Three-dimensional geometry studies the analytical representation, properties, and relations of points, lines, planes, and spheres in Euclidean space using three independent coordinates. For every point in space, three real numbers uniquely specify its position with respect to a fixed origin and set of orthogonal axes.

Coordinate Representation and Distance in Three-Dimensional Space

Let the origin be denoted by $O(0,0,0)$, and let the axes $OX$, $OY$, and $OZ$ be three mutually perpendicular lines in space. The position of any point $P$ is described by the coordinates $(x, y, z)$, where $x$ is the signed distance from $P$ to the $YZ$-plane, $y$ the signed distance to the $ZX$-plane, and $z$ the signed distance to the $XY$-plane.

The distance between the origin $O(0,0,0)$ and point $P(x, y, z)$ is given by the Euclidean metric:

$\displaystyle OP = \sqrt{x^2 + y^2 + z^2}$

Given two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, the distance between them is derived as follows. First, compute the differences along each axis:

$\Delta x = x_2 - x_1$

$\Delta y = y_2 - y_1$

$\Delta z = z_2 - z_1$

Now, apply the Pythagorean theorem in three dimensions:

$AB = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Section and Centroid Formulae in Space

Given two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, the coordinates of a point $P$ that divides the segment $AB$ in the ratio $m:n$ (internally) are:

$P(x, y, z) = \left( \dfrac{mx_2 + nx_1}{m+n},\, \dfrac{my_2 + ny_1}{m+n},\, \dfrac{mz_2 + nz_1}{m+n} \right)$

Given the vertices $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$ of a triangle in space, its centroid $G$ is located at:

$G = \left( \dfrac{x_1 + x_2 + x_3}{3},\, \dfrac{y_1 + y_2 + y_3}{3},\, \dfrac{z_1 + z_2 + z_3}{3} \right)$

Three Dimensional Geometry Important Questions cover mixed applications of these formulae.

Direction Cosines and Direction Ratios of a Line

Given a line $L$ in three-dimensional space, let $\alpha$, $\beta$, and $\gamma$ be the angles that $L$ makes with the positive $x$-, $y$-, and $z$-axes, respectively. The direction cosines $(l, m, n)$ of this line are defined as:

$l = \cos \alpha$

$m = \cos \beta$

$n = \cos \gamma$

These cosines satisfy the fundamental relation:

$l^2 + m^2 + n^2 = 1$

If a vector $\vec{d} = (a, b, c)$ is parallel to the line, its components $(a, b, c)$ are called the direction ratios of $L$. They are proportional to the direction cosines:

$\dfrac{l}{a} = \dfrac{m}{b} = \dfrac{n}{c} = \dfrac{1}{\sqrt{a^2 + b^2 + c^2}}$, provided not all $a, b, c$ are zero.

Given two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ on the line, the direction ratios are $(x_2 - x_1,\, y_2 - y_1,\, z_2 - z_1)$, and the corresponding direction cosines are:

$\left( \frac{x_2 - x_1}{D}, \frac{y_2 - y_1}{D}, \frac{z_2 - z_1}{D} \right)$ where $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

Equation of a Line in Cartesian and Vector Forms

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a nonzero vector $\vec{b}$ is:

$\vec{r} = \vec{a} + \lambda \vec{b},\ \lambda \in \mathbb{R}$

Let $\vec{a} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}$ and $\vec{b}$ have direction ratios $(a, b, c)$. The corresponding cartesian equation of the line is derived as follows.

The general point $\vec{r}$ can be expressed as $x\hat{i} + y\hat{j} + z\hat{k}$. Equate:

$x\hat{i} + y\hat{j} + z\hat{k} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} + \lambda(a\hat{i} + b\hat{j} + c\hat{k})$

Equate coefficients for each unit vector:

$x = x_1 + \lambda a$

$y = y_1 + \lambda b$

$z = z_1 + \lambda c$

Solving for $\lambda$ in each, we have:

$\lambda = \dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$ (provided $a,b,c \neq 0$).

Therefore, the symmetric form of the line’s equation is:

$\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$

If the line passes through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the line’s equation in parameter $t$ is:

$x = x_1 + t(x_2 - x_1),\, y = y_1 + t(y_2 - y_1),\, z = z_1 + t(z_2 - z_1),\ t \in \mathbb{R}$

Equation of a Plane: Normal Form, General Form, and Through Points

The general equation of a plane in space is given by:

$ax + by + cz + d = 0$

Here, $(a, b, c)$ are direction ratios of the normal to the plane, and $d$ is a real constant.

The normal form of the plane is:

$lx + my + nz = p$

where $(l, m, n)$ are the direction cosines of the normal vector and $p$ is the perpendicular distance from the origin to the plane.

