Three Dimensional Geometry

Imagine a world with no height, with just two dimensional objects like square, rectangle and circle, etc. It just feels meaningless and impossible!!! Right??? Now atleast let us think of games like Pubg and Free fire in two dimension, it also doesn't make any sense. These simple imaginations helps us understand the importance of three dimensional geometry. Every single thing around us is a three dimensional object. Even a piece of paper has some thickness.

This article is about the concept of three dimensional geometry class 11 and three dimensional geometry class 12. Three Dimensional Geometry chapter is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE, BCECE, and more.

Introduction to Three Dimensional Geometry

Three dimensional geometry is the one which defines the world. Three dimensional geometry is one of the fundamental topics in various domains like astronomy, engineering, architecture, and so on.

3D Coordinate System

The position of a point in two-dimension (2D) is given by two numbers $P(x, y)$ but in three-dimension geometry, the position of a point $P$ is given by three numbers $P(x, y, z)$.

Three mutually perpendicular lines intersect at one point, the point $O(0, 0, 0)$ is known as the origin in the space. These three mutually perpendicular lines form three planes namely $XY, YZ, ZX$ called coordinate planes. The $x$ and $y$ axis make $X Y$ plane, $y, z$ axis make $Y Z$ plane, similarly, ${x}, z$ axis make $XZ$ plane. This three-plane divides space into eight regions called octants.

The sign of the coordinates of a point determines the octant in which the point lies. The following table shows the signs of the coordinates in eight octants.

Three Dimensional Geometry Class 12 Formulas

Three dimensional geometry formulas include direction cosines, direction ratios, section formula, equation of line in 3D, angle between two lines, equation of plane and the angle between two planes.

Direction Cosines

Direction Cosine(DC) gives the relation of a vector or a line in a three-dimensional space, with each of the three axes. The direction cosine is the cosine of the angle subtended by this line with the $x$-axis, $y$-axis, and $z$-axis respectively. Let $r$ be the position vector of a point $P(x, y, z)$. Then, the direction cosines of vector $r$ are the cosines of angles $α, β,$ and $γ$ (i.e. $\cos α, \cos β,$ and $\cos γ$) that the vector $r$ makes with the positive direction of $X, Y,$ and $Z$ -axes respectively. Direction cosines are usually denoted by $l, m,$ and $n$ respectively.

$\alpha, \beta, \gamma$ are the angles that a vector makes with positive $X$ -axis, $Y$ -axis, and $Z$ -axis respectively then $\cos \alpha, \cos \beta, \cos \gamma$ are known as direction cosines, generally denoted by $(l,m,n)$.

$
\begin{aligned}
& l=\cos \alpha, m=\cos \beta, n=\cos \gamma \\
& l^2+m^2+n^2=1 \\
& \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1
\end{aligned}
$

Direction Cosines of the Line Passing Through Two Points

Let $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ and $\mathrm{Q}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)$ be two points on the line L .
Let $1, \mathrm{~m}$, and n be the direction cosines of the line $P Q$, and let it make angles $\alpha, \beta$, and $y$ with the $x$-axis, $y$-axis, and $z$-axis respectively.
The direction cosines of the line segment joining the points $P$ and $Q$ are given by

$
\left(\frac{x_2-x_1}{P Q}, \frac{y_2-y_1}{P Q}, \frac{z_2-z_1}{P Q}\right)
$

Properties of Direction Cosines

i) $\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2$
ii) If $\mathrm{OP}=\mathrm{r}$ then the co-ordinates of P will be ( $\mathrm{lr}, \mathrm{mr}, \mathrm{nr}$ )
iii) Direction cosines of the $X$ -axis are $(1,0,0)$
iv) Direction cosines of the $Y$ -axis are $(0,1,0)$
v) Direction cosines of the $Z$ -axis are $(0,0,1)$

