Introduction to Three Dimensional Geometry

3D Geometry involves the mathematics of shapes in 3-dimensional space and involves three coordinates, namely the x-coordinate, y-coordinate, and z-coordinate. In a 3D space, three parameters are required to accurately pinpoint the location of a point. For the JEE, three-dimensional geometry is an important topic, as many questions are included on the exam. Here, the basic concepts of geometry involving 3-dimensional coordinates are covered, which will help to understand different operations on a point in a 3D plane.

Coordinate System in 3D Geometry

In 3-dimensional geometry, a coordinate system is used to identify the position or location of a point in the coordinate plane. To gain a better understanding of coordinate planes and systems, take a look at the coordinate geometry lesson which covers all the basic concepts, theorems, and formulas related to coordinate or analytic geometry.

Rectangular Coordinate System

Three lines perpendicular to each other pass through a common point, known as the origin. These three lines are known as the axes and are referred to as the x-axis, y-axis, and z-axis, respectively. An observer, referred to as O, uses these axes to measure the position of any other point relative to his position. The coordinates of any point in 3D space can be measured by how much it has moved along the x, y, and z-axes, respectively. For example, if a point has a position of (3, -4, 5), it has moved 3 units along the positive x-axis, 4 units along the negative y-axis, and 5 units along the positive z-axis.

Rectangular coordinate system – 3D Geometry

Distance from the Origin

Distance from the Origin in 3D Space – 3D Geometry

The distance of P(x, y, z) from the origin (0, 0, 0) is given by the Pythagorean theorem: (\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}).

Distance Between Two Points

Distance between 2 points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is: $\sqrt{({x_2}-{x_1})^2 + ({y_2}-{y_1})^2 + ({z_2}-{z_1})^2}$

Division of a Line Joining Two Points

Let P(x1, y1, z1) and Q(x2, y2, z2) be 2 points. R internally derives the line segment PQ in ratio and has coordinates.

(\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))

Projection in 3D Space

Projection in 3D Space

3D Geometry

Let AB be a line segment. Its projection on a line PQ is AB cos $\theta$, where $\theta$ is the angle between AB and PQ or CD.

Direction Cosines and Direction Ratios of a Line in a Cartesian Plane

The direction cosines of a line are the cosines of the angles the line makes with the positive x, y, and z axes, respectively.

Explore Further: Direction Cosines & Direction Ratios Of A Line

The direction cosines of the line given by the angles α, β and γ are denoted by l, m and n, respectively, and are equal to cos α, cos β, and cos γ.

Proof:

Let (l=\cos \theta ,m=\sin \theta \cos \phi ,n=\sin \theta \sin \phi )

Then,

\begin{align*} l^2 + m^2 + n^2 &= \cos^2 \theta + \sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi \ &= \cos^2 \theta + \sin^2 \theta (\cos^2 \phi + \sin^2 \phi) \ &= \cos^2 \theta + \sin^2 \theta \ &= 1 \end{align*}

Any three numbers, a, b, and c, which are proportional to direction cosines, are called direction ratios.

Hence, $$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=\frac{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\frac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$$

(\therefore l=\frac{a}{\sqrt{\sum{{{a}^{2}}}}},\ m=\frac{b}{\sqrt{\sum{{{a}^{2}}}}},\ n=\frac{c}{\sqrt{\sum{{{a}^{2}}}}})

Direction Cosine of Line Joining Two Given Points

Let P$(x_1, y_1, z_1)$ and Q$(x_2, y_2, z_2)$ be 2 points. Then, direction cosines will be

$$\begin{array}{l} l=\frac{{x_2}-{x_1}}{\left| PQ \right|}, m=\frac{{y_2}-{y_1}}{\left| PQ \right|}, n=\frac{{z_2}-{z_1}}{\left| PQ \right|} \end{array}$$

Projection of Line Segment Joining Two Points onto Another Line

Consider P(x1, y1, z1) and Q(x2, y2, z2).

