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If you have been asked the position of a butterfly flying in space around at an instant of time, what distance should you measure from a reference point? To answer this, you need to understand the concept of three-dimensional geometry. Three-dimensional geometry talks about the position of an object in space, and for this, you must know three dimensions, ie, x-axis, y-axis, and z-axis. Class 12 maths chapter 11 notes help you to understand the relation between two lines in space with the help of direction cosines, direction ratios. Also, these notes contain the equation of a straight line in different forms and the angle between them. In real life, 3D geometry is used in architecture, aviation, robotics, and 3D modeling.
NCERT class 12th Math chapter 11 notes contain standard formulas and detailed information that are used to solve the questions with the help of suitable examples. Our subject matter experts ensure that all concepts are covered as per the latest syllabus of CBSE in NCERT Notes for Class 12 Math chapter 11.
A line OA makes angles α,β,γ with the x,y, and z-axis, respectively. Then cosα, cosβ, and cosγ are called direction cosines.
l = cosα, m = cosβ, n = cosγ
$\begin{aligned} & l^2+m^2+n^2=1 \\ & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\end{aligned}$
Note: Direction cosines are always unique.
Direction ratios: The values that are proportional to directional cosines are called direction ratios.
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are ratios, then $\frac{l}{a}=\frac{m}{b}=\frac{n}{c}$
Direction cosines of ratios a,b,c :
$\begin{aligned} & l= \pm \frac{a}{\sqrt{a^2+b^2+c^2}} \\ & m= \pm \frac{b}{\sqrt{a^2+b^2+c^2}} \\ & n= \pm \frac{c}{\sqrt{a^2+b^2+c^2}}\end{aligned}$
Direction cosines pass through two points:
$P\left(x_1, y_1, z_1\right)$ and $\mathrm{Q}\left(x_2, y_2, z_2\right)$
$\cos \alpha=\frac{x_2-x_1}{O A}, \cos \beta=\frac{y_2-y_1}{O A}, \cos \gamma=\frac{z_2-z_1}{O A}$
Direction cosines: $\frac{x_2-x_1}{O A}, \frac{y_2-y_1}{O A}, \frac{z_2-z_1}{O A}$
Direction ratios: $x_2-x_1, y_2-y_1, z_2-z_1$
NOTE: Directions ratios need not be unique
Equation of Straight Line:
Equation of a Line passing through the Point and parallel to vector b
$\vec{r}=\vec{a}+\lambda \vec{b}$
where, $\vec{a}$ is a position vector. $\vec{b}$ is a vector that is parallel to the line
Cartesian form:
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
Where, $\left(x_1, y_1, z_1\right)$ be the point that line passes through and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the direction ratios.
l,m,n are the direction cosines, then the equation is:
$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$
Equation of line that passes through 2 points:
Points are : $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$
$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
$\vec{a}, \vec{b}$ are a position vectors.
Cartesian form of two points:
Points are : $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$
$\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}=\frac{z-z_1}{z_1-z_2}$
Vector equations of 2 lines:
$\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda\left(\overrightarrow{b_1}-\overrightarrow{a_1}\right)$ and $\overrightarrow{r_2}=\overrightarrow{a_2}+\lambda\left(\overrightarrow{b_2}-\overrightarrow{a_2}\right)$
The angle between 2 lines:
$\cos \theta=\frac{b_1 \cdot b_2}{\left|b_1\right| \cdot\left|b_2\right|}$
Cartesian form: Let be an angle between the lines below:
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$
Then,
$\begin{aligned} & \cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2{ }^2+b_2{ }^2+c_2^2}}\right| \\ & \sin \theta=\sqrt{1-\cos ^2 \theta} \\ & \sin \theta=\frac{\sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2}}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\end{aligned}$
Direction cosines with angles are:
$\begin{aligned} & \cos \theta=l_1 l_2+m_1 m_2+n_1 n_2 \\ & \sin \theta=\sqrt{\left(m_1 n_2-m_2 n_1\right)^2+\left(n_1 l_2-n_2 l_1\right)^2+\left(l_1 m_2-l_2 m_1\right)^2}\end{aligned}$
Few conditions:
When lines are perpendicular,$\theta=90^{\circ}$
Then Cartesian form:
$\begin{aligned} & a_1 a_2+b_1 b_2+c_1 c_2=0 \\ & l_1 l_2+m_1 m_2+n_1 n_2=0\end{aligned}$
When lines are parallel $\theta=0^{\circ}$.
Then Cartesian form:
$\begin{aligned} & \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ & \frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2}\end{aligned}$
Shortest path:
Vector equations of 2 lines:
$\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r_2}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$
Shortest distance
$d=\left|\frac{\left(b_1 \times b_2\right) \cdot\left(a_2 \times a_1\right)}{\left(b_1 \times b_2\right)}\right|$
where $\overrightarrow{a_1}, \overrightarrow{a_2}$ are position vectors and $\overrightarrow{b_1}, \overrightarrow{b_2}$ are vectors in the direction of a line.