Direction Ratios

Direction Ratios are any set of three numbers that are proportional to the Direction cosines.
If $\mathrm{l}, \mathrm{m}, \mathrm{n}$ are DCs of a vector then $\lambda l, \lambda m, \lambda n$ are DRs of this vector, where $\lambda$ can take any real value.
DRs are also denoted as $\mathrm{a}, \mathrm{b}$, and c respectively.
A vector has only one set of DCs, but infinite sets of DRs.
Note:
The coordinates of a point equal Ir, mr, and nr, which are proportional to the direction cosines. Hence the coordinates of a point are also its DRs.
If $\vec{r}=a i+b \hat{j}+c \hat{k}$, then $a, b$ and $c$ are one of the direction ratios of the given vector. Also, if $a^2+b^2+c^2=1$, then $a, b$ and $c$ will be direction cosines of given vector.

Properties of Directions Ratios

(i) If $a, b,$ and $c$ are direction ratios then direction cosines will be

$
l=\frac{ \pm a}{\sqrt{a^2+b^2+c^2}}, m=\frac{ \pm b}{\sqrt{a^2+b^2+c^2}}, n=\frac{ \pm c}{\sqrt{a^2+b^2+c^2}}
$

(ii) Direction ratios of a line joining two given points
$A\left(x_1, y_1, z_1\right)$ and $B\left(x_2, y_2, z_2\right)$ is given by

$
\left(x_2-x_1, y_2-y_1, z_2-z_1\right)
$

(iii) If $r=a \hat{i}+b \hat{j}+c \hat{k}$ be a vector with direction cosines $\mathrm{I}, \mathrm{m}, \mathrm{n}$ then

$
l=\frac{a}{|r|}, m=\frac{b}{|r|}, n=\frac{c}{|r|}
$

Section Formula in Three Dimensional Geometry

If the point $R(x, y, z)$ divides the line joining the points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ internally in the ratio $m: n$, then the coordinate of $R(x, y, z)$ is given by :

$\mathrm{R}(\mathrm{x}, \mathrm{y} . \mathrm{z})=\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)$

If the point $R$ divides $P Q$ externally in the ratio $m$ : $n$, then its coordinates are obtained by replacing $n$ with $(-n)$, so that the coordinates of point $R$ will be

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)$

Equation of Line in 3D

The equation of the straight line in 3D requires two points that are located in space. The point in three coordinates is expressed as ( $x, y, z)$.

Equation of a line through a given point and parallel to a given vector in Vector Form

Let $L$ be $a$ line in space passing through point $P\left(x_0, y_0, z_0\right)$. Let $\overrightarrow{\mathbf{b}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$ be a vector parallel to $L$ . Then, for any point on line $\mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{z})$, we know that vector $P Q$ is parallel to vector $b$ . Thus, there is a scalar, $\lambda$, such that $\overrightarrow{P Q}=\lambda \overrightarrow{\mathbf{b}}$, which gives,

$
\begin{gathered}
\overrightarrow{P Q}=\lambda \tilde{\mathbf{b}} \\
\left(x-x_0\right) \hat{i}+\left(y-y_0\right) \hat{j}+\left(z-z_0\right) \hat{k}=\lambda(a \hat{i}+b \hat{j}+c \hat{k})
\end{gathered}
$

Using vector operations, we can rewrite,

$
\begin{aligned}
(x \hat{i}+y \hat{j}+z \hat{k})-\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right) & =\lambda(a \hat{i}+b \hat{j}+c \hat{k}) \\
(x \hat{i}+y \hat{j}+z \hat{k}) & =\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right)+\lambda(a \hat{i}+b \hat{j}+c \hat{k})
\end{aligned}
$

Setting $\overrightarrow{\mathbf{r}}=x \hat{i}+y \hat{j}+z \hat{k}$ and $\overrightarrow{\mathbf{r}}_0=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ we now have the vector equation of a line:

$
\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}
$

Equation of a line through a given point and parallel to a given vector in Cartesian Form

The Equation of a line through a given point and parallel to a given vector in Cartesian Form is $
\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}
$

Equation of a line passing through two given points in Cartesian Form

We can also find parametric equations for the line segment $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{p}}+\lambda(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$.

