } & =\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]
\end{aligned}
$
or
or

Which is required equation of plane.

Cartesian Form

From $(\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0$ we have, $[\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$
\Rightarrow \quad\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3
\end{array}\right|=0
$

Which is required equation of plane in cartesian form where $\overrightarrow{\mathbf{b}}=x_2 \hat{\mathbf{i}}+y_2 \hat{\mathbf{j}}+z_2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=x_3 \hat{\mathbf{i}}+y_3 \hat{\mathbf{j}}+z_3 \hat{\mathbf{k}}$.