Three Dimensional Geometry

Three-dimensional geometry encompasses the mathematical study of shapes and figures in 3D space, involving three coordinates: the x-coordinate, y-coordinate, and z-coordinate. These coordinates define the precise location of a point in three-dimensional space, requiring three parameters for accurate positioning. Three-dimensional geometry is crucial for exams like JEE, as it features prominently and includes various operations on points within a 3D plane. It covers fundamental concepts such as direction cosines and Direction Ratios of a line, as well as the Direction cosines of a line passing through two points. 

1.0Introduction to Three Dimensional Geometry

Three-dimensional geometry, also known as 3D geometry, is a branch of mathematics that deals with the study of shapes and figures in three-dimensional space. In three-dimensional space, objects not only have length and width (like in two-dimensional geometry) but also depth or height, creating a three-dimensional aspect.

2.0Section Formula in 3D

The section formula in 3D geometry helps determine the coordinates of a point that divides a line segment between two specified points in a given ratio. It is an extension of the section formula used in 2D geometry.

Let's consider two points in three-dimensional space:

• Point A with coordinates (x₁, y₁, z₁)
• Point B with coordinates (x₂, y₂, z₂)

Now, we want to find the coordinates of a point P that divides the line segment AB in the ratio m: n. The section formula in 3D can be expressed as follows:

$\mathrm{P}(\mathrm{x},\mathrm{y},\mathrm{z})=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{mz}}_2+{\mathrm{nz}}_1}{\mathrm{m}+\mathrm{n}}$

3.0Direction Cosines and Direction Ratios of a Line

Direction Cosines

If 𝛼, 𝛽, 𝛾 are the angles made by a line with 𝑥 −axis, 𝑦 −axis, 𝑧 −axis, respectively, then cos𝛼, cos𝛽 & cos𝛾 are called direction cosines of a line, denoted by ℓ, 𝑚 & 𝑛 respectively.

(i) Direction cosines of a line

Take a vector $\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}$ parallel to a line whose 𝐷. 𝐶’𝑠 are to be found out. 

$\overrightarrow{\mathrm{A}}.\hat{\mathrm{i}}=\mathrm{a}|\vec{A}|\cos \alpha =a$

$\cos \alpha =\frac{a}{|\vec{A}|}\text{ similarly, }\cos \beta =\frac{b}{|\vec{A}|};\cos \gamma =\frac{c}{|\vec{A}|}$

⇒ cos² α + cos² β + cos² γ = 1 ⇒  l² + m² + n² =1

$a\hat{i}+b\hat{j}+c\hat{k}$

(ii) Direction cosines of axes

Since the positive 𝑥 −axes makes angle 0°, 90°, 90° with axes of 𝑥, 𝑦 and 𝑧 respectively,

∴ 𝐷. 𝐶. ’𝑠 of 𝑥 − axis are 1, 0, 0. 

    𝐷. 𝐶. ’𝑠 of 𝑦 − axis are 0, 1, 0. 

    𝐷. 𝐶. ’𝑠 of 𝑧 − axis are 0, 0, 1. 

Direction Ratios

Any three numbers 𝑎, 𝑏, 𝑐, proportional to direction cosines ℓ, 𝑚, 𝑛 are called direction ratios of the line. i.e. $\frac{1}{a}=\frac{m}{b}=\frac{n}{c}$.

There can be infinitely many sets of direction ratios for a given line.

4.0Relation Between Direction Cosines and Direction Ratios:

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

5.0Direction Ratios and Direction Cosines of the Line Joining Two Points

Let us assume 𝐴 (𝑥₁, 𝑦₁, 𝑧₁) and 𝐵(𝑥₂, 𝑦₂, 𝑧₂) be two points, then Direction Ratios of AB are 𝑥₂ –  𝑥₁, 𝑦₂ – 𝑦₁, 𝑧₂  –  𝑧₁ and the  Direction Cosines of 𝐴𝐵 are 

$\frac{\mathrm{l}}{\mathrm{r}}\left({\mathrm{x}}_2-{\mathrm{x}}_1\right),\frac{1}{\mathrm{r}}\left({\mathrm{y}}_2-{\mathrm{y}}_1\right),\frac{1}{\mathrm{r}}\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)$

where $r=\sqrt{\left[\sum {\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2\right]}$

6.0Projection Of Line Segment Joining Two Points On Another Line

Consider A (x₁, y₁, z₁) and B (x₂, y₂, z₂).