As the position vector of point $\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is $\overrightarrow{\mathbf{r}}=(x \hat{i}+y \hat{j}+z \hat{k})$.

Angle Between Two Lines

The intersection of two straight lines forms an angle. For two intersecting lines, there are two types of angles between the lines, the acute angle and the obtuse angle. The angle between two lines generally gives the acute angle between the two lines.

Angle Between Two Lines in Vector Form

Let the given lines be,
$
\begin{aligned}
& \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}} \\
& \overrightarrow{\mathbf{r}}={\overrightarrow{\mathbf{r}^{\prime}}}_0+\lambda \overrightarrow{\mathbf{b}}^{\prime}
\end{aligned}
$
As equation (i) and equation (ii) are straight lines in the directions of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{b}}^{\prime}$, respectively.
Let $\theta$ be the angle between the vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{b}}^{\prime}$
Using the dot product,

$
\begin{aligned}
\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}^{\prime} & =|\overrightarrow{\mathbf{b}}|\left|\overrightarrow{\mathbf{b}}^{\prime}\right| \cos \theta \\
\Rightarrow \quad \cos \theta & =\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}^{\prime}}}{|\overrightarrow{\mathbf{b}}|\left|\overrightarrow{\mathbf{b}^{\prime}}\right|}
\end{aligned}
$

Angle Between Two Lines in Cartesian Form

The equation of a straight line in cartesian form is
$
\begin{aligned}
& \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \\
& \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}
\end{aligned}
$

Then,
$
\overrightarrow{\mathbf{b}}=a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k} \quad \text { and } \quad \overrightarrow{\mathbf{b}}^{\prime}=a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}
$
So that,
$
\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}^{\prime}=a_1 a_2+b_1 b_2+c_1 c_2
$
$
\begin{aligned}
|\overrightarrow{\mathbf{b}}| & =\sqrt{a_1^2+b_1^2+c_1^2}, \quad \text { and } \quad\left|\overrightarrow{\mathbf{b}}^{\prime}\right|=\sqrt{a_2^2+b_2^2+c_2^2} \\
\cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}
\end{aligned}
$

Ange between two lines in terms of direction cosines and direction ratios

If two lines having direction ratios $\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1$ and $\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2$ then the angle between them is given by

$
\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}
$
If two lines have direction ratios as $\mathrm{l}_1, \mathrm{~m}_1, \mathrm{n}_1$ and $\mathrm{l}_2, \mathrm{~m}_2, \mathrm{n}_2$ then the angle between them is given by$
\cos \theta=l_1 l_2+m_1 m_2+n_1 n_2
$

Shortest Distance Between Two Lines

The shortest distance between two lines in three-dimensional space is the length of the perpendicular segment drawn from a point on one line to the other line.

Shortest Distance between two skew lines

When two lines are neither parallel nor intersecting at a point they are referred to as Skew Lines.

If $\vec{r}=\vec{a}+\lambda \vec{b}$ and $\vec{r}=\overrightarrow{a_1}+\mu \vec{b}$ are skew lines then the shortest distance between them is given by

$
\left|\frac{\left(\vec{b} \times \overrightarrow{b_1}\right) \cdot\left(\vec{a}-\overrightarrow{a_1}\right)}{\left|\vec{b} \times \overrightarrow{b_1}\right|}\right|
$

Distance between Intersecting lines

The shortest distance between intersecting lines is $0$.

$\left(\vec{b} \times \overrightarrow{b_1}\right) \cdot\left(\overrightarrow{\mathbf{r}}_0^{\prime}-\overrightarrow{\mathbf{r}}_0\right)=0$

Shortest Distance between parallel lines

If the direction vectors of two lines are parallel and the two lines never meet they are referred as Parallel Lines.

If $\vec{r}=\overrightarrow{p_1}+\lambda \vec{\nu}$ and $\vec{r}=\overrightarrow{p_2}+\mu \vec{\nu}$ are Parallel Lines then the shortest distance between them is given by $
\frac{\left|\left(\overrightarrow{p_2}-\overrightarrow{p_1}\right) \times \vec{\nu}\right|}{|\vec{\nu}|}
$

What is a Plane?