Projection of PQ on a line whose direction cosines are (l), (m), (n) is (\begin{array}{l}l\left( {{x}_{2}}-{{x}_{1}} \right)+m\left( {{y}_{2}}-{{y}_{1}} \right)+n\left( {{z}_{2}}-{{z}_{1}} \right)\end{array})

Angle Between Two Lines in Three-Dimensional Space

The angle between two lines having direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $$\theta = \cos^{-1}\left(l_1l_2 + m_1m_2 + n_1n_2\right).$$

Projection of a Plane Area onto Three Coordinate Planes

Let $\bar{A}$ be the vector with direction cosines cos $\alpha$, cos $\beta$, and cos $\gamma$. Then the projections of $\bar{A}$ are ${A}{1}=A\cos \alpha$, ${A}{2}=A\cos \beta$, and ${A}_{3}=A\cos \gamma$.

(\therefore A^2 = A_1^2 + A_2^2 + A_3^2)

Area of a Triangle:

#Using the Projection Formula to Find the Area of a Triangle

(\begin{array}{l}=\frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \ {{x}_{2}} & {{y}_{2}} & 1 \ {{x}_{3}} & {{y}_{3}} & 1 \ \end{matrix} \right|}^{2}} + \frac{1}{4}{{\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \ {{y}_{2}} & {{z}_{2}} & 1 \ {{y}_{3}} & {{z}_{3}} & 1 \ \end{matrix} \right|}^{2}} + \frac{1}{4}{{\left| \begin{matrix} {{x}_{1}} & {{z}_{1}} & 1 \ {{x}_{2}} & {{z}_{2}} & 1 \ {{x}_{3}} & {{z}_{3}} & 1 \ \end{matrix} \right|}^{2}}\end{array})

Discover more about the area of a triangle in coordinate geometry, its derivation and problem solving strategies here!

Concept of Plane in 3-Dimensional Geometry

A first degree equation in x, y, and z represents a plane in three-dimensional space.

The equation (\begin{array}{l}ax+by+cz=0,{{z}^{2}}{{b}^{2}}+{{c}^{2}}\ne 0\end{array} ) represents a plane.

Normal Form of a Plane

The equation of the plane can be expressed as lx + my + nz = P, where P is the length of the normal from the origin to the plane and l, m, and n are the direction cosines of that normal.

Intercept Form

Let a plane cut lengths a, b, and c from the coordinate axis.

The equation of the plane is: $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$

Planes Passing Through Three Given Points

A plane passing through $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is:

(\left| \begin{matrix} x & y & z & 1 & {{x}_{1}} & {{y}_{1}} & {{z}_{1}} & 1 \ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} & 1 & {{x}_{3}} & {{y}_{3}} & {{z}_{3}} & 1 \ \end{matrix} \right|=0)

Angle Between Two Planes

The equation a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given.

(\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}}})

Two Sides of a Coin

Points A(x1, y1, z1) and B(x2, y2, z2) lie on the same side or opposite sides of a plane ax + by + cz + d = 0, depending on whether the values of a*x1 + b*y1 + c*z1 + d and a*x2 + b*y2 + c*z2 + d are the same sign or opposite sign.

Distance from a Point to a Plane

The distance of a point $(x_1, y_1, z_1)$ from a plane.

Distance of $(x_1, y_1, z_1)$ from $ax + by + cz + d$ is $\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$

Equation for the Plane Bisecting the Angle Between Two Planes

The equation of the plane bisecting the angles between the two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is

(\begin{array}{l}\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\end{array} )

Origin Position

The origin lies in an acute angle between a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 if a1a2 + b1b2 + c1c2 > 0, and in an obtuse angle if a1a2 + b1b2 + c1c2 < 0, provided d1 and d2 are both positive.

Two Planes Intersecting

The plane passing through the line of intersection of U = 0 and V = 0 is given by U + λV = 0, where λ is to be determined from the given condition.

Straight Lines in 3D

Two intersecting planes together represent a straight line.