The Projection of line segment AB onto a line with direction cosines l, m, n is given as:

l (x₂ − x₁) + m(y₂ − y₁) + n(z₂ − z₁)

7.0Angle Between Two Lines in 3 Dimensional Space

Angle between two lines in Three Dimensional Space whose direction cosines are (l₁, m₁, n₁) and (l₂, m₂, n). Then angle between them is

θ = cos^–1 (l₁l₂ + m₁m₂ + n₁n₂)

8.0Equation Of A Line In Space

A line is uniquely determined in three-dimensional space if:

(i) It passes through a given point and has a given direction vector, or

(ii) It passes through two distinct given points.

Equation of a line passing through a given point and parallel to a given vector b.

Let a be the position vector of a given point A w.r.t origin O and l be the line that passes through point A and is parallel to a given vector b. Let r be the position vector of an arbitrary point P on the line.

So, the equation of a line in a Vector Form is given by: r = a + λ b

If l, m, and n represent the direction cosines of the line, the equation of the line in Cartesian coordinates can be expressed as:

$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$

9.0Angle between Two Lines

Let 𝜃 represent the angle between the lines with direction cosines ℓ₁, 𝑚₁, 𝑛₁ and ℓ₂, 𝑚₂, 𝑛₂. Then, the cosine of 𝜃 is given by 𝜃 = ℓ₁ℓ₂ + 𝑚₁𝑚₂ + 𝑛₁𝑛₂. If direction ratios are 𝑎₁, 𝑏₁, 𝑐₁ and 𝑎₂, 𝑏₂, 𝑐₂ of two lines, then the angle 𝜃 between them is given by: 

$\cos \theta =\frac{\left(a_1{a}_2+b_1{b}_2+c_1{c}_2\right)}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}\sqrt{\left(\sqrt{a_2^1+b_2^2+c_2^2}\right)}}$ 

10.0Angle Between a Line and a Plane

Let equations of the line and plane be $\frac{\mathrm{x}-{\mathrm{x}}_1}{\ell }=\frac{\mathrm{y}-{\mathrm{y}}_1}{\mathrm{m}}=\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{n}}$ and 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0, respectively, and 𝜃 be the angle which the line makes with the plane. Then

$\left(\frac{\pi }{2}-\theta \right)$

is the angle between the line and the normal to the plane.

 So, $\sin \theta =\frac{al+bm+cn}{\sqrt{\left(a^2+b^2+c^2\right)}\sqrt{\left(l^2+m^2+n^2\right)}}$

Line is parallel to the plane

If 𝜃 = 0, i.e. if 𝑎ℓ + 𝑏𝑚 + 𝑐𝑛 = 0. 

Line is perpendicular to the plane-

If the line is parallel to the normal of the plane i.e., if  $\frac{a}{l}=\frac{b}{m}=\frac{c}{n}$

11.0Shortest Distance Between Two Lines

If two lines in space intersect at a common point, then their shortest distance between them is zero. Also, if two lines are parallel to each other, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line.

12.0Distance Between Two Skew Lines

The length of the line intercepted between two lines, which is perpendicular to both the lines, is the shortest distance between them. Let the equations of the lines 𝐴𝐵 and 𝐶𝐷 be:

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$

And $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$        

$\left|\frac{\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{a}}_1& {\mathrm{b}}_1& c_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\end{array}\right|}{\sqrt{{\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)}^2+{\left({\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{c}}_2{\mathrm{a}}_1\right)}^2+{\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)}^2}}\right|$

13.0Three Dimensional Geometry Solved Examples

Example 1: A line OP makes with the x −axis an angle of measure 120° and with y −axis an angle of measure 60°. Find the angle made by the line with the z −axis.

Solution: 𝛼 = 120° and 𝛽 = 60°

∴ cos𝛼 = cos120° = $-\frac{1}{2}$ and cos𝛽 = cos 60° = $\frac{1}{2}$

  but cos² 𝛼 + cos² 𝛽 + cos² 𝛾 = 1

$\therefore {\left(\frac{-1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\cos }^2\gamma =1$

$\Rightarrow {\cos }^2\gamma =1-\frac{1}{4}-\frac{1}{4}=\frac{1}{2}$

$\Rightarrow \cos \gamma =\pm \frac{1}{\sqrt{2}}$

∴ 𝛾 = 45° or 135°

Example 2: Determine the length of the projection of the line segment that connects the points (-1, 0, 3) and (2, 5, 1) onto the line with direction ratios 6, 2, 3.