Cartesian form for two lines:
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$
Cartesian form :
Shortest distance between the lines:
$\begin{aligned} & l_1: \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \\ & l_2: \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\end{aligned}$
$\left|\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2} \mid}\right|$
Distance between two Parallel Lines:
Lines are said to be coplanar when they are parallel.
Vector equations of 2 lines:
$\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r_2}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$
Distance between two Parallel Lines:
$\left|\frac{\left(b_1\right) \times\left(a_2-a_1\right)}{\left|b_2\right|}\right|$
Note: If lines are parallel, they always have the same direction ratios.
Distance between two points:$P\left(x_1, y_1, z_1\right)$ and $\mathrm{Q}\left(x_2, y_2, z_2\right)$
$|P Q|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$
The midpoint of two points: $P\left(x_1, y_1, z_1\right)$ and $\mathrm{Q}\left(x_2, y_2, z_2\right)$
$P Q=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$
PLANE:
The plane is found unique, only if it satisfies any one of the following:
Equation of a plane in normal form:
$\vec{r} \cdot \vec{n}=d$
Cartesian form: one equation of the plane is ax + by + cz = d, and another equation of the plane is lx + my + nz = p
Formula: Foot of perpendicular (ld, md, nd).
Equation of a plane perpendicular to a vector and passing through a point:
Vector equation: $(\vec{r}-\vec{a}) \cdot \vec{n}=0$
Cartesian form: Equation of a plane that passes through the point $\left(x_1, y_1, z_1\right)$
$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$
Equation of a plane passing through non-collinear points:
Vector form: $(\overrightarrow{r-a}) \cdot\{(\overrightarrow{b-a}) \times(\overrightarrow{c-} \vec{a})\}=0$
Cartesian form: $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ are non-collinear points
Equation: $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{array}\right|=0$
If they are collinear: $\left|\begin{array}{lll}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3\end{array}\right|=0$
Intercept Form:
Let a, b, and c be the x-intercept, y-intercept, and z-intercept, respectively, then
Equation of intercept: xa + by + zc = 1
Equation of Plane Passing through the intersection of two Planes: The equation of the planes is
$\overrightarrow{r_1} \cdot \overrightarrow{n_1}=d_1$ and $\overrightarrow{r_2} \cdot \overrightarrow{n_2}=d_2$
Equation of the plane passing through the intersection :
$r\left(n_1+\lambda n_2\right)=d_1+\lambda d_2$
Cartesian form: equation of planes are $a_1 x+b_1 y+c_1 z=d_1$ and $a_2 x+b_2 y+c_2 z=d_2$
Equation of the plane passing through the intersection :
$a_1 x+b_1 y+c_1 z-d_1+\lambda\left(a_2 x+b_2 y+c_2 z-d_2\right)=0$
Coplanarity:
Lines $: \overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r_2}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ are coplanar then Vectorform : $\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0$
Cartesian form:
For Two lines
$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar then
$\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=0$
The angle between Two Planes: θ is the angle between two planes.
$n_1, n_2$ are normals, the angle between $\overrightarrow{r_1} \cdot \overrightarrow{n_1}=d_1$ and $\overrightarrow{r_2} \cdot \overrightarrow{n_2}=d_2$
$\cos \theta=\left|\frac{n_1 \cdot n_2}{\left|n_1\right| \cdot\left|n_2\right|}\right|$
Cartesian form: planes are $a_1 x+b_1 y+c_1 z=d_1$ and $a_2 x+b_2 y+c_2 z=d_2$,
$\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1{ }^2+b_1^2+c_1{ }^2} \sqrt{a_2{ }^2+b_2{ }^2+c_2{ }^2}}$
Angle Between Line and Plane:
Vector form: The equation of a line is $\vec{r}=(\vec{a}+\lambda \vec{b})$
angle θ between the line and the normal to the plane is
$\cos \theta=\frac{\overrightarrow{b \cdot n}}{|\vec{b} \cdot| \vec{n} \mid}$
Cartesian form: a, b, and c are direction ratios, and lx + my + nz + d = 0 is the equation of the plane.
$\sin \theta=\frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}$
NCERT Class 12 Maths Chapter 11 Notes |
Subject-wise NCERT Exampler solutions
NCERT problems are very important in order to perform well in exams. Students must try to solve all the NCERT problems, including miscellaneous exercises, and if needed, refer to the NCERT Solutions for Class 12 Maths Chapter 11 Three-Dimensional Geometry.
Students are advised to go through the NCERT Class 12 Maths Chapter 11 Notes before solving the questions.
To boost your exam preparation as well as for quick revision, these NCERT notes are very useful.
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