Equation of a plane perpendicular to a given vector and passing through a given point - Practice Questions & MCQ

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. 

Let $\overrightarrow{\mathbf{n}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$ be a vector and $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0\right)$ be a point. Then the set of all point $\mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ such that $\overrightarrow{P Q}$ orthogonal to $\overrightarrow{\mathbf{n}}$ forms a plane.
We say that $\overrightarrow{\mathbf{n}}$ is a normal vector, or perpendicular to the plane. Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane:
$$
\overrightarrow{\mathbf{n}} \cdot \overrightarrow{P Q}=0
$$

If position vector of point $P$ is $\overrightarrow{\mathbf{P}}$ and position vector of point $Q$ is $\overrightarrow{\mathbf{q}}$, then
$$
(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}}) \cdot \overrightarrow{\mathbf{n}}=0
$$
$($ As $\overrightarrow{P Q}=\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$

This is the vector equation of the plane.

Cartesian form

Position vector of point P and point Q is $\overrightarrow{\mathbf{p}}=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ and $\overrightarrow{\mathbf{q}}=x \hat{i}+y \hat{j}+z \hat{k}$ respectively and vector $\overrightarrow{\mathbf{n}}$ is $a \hat{i}+b \hat{j}+c \hat{k}$
Then,
$$
\begin{array}{lc} 
& (\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}}) \cdot \overrightarrow{\mathbf{n}}=\mathbf{0} \\
\Rightarrow & \left((x \hat{i}+y \hat{j}+z \hat{k})-\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right)\right) \cdot(a \hat{i}+b \hat{j}+c \hat{k})=0 \\
\Rightarrow & {\left[\left(x-x_0\right) \hat{i}+\left(y-y_0\right) \hat{j}+\left(z-z_0\right) \hat{k}\right] \cdot(a \hat{i}+\mathrm{b} \hat{j}+c \hat{k})=0} \\
\text { i.e. } & \mathbf{a}\left(x-x_0\right)+\mathbf{b}\left(y-y_0\right)+\mathbf{c}\left(z-z_0\right)=\mathbf{0}
\end{array}
$$

Thus, the coefficients of x, y, and z in the cartesian equation of a plane are the direction ratios of the normal to the plane.