We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line.

Equation of Plane

The equation of a plane in a cartesian coordinate system can be computed through different methods based on the available input values about the plane.

Equation of plane in normal form in vector form

The vector equation of a plane normal to unit vector $\hat{\mathbf{n}}$ and at a distance d from the origin is $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$.

Equation of plane in normal form in cartesian form

If $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is any point in the plane and $\hat{\mathbf{n}}$ is the unit vector normal to the plane.

Let $\mathrm{l}, \mathrm{m}, \mathrm{n}$ be the direction cosines. Then Cartesian equation of the plane in the normal form is given by

$\begin{array}{r}(x \hat{i}+y \hat{j}+z \hat{k}) \cdot({l} \hat{i}+m \hat{j}+n \hat{k})=d \\ \mathbf{l x}+\mathbf{m y}+\mathbf{n z}=\mathbf{d}\end{array}$

Equation of a plane perpendicular to a given vector and passing through a given point in Vector Form

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 points $Q(\mathrm{x}, \mathrm{y}, \mathrm{z})$ 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 the position vector of point $P$ is $\overrightarrow{\mathbf{P}}$ and the plosition vector of point $Q$ is $\overrightarrow{\mathbf{q}}$ , then

$(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}}) \cdot \overrightarrow{\mathbf{n}}=\mathbf{0} \quad($ As $\overrightarrow{P Q}=\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$

This is the vector equation of the plane.

Equation of a plane perpendicular to a given vector and passing through a given point in 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(\mathrm{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.

Equation of a plane passing through three non-collinear points in Vector Form

Let $\mathrm{A}, \mathrm{B}$, and C be three non-collinear points on the plane with position vectors $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ respectively.

The vectors, $\overrightarrow{A B}=\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}$ and $\overrightarrow{A C}=\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}$ are in the given plane. Therefore, the vector $\overrightarrow{A B} \times \overrightarrow{A C}$ is perpendicular to the plane containing points $A, B$, and $C$.

Let $P$ be any point in the plane with a position vector $\overrightarrow{\mathbf{r}}$

Therefore, the equation of the plane passing through $OP$ and perpendicular to the vector $\overrightarrow{A B} \times \overrightarrow{A C}$ is

$
\begin{aligned}
(\vec{r}-\vec{a}) \cdot(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}) & =0 \\
\text { or } \quad(\tilde{\mathbf{r}}-\tilde{\mathbf{a}}) \cdot[(\tilde{\mathbf{b}}-\tilde{\mathbf{a}}) \times(\tilde{\mathbf{c}}-\tilde{\mathbf{a}})] & =0
\end{aligned} \quad(\because \overrightarrow{A R}=(\vec{r}-\vec{a}))
$
This is the equation of the plane in vector form passing through three non-collinear points.

Equation of a plane passing through three non-collinear points in 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 points $A$ , $B$ and $C$ respectively.

Let $P(x, y, z)$ be any point on the plane.
Then, the vectors $\overrightarrow{P A}, \overrightarrow{B A}$ and $\overrightarrow{C A}$ are coplanar.
which is required equation of the plane.

$
\begin{aligned}
& {\left[\begin{array}{lll}
\overrightarrow{\mathrm{PA}} & \overrightarrow{\mathrm{BA}} & \overrightarrow{\mathrm{CA}}
\end{array}\right]=0} \\
& \left|\begin{array}{ccc}
\mathrm{x}-\mathrm{x}_1 & \mathrm{y}-\mathrm{y}_1 & \mathrm{z}-\mathrm{z}_1 \\
\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{x}_3-\mathrm{x}_1 & \mathrm{y}_3-\mathrm{y}_1 & \mathrm{z}_3-\mathrm{z}_1
\end{array}\right|=0
\end{aligned}
$

which is required equation of the plane.