Slope-Intercept Form:

$$y = mx + b$$

Equation of a Straight Line in Symmetrical Form

A straight line passing through $(x_1, y_1, z_1)$ and having direction cosines $(l, m, n)$ is given by: $$\vec{r} = (x_1, y_1, z_1) + t(l, m, n)$$

(\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n})

Two-point form

Equation of a straight line passing through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$

Two planes transformed into symmetrical form

Eliminate x from the equations a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 to get a relation between y and z. Eliminate y from the same equations to get a relation between x and z. Then find x in terms of z and find y in terms of z. Finally, equate the two expressions for x and y.

Intersection of a Plane and a Line

Let (\begin{array}{l} ax + by + cz + d = 0 \\ \frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}.\end{array} ) be intersected.

Find the intersection point by letting.

(\begin{array}{l} \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} = t \end{array})

Substituting the values of (x), (y) and (z) in the equation of the plane (ax + by + cz + d = 0), we get:

(ax_{1} + b(y_{1} + mt) + c(z_{1} + nt) + d = 0)

Passing a Plane Through a Given Straight Line

Let the line be (\begin{array}{l}\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\end{array} ) of the plane through this line be ax + by + cz + d = 0, then

(ax_1 + by_1 + cz_1 + d = 0)

and

(al + bm + cn = 0)

and from other given conditions a, b, c are determined.

Coplanarity of Two Lines in 3D Geometry

(\begin{array}{l}\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}},,\text{and},,\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\end{array} )

Two lines are coplanar iff $$\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \ \end{matrix} \right|=0$$

Distance of a Point from a Straight Line

The Distance of a Point from a Straight Line - 3D Geometry is illustrated in the image above.

(\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n})

AQ = projection of AP on the straight line $$=l\left( {{x}_{1}}-\alpha \right)+m\left( {{y}_{1}}-\beta \right)+n\left( {{z}_{1}}-\gamma \right)$$

(\therefore PQ=\sqrt{A{{P}^{2}}-A{{Q}^{2}}})

Shortest Distance Between Two Skew Lines

Let the two skew lines be $$\begin{array}{l}\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\end{array} $$ and $$\begin{array}{l}\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\end{array} $$

The shortest distance is

(\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \ \end{matrix} \right|}{\sqrt{\sum{{{\left( {{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}} \right)}^{2}}}})

The equation of the shortest distance between two points is:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

(\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \ l & m & n \ \end{matrix} \right|=0) is the only conjunction used

“I like apples, oranges, and bananas”

I like apples, oranges, and bananas.

(\left| \begin{matrix} x-{{x}_{2}} & y-{{y}_{2}} & z-{{z}_{2}} \ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \ l & m & n \ \end{matrix} \right|=0)

Problems on 3D Geometry

Problem 1. Find the value of u such that if a variable plane forms a tetrahedron of constant volume 64 K³ with the coordinate planes then the location of the centroid of the tetrahedron is xyz = uK³.

Answer: The equation of the plane can be written as (\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1)

The centroid of the tetrahedron is $(\frac{a}{4}, \frac{b}{4}, \frac{c}{4})$.

Volume of the tetrahedron = $\frac{abc}{6} = 64\mathrm{K}^3$.

So letting $$\frac{a}{4}=x, \frac{b}{4}=y, \frac{c}{4}=z$$

We have $$\frac{abc}{6} = \frac{{4}^{3}xyz}{6} = 64{K}^{3}.$$

xyz = 6,000

We have u = 6 when compared.

Problem 2. What is the ratio in which the yz plane divides the line joining the points (2, 4, 5) and (3, 5, 7)?

Answer: The ratio is $\lambda$: 1.

x-coordinate = 0

(\frac{3\lambda +2}{\lambda +1} = 0 \Leftrightarrow \lambda = -\frac{2}{3})

(\lambda = \frac{-2}{3})

The ratio is 2:3

Problem 3. (\sum{{{\cos }^{2}}\alpha =\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta )

I love cats!

Answer: I love cats!

Let the direction cosine of that line be $\left(l, m, n\right)$.

The direction cosine of the first diagonal is (1/√3, 1/√3, 1/√3).