The equation of a plane passing through a fixed point $(x_1, y_1, z_1)$ with normal vector having direction ratios $(a, b, c)$ is:

$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$

The equation of a plane passing through three non-collinear points $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$ is given as the vanishing of the determinant:

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$

This determinant expands to a linear equation in $x, y, z$ for the required plane.

Angle Between Two Lines and Coplanarity Criteria

Let two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively. The angle $\theta$ between the lines is given by:

$\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$

Alternatively, if lines have direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$, then

$\cos \theta = \dfrac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$

Two lines are coplanar if the volume of the parallelepiped formed by their direction vectors and the vector joining points on the two lines vanishes, i.e., the scalar triple product is zero.

Angle and Distance Between Two Planes

Consider two planes with equations $a_1 x + b_1 y + c_1 z + d_1 = 0$ and $a_2 x + b_2 y + c_2 z + d_2 = 0$. The angle $\phi$ between them is the angle between their normal vectors:

$\cos \phi = \dfrac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$

If the planes are parallel, the perpendicular distance between them is:

$d = \left| \dfrac{d_2 - d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} \right|$ if normal coefficients are proportional.

Distance-related results such as the distance between two parallel lines are covered in detail.

Angle Between a Line and a Plane

Given a line with direction ratios $(a, b, c)$ and a plane with normal direction ratios $(A, B, C)$, let $\theta$ be the angle between them. The sine of the angle $\theta$ is expressed as:

$\sin \theta = \dfrac{|aA + bB + cC|}{\sqrt{a^2 + b^2 + c^2}\sqrt{A^2 + B^2 + C^2}}$

Shortest Distance Between Two Skew Lines

Let $L_1$ and $L_2$ be two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$, and points $P_1(x_1, y_1, z_1)$ on $L_1$ and $P_2(x_2, y_2, z_2)$ on $L_2$. The shortest distance $D$ between them is given by:

$D = \dfrac{\left| \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \right|}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}$

This expression in the numerator evaluates the scalar triple product, representing the volume of the parallelepiped formed by the two direction vectors and the line joining the two points. The denominator gives the magnitude of their cross product.

Equation of a Sphere and Geometric Characteristics

A sphere with center at $(h, k, l)$ and radius $r$ is defined by:

$(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$

Geometric properties of a sphere include:

Surface area: $S = 4\pi r^2$

Volume: $V = \dfrac{4}{3} \pi r^3$

Intersection Problems: Line and Plane, Line and Sphere

The coordinates of the intersection of a line $\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$ and a plane $Ax + By + Cz + D = 0$ are found by parametrizing the line, expressing $x = x_1 + ta,\ y = y_1 + tb,\ z = z_1 + tc$ and substituting these values in the plane’s equation.

Explicitly, the procedure is as follows:

Substitute $x, y, z$ into $Ax + By + Cz + D$ to get:

$A(x_1 + ta) + B(y_1 + tb) + C(z_1 + tc) + D = 0$

Expanding:

$A x_1 + A t a + B y_1 + B t b + C z_1 + C t c + D = 0$

Combine terms of $t$:

$[A a + B b + C c] t + [A x_1 + B y_1 + C z_1 + D] = 0$

Solve for $t$:

$t = -\dfrac{A x_1 + B y_1 + C z_1 + D}{A a + B b + C c}$, provided denominator is not zero.

Substituting this value into the parametric expressions lets one find the intersection point.

For the intersection of a line and a sphere, substitute the parametric coordinates of the line into the sphere’s equation and expand. This results in a quadratic equation in the parameter. Real and distinct roots correspond to two intersection points; a single root, to tangency; and no real roots, to non-intersection.

Perpendicular and Parallel Criteria, and Projections

Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, and parallel if their direction ratios are proportional.

The length of the projection of a line segment from $A(x_1, y_1, z_1)$ to $B(x_2, y_2, z_2)$ on the $x$-axis is $|x_2 - x_1|$, on $y$-axis is $|y_2 - y_1|$, on $z$-axis is $|z_2 - z_1|$.

For exam practice, refer to the Three Dimensional Geometry Practice Paper.

Worked Example: Distance from a Point to a Plane

Given: Point $P(x_1, y_1, z_1)$ and the plane $a x + b y + c z + d = 0$.

Substitution: Insert the coordinates of $P$ into the plane’s equation:

$a x_1 + b y_1 + c z_1 + d$

Simplification: Modulus is taken to ensure a non-negative distance, and this value must be normalized by the length of the normal vector to the plane, which is $\sqrt{a^2 + b^2 + c^2}$.

Final result: The perpendicular distance from $P$ to the plane is $\displaystyle \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$.

Foundational concepts such as these are essential to analytic geometry and are systematically extended throughout advanced three-dimensional problems. Advanced loci and formulae may be cross-referenced under Coordinate Geometry.