Three Dimensional Geometry

# Three Dimensional Geometry

Edited By Team Careers360 | Updated on May 07, 2022 01:52 PM IST

Just Imagine!!! You are living in a two-dimensional plane, and in this world has no height. You could still travel around, visit your friends. You could measure distances and angles. You could move fast or slow. You could go forward and backward or sideways. You could move in straight lines, circles, or anything so long as you never go up or down. What would be your life like living in two dimensions plane? Well, for me it's impossible to imagine. And that is the reason why Three Dimension Geometry is important and necessary to learn their properties. In the real world, everything you see is in a three-dimensional shape, it has length, breadth, and height. Just simply look around and observe. Even a thin sheet of paper has some thickness.

The next time you play a mobile game like PUBG, thank three-dimension geometry for the realistic look to the landscape and the characters that inhabit the game’s virtual world.

Applications of geometry in the real world include the computer-aided design (CAD) for construction blueprints, the design of assembly systems in manufacturing such as automobiles, nanotechnology, computer graphics, visual graphs, video game programming, and virtual reality creation. Geometry plays a very important role in calculating the location of galaxies, solar systems, planets, stars, satellites and other moving bodies in space. Geometry helps in calculations between coordinates also help to chart a trajectory for a space vehicle’s journey and its entry point into a planet’s atmosphere.

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The Tesla Roadster in space An animation tracking the orbit of Musk’s Tesla Roadster

That’s why it is necessary to learn three-dimension geometry their properties.

## Important Topics of Three Dimensional Geometry Class 11 and 12

• Coordinate of a point in space, distance formula

• Direction cosine and Direction ratio

• The angle between two intersecting lines

• Skew lines and the shortest distance between two lines

• Equation of line and plane

• The intersection of line and plane

Overview of Three Dimensional Geometry

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.

Coordinate of a Point in Space, Distance Formula

are unit vectors along OX, OY, and OZ respectively. Point P(x, y, z) be any point in the space. Then, the position vector of a point is given by $\mathrm{\mathbf{x\hat i+y\hat j+z\hat k}}$.

The distance between two points in the space $\mathrm{P(x_1,y_1,z_1)}$ and $\mathrm{Q(x_2,y_2,z_2)}$ is given by $\mathrm{\mathbf{PQ}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}$.

Direction Cosine and Direction Ratio

If $\alpha,\;\beta$ and $\gamma$ are the angle which a vector OP makes with the positive direction of the coordinate axis OX, OY and OZ respectively. Then $\cos \alpha,\;\cos\beta$and $\cos \gamma$ are known as Direction Cosine of vector OP and denoted by 'l', 'm', and 'n' respectively.

Let 'l', 'm', and 'n' are Direction Cosine of a vector 'r' and 'a', 'b', and 'c' be three number such that 'a', 'b', and 'c' are proportional to 'l', 'm', and 'n'

$\mathrm{\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k }$

Then, (a,b,c) are direction ratios.

Equation of a Straight Line in Space

• Vector Equation of a Line Passing Through a Given Point and Parallel to Given Vector 'A' is any given point, 'BC' is the given line and 'AP' is parallel to given line 'BC'. Then, the equation of a line parallel to a given vector 'BC'and passing through a point 'A' is given by $\mathrm{\mathbf{r=a+\lambda b}}$. (Where $\mathrm{AP=\lambda b}$.

• Cartesian Equation of a Line Passing Through a Given Point and given Direction Ratio The coordinate of a given point $\mathrm{ A(x_1,y_1,z_1)}$ and the direction ratio of the given line be 'a', 'b' and 'c'. Then the equation of any line is given by $\mathrm{ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}}$ .
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Angle Between Two Lines

The equation of two straight lines in space is given as $$r=a+\lambda b$$ and $$r=a'+\mu b'$$. Then, the angle between the two lines is $\cos\theta=\frac{b\cdot b'}{|b||b'|}$.

If the equations of lines are in cartesian form, $\\\mathrm{ \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\;\;and\;\;\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}}$ . Then, the angle between the two lines is given as $\mathrm{\cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a^2_1+b_1^2+c_1^2}\sqrt{a^2_2+b_2^2+c_2^2}}}$.

Skew lines and the shortest distance between two lines

In space, if two lines intersect, then the shortest distance between them is zero. Also, if two lines are parallel in space, then the shortest distance between them is perpendicular distance. Further, there are such lines in space which are neither intersecting nor parallel. Such pair of lines are non-coplanar and known as skew lines.

The shortest distance between two skew lines $r=a+\lambda b$ and $r=a'+\mu b'$ is $\mathrm{PQ=\frac{\left ( b\times b' \right )\cdot(a-a')}{|b\times b'|}}$

In cartesian form, equation of two skew lines are $\\\mathrm{ \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\;\;and\;\;\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}}$ .

Then, the shortest distance between them is

$\mathrm{PQ=\frac{\begin{vmatrix} x_2-x_1 &y_2-y_1 &z_2-z_1 \\a_ 1 &b_1 &c_1 \\a_2 &b_2 &c_2 \end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1+b_2-a_2b_1)}}}$.

Equation of Plane

A plane is a surface such that if any two points are taken on it, the line segment joining these two points lies completely on the surface.