The direction cosine of the 2nd diagonal is (+1/√3, -1/√3, 1/√3).

The direction cosine of the 3rd diagonal is (1/√3, +1/√3, -1/√3).

The direction cosine of the 4th diagonal is (1/√3, -1/√3, -1/√3).

(\therefore \cos \alpha = \frac{l + m + n}{\sqrt{3}})

(\begin{array}{l} \cos \beta = \frac{l - m + n}{\sqrt{3}}, \ \cos \gamma = \frac{l + m - n}{\sqrt{3}}, \ \cos \delta = \frac{l - m - n}{\sqrt{3}} \end{array})

(\therefore \sum{{{\cos }^{2}}\alpha } = \frac{4}{3}.)

Problem 4. The angle between the lines (\begin{array}{l}\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0} \ \text{and} \ \frac{x-1}{1}=\frac{2y+3}{3}=\frac{z+5}{2}\end{array} ) is equal to

Answer: (\cos \theta = \frac{3\times1 - 2\times \frac{3}{2} + 0\times2}{\sqrt{3^2+(-2)^2+0^2}\sqrt{1^2+(\frac{3}{2})^2+2^2}})

$$\therefore \theta =\frac{\pi }{2}$$

Problem 5. If lines (\begin{array}{l}\frac{x-1}{2}=\frac{y-2}{{{x}_{1}}}=\frac{z-3}{{{x}_{2}}}\ \text{and}\ \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\end{array} ) lies in the same plane, then prove that for the equation (\begin{array}{l}{{x}_{1}}{{t}^{2}}+\left( {{x}_{2}}+2 \right)t+a=0\end{array} ) the sum of its roots is equal to -2.

Answer: Let the plane be $$ax+by+cz+d=0$$

(\begin{array}{l}a+2b+3c+d=0 \rightarrow \left( i \right)\end{array})

(\begin{array}{l}2a+3b+4c+d=0 \rightarrow{{}} \left( ii \right)\end{array})

(\begin{array}{l}2a + {{x}_{1}}b + {{x}_{2}}c = 0 \rightarrow \left( iii \right)\end{array})

(\begin{array}{l}3a+4b+5c=0 \\rightarrow \\left( iv \right): \\quad 5c=-3a-4b \end{array})

(\begin{array}{l}\text{Sum of the roots}=\frac{-(x_1 + 2)}{x_2}\end{array})

From (i) and (ii),

a + b + c = 0

From (iii) and (iv),

a + b + c - d = 0

d = 0

a + 2b + 3c = 0

b + 2c = 0

2a + 3b + 4c = 0

b = 2c

3a + 4b + 5c = 0

2a - bc + 4c = 2c

2a = 2c

a = c

The equation of the plane can be written as either:

• ax + by + cz = 0
• cx + 2cy + cz = 0

or

x - 2y = 0

and

z = 0

(\therefore x_1 = 3,\ x_2 = 4)

(\begin{array}{l}\text{Sum of roots} = \frac{{{x}_{1}} \cdot \left( {{x}_{2}}+2 \right)}{-1}\end{array})

-2

-2 =

Problem 6. What is the value of K if an isosceles triangle is formed with the line x/K = y/2 = z/-12 and the planes 2x + y + 3z – 1 = 0 and x + 2y – 3z – 1 = 0?

Answer: Equation of the bisector planes are $$\frac{2x+y+3z-1}{\sqrt{14}}=\pm \frac{x+2y-3z-1}{\sqrt{14}}$$

2x + y + 3z - 1 = x + 2y - 3z - 1

or $x - y + 6z = 0 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;(i)$

2x + y + 3z - 1 = -x - 2y + 3z + 1

(ii) 3x + 3y = 2

So the given line must be parallel to either (i) or (ii).

The coefficient of x and y for the required straight line equation are -1 and 1, respectively.

Therefore, the value of K is -2.

Problem 7. The direction cosines of the normal to the plane containing the lines $x=y=z$ and $x-1=y-1=(z-1)/d$ are

This sentence is written in bold.

Answer: This sentence is written in bold.