Intercept form of the equation of a plane in Cartesian Form

The equation of a plane having intercepting lengths $a, b$, and $c$ with $X$-axis, $Y$-axis, and $Z$-axis, respectively is

$
\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1
$

Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors in Vector Form

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

Which is the required equation of plane.

Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors in Cartesian Form

$
\begin{aligned}
& \text { From }(\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0 \text { 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
\end{aligned}
$

Which is the required equation of a plane in the cartesian form where

$
\overrightarrow{\mathbf{b}}=x_2 \hat{\mathbf{i}}+y_2 \hat{\mathbf{j}}+z_2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{c}}=x_3 \hat{\mathbf{i}}+y_3 \hat{\mathbf{j}}+z_3 \hat{\mathbf{k}}
$

Angle Between Two Planes

The angle between two planes is defined as the angle between the normals of the two planes. It is also called dihedral angles.

Equation of the Angle Between Two Planes in Vector Form

Let $\theta$ be the angle between two planes $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=\mathbf{d}_1$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_2=\mathbf{d}_2$ then,

$
\cos \theta=\frac{\overrightarrow{\mathbf{n}}_1 \cdot \overrightarrow{\mathbf{n}}_2}{\left|\overrightarrow{\mathbf{n}}_1\right|\left|\overrightarrow{\mathbf{n}}_1\right|}
$

Equation of the Angle Between Two Planes in Cartesian Form

Let $\theta$ be the angle between the planes, $a_1 x+b_1 y+c_1 z+d_1=0$ and $a_2 x+$ $b_2 y+c_2 z+d_2=0$

Then,

$
\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right|
$

The direction ratios of the normal to the planes are $a_1, b_1, c_1$, and $a_2, b_2$ $\mathrm{c}_2$ respectively.

List of Topics According to NCERT/JEE MAIN

Importance of Class 12 Three Dimensional Geometry

Three Dimensional Geometry have a significant weighting in the IIT JEE test, which is a national level exam for 12th grade students that aids in admission to the country's top engineering universities. It is one of the most difficult exams in the country, and it has a significant impact on students' futures. Several students begin studying as early as Class 11 in order to pass this test. You may begin and continue your studies with the standard books and these revision notes, which will ensure that you do not miss any crucial ideas and can be used to revise before any test or actual examination.

How to Study Class 12 Three Dimensional Geometry?

Start preparing by understanding the 3D coordinate system. Try to be clear on important formulas from three dimensional geometry like the direction cosines and direction ratios, equation of line in 3D, angle between tow lines, shortest distance and the equation plane. Practice many problems from each topic for better understanding.

If you are preparing for competitive exams then solve as many problems as you can. Do not jump on the solution right away. Remember if your basics are clear you should be able to solve any question on this topic.

NCERT Notes Subject wise link:

• NCERT Notes for class 12 Physics.
• NCERT Notes for class 12 Chemistry.
• NCERT Notes for class 12 Mathematics.
• NCERT Notes for class 12 Biology

Important Books for Class 12 Three Dimensional Geometry

Start from NCERT Books, the illustration is simple and lucid. You should be able to understand most of the things. Solve all problems (including miscellaneous problems) of NCERT. If you do this, your basic level of preparation will be completed.

Then you can refer to the book Vectors and 3D Geometry of Arihant Publication written by Amit M. Agarwal. Three dimensional Geomery is explained very well in these books and there are an ample amount of questions with crystal clear concepts. Choice of reference book depends on person to person, find the book that best suits you the best, depending on how well you are clear with the concepts and the difficulty of the questions you require.

NCERT Solutions Subject wise link:

• NCERT solutions for class 12 Physics.
• NCERT solutions for class 12 Chemistry.
• NCERT solutions for class 12 Mathematics.
• NCERT solutions for class 12 Biology.

NCERT Exemplar Solutions Subject wise link:

• NCERT exemplar solutions for class 12 Physics.
• NCERT exemplar solutions for class 12 Chemistry.
• NCERT exemplar solutions for class 12 Mathematics.
• NCERT exemplar solutions for class 12 Biology.