Equation of plane in Normal form

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

To get the equation of the plane in cartesian form. let P(x, y, z) be any point on the plane then, position vector $\mathrm{\mathbf{OP=r=x\hat i+y\hat j+z\hat k}}$. let l, m and n be the direction cosine of $\mathrm{\mathbf{\hat n}}$.

Then, $\mathrm{\mathbf{\hat n=(\textit{l}\hat i+\textit{m}\hat j+\textit{n}\hat k)}}$

From the vector equation of the plane

$\\\mathrm{\mathbf{(x\hat i+y\hat j+z\hat k)\cdot(\textit{l}\hat i+\textit{m}\hat j+\textit{n}\hat k)}=d}\\\mathrm{\textit{l}x+\textit{m}y+\textit{n}z=d}$

This is the cartesian equation of the plane in normal form.

Intercept form of a Plane

The equation of a plane having intercepting lengths a, b and c with X-axis, Y-axis and Z-axis respectively is given by $\\\mathrm{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$.

The important formulas of Three Dimension Geometry

$\\\mathrm{\bullet Two\;lines\;with\;direction\;cosine\;\mathit{l}_1,\mathit{m}_1,\mathit{n}_1\;and\;\mathit{l}_2,\mathit{m}_2,\mathit{n}_2\;are\;perpendicular }\\\mathrm{\;\;to\;each\;other\;then}\\\mathrm{\;\;\Rightarrow \;\;\mathbf{\mathit{l}_1\mathit{l}_2+\mathit{m}_1\mathit{m}_2}+\mathit{n}_1\mathit{n}_2=0}\\\mathrm{\bullet Two\;lines\;with\;direction\;cosine\;\mathit{l}_1,\mathit{m}_1,\mathit{n}_1\;and\;\mathit{l}_2,\mathit{m}_2,\mathit{n}_2\;are\;parallel \;to\;each}\\\mathrm{\;\;other\;then\;\;}\\\mathrm{\;\;\Rightarrow \;\;\frac{\mathit{l}_1}{\mathit{l}_2}=\frac{\mathit{m}_1}{\mathit{m}_2}=\frac{\mathit{n}_1}{\mathit{n}_2}}$

$\\\mathrm{\bullet \;Angle\;between\;two\;planes\;\;\mathbf{r\cdot \hat n_1}=\mathit{d}_1\;and\;;\mathbf{r\cdot \hat n_2}=\mathit{d}_2\;is\cos\theta=\frac{\mathbf{n_1\cdot n_2}}{|\mathbf{n_1}||\mathbf{n_2}|}}\\\mathrm{\bullet \;If\;the\;plane\;\mathbf{r\cdot \hat n_1}=\mathit{d}_1\;and\;;\mathbf{r\cdot \hat n_2}=\mathit{d}_2\;are\;perpendicular,\;then\;\mathbf{n_1}\cdot \mathbf{n_2}=0.}\\\mathrm{\bullet \;If\;the\;plane\;\mathbf{r\cdot \hat n_1}=\mathit{d}_1\;and\;;\mathbf{r\cdot \hat n_2}=\mathit{d}_2\;are\;parallel,\;then\;\mathbf{n_1}=\lambda \mathbf{n_2}.(\lambda\;is\;scalar)}$

$\\\mathrm{\bullet\;In\;cartesian\;form,\;angle\;between\;the\;planes}\\\mathrm{\;\;\;\mathit{a_1x+b_1y+c_1z+d_1=0}\;and\;\mathit{a_2x+b_2y+c_2z+d_2=0}\;\;is}\\\mathrm{\;\;\;\cos\theta=\mathbf{\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}}}\\\mathrm{\bullet \;Condition\;of\;perpendicularity:}\\\mathrm{\;\;\;\mathit{a_1a_2+b_1b_2+c_1c_2=0}}\\\mathrm{\bullet \;Condition\;of\;parallellism:}\\\mathrm{\;\;\;\mathit{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}}}$

$\\\mathrm{\bullet \;Angle\;between\;the\;line,\;\mathbf{r=a+\lambda b}\;\;and\;the\;plane,\;\mathbf{r\cdot n}=\mathit{d}\;\;is\;\:\theta=\sin^{-1}\left | \frac{\mathbf{b\cdot n}}{\mathbf{|b||n|}} \right |.}$

$\\\mathrm{\bullet \;Length\;of\;perpendicular\;distance\;between\;a\;point\;having\;position\;vector\;\mathbf{a}\;to\;the\;}\\\mathrm{\;\;\;the\;plane\;\mathbf{r\cdot n}=\mathit{d}\;\;is\;\;\frac{|\mathbf{a\cdot n}-\mathit{d}|}{|\mathbf{n}|}.}$

## Tips that will help you in preparing Three Dimension Geometry in the best possible way:

• As soon as you are done with the concepts and numerical you must do the previous year’s questions. With the previous year questions, you would totally understand where you are lacking and you will be able to improve accordingly.

• Take online mock test regularly in a time-bound manner to increase your speed and accuracy. This activity will particularly help you in JEE Mains.

• Know your strength and weakness and try to improve both.

• While practicing if you feel that any question which appears to be important then make a note of that. Later while revising this chapter, you must solve that question again, this will help you to brush up your concepts.

• You must create small formula notebook/flashcards for this chapter and then revise them on a weekly basis to keep them in your mind always.