Problems in 3D Geometry

Let the direction cosines be l, m, and n.

l + m + n = 0 and l + m + dn = 0

n = 0

l = \sqrt{1 - m^2 - n^2}

We have $$2l2 = 1$$

\(\therefore l = \pm \frac{1}{\sqrt{2}}\)

Therefore, ((l, m, n)) is either (\left( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0 \right)) or (\left( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0 \right)).

Problem 8. If the lines $$x = y = z$$ and $$x \sin A + y \sin B + z \sin C = 2d^2, \quad x \sin 2A + y \sin 2B + z \sin 2C = d^2,$$ where $(A + B + C = \pi)$, then

\(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)

Answer: Let the point of intersection be (t, t, t).

(\therefore \left( \sin A + \sin B + \sin C \right)t = 2d^2)

(\left( \sin 2A + \sin 2B + \sin 2C \right)t = {{d}^{2}})

(\begin{array}{l}\sin(2A) + \sin(2B) - \sin\left(2A + 2B\right) = \frac{{d}^{2}}{t}\end{array})

(\begin{array}{l}\Rightarrow 2\sin \left( A+B \right).\cos \left( A-B \right) - 2\sin \left( A+B \right).\cos \left( A+B \right) = \frac{{{d}^{2}}}{t}\end{array})

(\begin{array}{l}\Rightarrow \frac{4\sin C,,\sin A.,,\sin B}{t}={{d}^{2}}\rightarrow{{}}\left( i \right)\end{array} )

Again, $\sin A + \sin B + \sin C$

(\begin{array}{l}2\sin \frac{A+B}{2}\cos \frac{A-B}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}\end{array})

(\begin{array}{l}2\cos \frac{C}{2}\cos \frac{A-B}{2} + 2\sin \frac{C}{2}\sin \frac{C}{2}\end{array})

(\begin{array}{l}2\cos \frac{C}{2}\left(\cos \frac{A-B}{2} + \cos \frac{A+B}{2}\right)\end{array})

(\begin{array}{l}2\cos \frac{C}{2} \cdot 2\cos\frac{A}{2} \cdot \cos \frac{B}{2} = \frac{2{{d}^{2}}}{t} \rightarrow \left( ii \right)\end{array})

Dividing (i) by (ii)

\(\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{1}{16}\)

Problem 9. The equation of the sphere with center at $(3, 6, -4)$ and touching the plane $\bar{r}.\left( 2\hat{i}-2\hat{j}-\hat{k} \right)=10$ is given by $${{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}+{{\left( z+4 \right)}^{2}}={{K}^{2}}$$ then $K =$

Answer: The equation of the plane is:

$$2x - 2y - z - 10 = 0$$

The distance from $(3, 6, -4)$ to $(12, 10, -10)$ is $\left| \frac{12}{3} \right| = 4$

Therefore, k = 4.

Vector Algebra and 3D Geometry: Important Topics

Important Questions on Vector Algebra and 3D Geometry

Vector Algebra and 3D Geometry - Part 2: Important Topics

Important Questions on Vector Algebra and 3D Geometry - Part 2

Top 10 Most Important and Expected JEE Questions on 3D Geometry

3D Geometry - JEE Advanced Problems

Frequently Asked Questions

Direction Cosines of a line refer to the cosines of the angles that the line makes with the three coordinate axes.

The direction cosine of a line is the cosine of the angle subtended by this line with the x-axis, y-axis, and z-axis respectively.

What are the coordinates of a point in 3D Geometry?

The vector form of representation of any point in 3D geometry is $\vec{p} = \langle x, y, z \rangle$ and is with reference to the x-axis, y-axis, and z-axis respectively.

The formula to find the distance of a point P(x, y, z) from the origin is: $\sqrt{x^2 + y^2 + z^2}$

The distance of a point P(x, y, z) from the origin (0, 0, 0) is given by √(x^2 + y^2 + z^2).

What is the Relationship between Direction Cosines of a Line?

If l, m, and n denote the direction cosines of a line, then l² + m² + n² = 1.