Solution: The direction cosines ℓ, 𝑚, 𝑛 of the line are given by  

The direction cosines l m , n of the line are given by

$\frac{l}{6}=\frac{m}{2}=\frac{n}{3}=\sqrt{\frac{{\ell }^2+m^2+n^2}{6^2+2^2+3^2}}=\frac{1}{\sqrt{49}}=\frac{1}{7}$ 

∴ $\ell =\frac{6}{7},\mathrm{m}=\frac{2}{7},\mathrm{n}=\frac{3}{7}$

The required length of projection is given by

$=\left|\ell \left({\mathrm{x}}_2-{\mathrm{x}}_1\right)+\mathrm{m}\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)+\mathrm{n}\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)\right|$

$=\left|\frac{6}{7}\left[2-(-1)+\frac{2}{7}(5-0)+\frac{3}{7}(1-3)\right]\right|$

$=\left|\frac{6}{7}\times 3+\frac{2}{7}\times 5+\frac{3}{7}\times -2\right|=\left|\frac{18}{7}+\frac{10}{7}-\frac{6}{7}\right|=\left|\frac{18+10-6}{7}\right|=\frac{22}{7}$

Example 3. If a line makes an angle 90°, 60°, and 30° with the positive direction of the x, y, and z-axis, respectively, find its direction cosines.

Solution: Let the d. c. 's of the lines be l, m, n. Then l = cos 90° = 0, m = cos 60° = $\frac{1}{2}$, n = $cos30^{\circ }$$=\frac{\sqrt{3}}{2}$

Example 4. Calculate the shortest distance between the lines l₁ and l₂, whose vector equations are  

$\vec{r}=\hat{i}+\hat{j}+\lambda (2\hat{i}-\hat{j}+\hat{k})$ ….(1)

$\vec{r}=2\hat{i}+\hat{j}-\hat{k}+\mu (3\hat{i}-5\hat{j}+2\hat{k})$     ….(2)

Solution Comparing (1) and (2) with

$\overrightarrow{\mathrm{r}}={\overrightarrow{\mathrm{a}}}_1+\lambda {\overrightarrow{\mathrm{b}}}_1\text{ and }\overrightarrow{\mathrm{r}}={\overrightarrow{\mathrm{a}}}_2+\mu {\overrightarrow{\mathrm{b}}}_2$ respectively.   

We get

${\vec{a}}_1=\hat{\mathrm{i}}+\hat{\mathrm{j}},{\overrightarrow{\mathrm{b}}}_1=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

${\overrightarrow{\mathrm{a}}}_2=2\hat{\mathbf{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\text{ and }{\overrightarrow{\mathrm{b}}}_2=3\hat{\mathrm{i}}-5\hat{\mathrm{j}}+2\hat{\mathrm{k}}$

Therefore $\overrightarrow{{\mathrm{a}}_2}-{\overrightarrow{\mathrm{a}}}_1=\hat{\mathrm{i}}-\hat{\mathrm{k}}$ 

and  $\overrightarrow{{\mathrm{b}}_1}\times \overrightarrow{{\mathrm{b}}_2}=(2\hat{\mathrm{i}}-\hat{\mathbf{j}}+\hat{\mathrm{k}})\times (3\hat{\mathrm{i}}-5\hat{\mathrm{j}}+2\hat{\mathrm{k}})$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ 3& -5& 2\end{array}\right|=3\hat{i}-\hat{\mathbf{j}}-7\hat{k}$

So $\left|{\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_2\right|=\sqrt{9+1+49}=\sqrt{59}$ 

Hence , the shortest distance between the given lines is given by

$\mathrm{d}=\left|\frac{\left({\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_2\right)\cdot \left({\overrightarrow{\mathrm{a}}}_2-{\overrightarrow{\mathrm{a}}}_1\right)}{\left|{\overrightarrow{\mathrm{b}}}_1\times {\overrightarrow{\mathrm{b}}}_1\right|}\right|=\frac{|3-0+7|}{\sqrt{59}}=\frac{10}{\sqrt{59}}$

14.0Solved Questions on Three dimensional Geometry

Q. How do you calculate distances in 3D space?

Ans: The distance between two points represented as (x₁, y₁, z₁) & (x₂, y₂, z₂) in 3D space can be found using the distance formula derived from the Pythagorean theorem:

d = (x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²​

Q. How do you find the angle between two vectors in 3D space?

Ans: The angle (θ) between two vectors A = (a₁, a₂, a₃) and B = (b₁, b₂, b₃) in 3D space can be calculated using the dot product formula: $\cos \theta =\frac{\mathrm{A}\cdot \mathrm{B}}{|\mathrm{A}||\mathrm{B}|}$