4.2 Equation of a plane perpendicular to a given vector and passing through a given point

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

The vector equation of plane passing through a point having position vector $\overrightarrow a $ and normal to the vector $\overrightarrow n $ is $$\left( {\overrightarrow r - \overrightarrow a } \right).\overrightarrow N = 0$$

Proof: Let us assume a plane passes through a point $A$ with position vector ${\overrightarrow a }$ and perpendicular to the vector $\overrightarrow N $.

Let us consider any arbitrary point on a plane $P(x,y,z)$ with position vector ${\overrightarrow r }$. If the point $P$ lies in the plane then, $\overrightarrow {AP} $ is perpendicular to $\overrightarrow N $ i.e., $$\overrightarrow {AP} .\overrightarrow N = 0$$ but $${\overrightarrow {AP} = \overrightarrow r - \overrightarrow a }$$ Therefore, we can write $$\left( {\overrightarrow r - \overrightarrow a } \right).\overrightarrow N = 0$$ which is the vector equation of plane passing through a given point $A$ and perpendicular to a given vector $\overrightarrow N $.

Note: There can be many planes that are perpendicular to the given vector, but only one such plane exists which passes through a given point.

As we discussed earlier the general equation of plane can be written as $$ax+by+cz+d=0...(1)$$ where $a$, $b$ and $c$ are the direction ratios of the normal $\overrightarrow N $ to the given plane.

Let us assume the equation of plane passing through a given point $A({x_1},{y_1},{z_1})$ in general form is $ax+by+cz+d=0$. Since it passes through $A$, co-ordinates must satisfy the equation of plane i.e, $$a{x_1} + b{y_1} + c{z_1} + d = 0...(2)$$

On subtracting $(1)$ and $(2)$, we get $$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1})0$$ which is the Cartesian equation of plane passing through a given point.

Question 11. Find the vector and cartesian equations of the plane which passes through the point $(5,2,4)$ and perpendicular to the line with direction ratios $2$, $3$ and $-1$.

Solution: We can write the position vector of given point as $$\overrightarrow a = 5\widehat i + 2\widehat j + 4\widehat k$$ and the normal vector to the plane using the direction ratios given as $$\overrightarrow N = 2\widehat i + 3\widehat j - \widehat k$$ Therefore, the vector equation of plane is $$\left( {\overrightarrow r - \overrightarrow a } \right).\overrightarrow N = 0$$

$$\left[ {\overrightarrow r - \left( {5\widehat i + 2\widehat j + 4\widehat k} \right)} \right].\left( {2\widehat i + 3\widehat j - \widehat k} \right) = 0$$

Now, To find the cartesian equation, put the value of position vector of any arbitrary point i.e., $${\overrightarrow r = x\widehat i + y\widehat j + z\widehat k}$$ We get

$$\begin{equation} \begin{aligned} \left[ {\left( {x\widehat i + y\widehat j + z\widehat k} \right) - \left( {5\widehat i + 2\widehat j + 4\widehat k} \right)} \right].\left( {2\widehat i + 3\widehat j - \widehat k} \right) = 0 \\ \left[ {\left( {x - 5} \right)\widehat i + \left( {y - 2} \right)\widehat j + \left( {z - 4} \right)\widehat k} \right].\left( {2\widehat i + 3\widehat j - \widehat k} \right) = 0 \\ 2\left( {x - 5} \right) + 3\left( {y - 2} \right) - 1\left( {z - 4} \right) = 0 \\ 2x + 3y - z = 12 \\\end{aligned} \end{equation} $$