Three-dimensional Geometry revision notes

Key Concepts & Definitions

Introduction to 3D Geometry Analytical geometry in three dimensions is heavily simplified and made elegant by using vector algebra. As A. De Morgan stated, "The moving power of mathematical invention is not reasoning but imagination". A line in space can be uniquely determined if it passes through a given point with a given direction, or if it passes through two given points.

Direction Angles and Direction Cosines If a directed line $L$ passes through the origin and makes angles $\alpha, \beta$, and $\gamma$ with the positive $x, y$, and $z$-axes respectively, these are called the direction angles. The cosines of these angles ($\cos\alpha, \cos\beta, \cos\gamma$) are defined as the direction cosines of the directed line $L$, uniquely denoted by $l, m$, and $n$. For a line in space not passing through the origin, we draw a parallel line passing through the origin to determine its unique set of direction cosines.

Reversing Line Direction A given line in space can be extended in two opposite directions. If we reverse the direction of $L$, the direction angles are replaced by their supplements ($\pi - \alpha, \pi - \beta, \pi - \gamma$), which reverses the signs of the direction cosines (i.e., $-l, -m, -n$). → [JEE TIP]

Direction Ratios (Direction Numbers) Any three numbers $a, b, c$ which are proportional to the direction cosines $l, m, n$ of a line are called the direction ratios. Thus, $a = \lambda l, b = \lambda m$, and $c = \lambda n$ for any non-zero $\lambda \in \mathbb{R}$. Any two sets of direction ratios for a given line are proportional; hence, there are infinitely many sets of direction ratios for any line (i.e., $ka, kb, kc$ where $k \neq 0$ is also a valid set).

Skew Lines In three-dimensional space, lines that are neither intersecting nor parallel are called skew lines. They are strictly non-coplanar. A physical example is a room: a line traversing diagonally across the ceiling and a line passing through a corner directly below on the wall, going diagonally down, are skew lines.

Planes in Space (JEE Advanced Extension) A plane is a surface such that a line segment joining any two points on the surface lies entirely within it. The orientation of a plane is entirely determined by its normal vector ($\vec{n}$), which is perpendicular to every line lying in the plane.

Direction Cosines and Ratios

Axes Direction Cosines The positive $x, y$, and $z$-axes make angles of $0^\circ, 90^\circ, 90^\circ$ with themselves and the other axes, giving direction cosines of $(1,0,0), (0,1,0)$, and $(0,0,1)$ respectively.

Relation Between Direction Cosines and Ratios Given direction ratios $a, b, c$, the direction cosines are obtained by dividing by $\sqrt{a^2+b^2+c^2}$. Specifically, $l = \pm a/\sqrt{a^2+b^2+c^2}$, $m = \pm b/\sqrt{a^2+b^2+c^2}$, and $n = \pm c/\sqrt{a^2+b^2+c^2}$. The positive or negative sign is chosen collectively depending on the desired orientation of the directed line.

Line Passing Through Two Points For a line joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, the direction ratios are $x_2 - x_1, y_2 - y_1, z_2 - z_1$. The direction cosines are $(x_2 - x_1)/PQ, (y_2 - y_1)/PQ$, and $(z_2 - z_1)/PQ$, where the distance $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. The direction ratios can also be taken as $x_1-x_2, y_1-y_2, z_1-z_2$ if the direction is reversed. → [JEE TIP]

Equations of a Line in Space

Line Through a Given Point and Parallel to a Given Vector Let a line pass through a point $A$ with position vector $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and be parallel to vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ be an arbitrary point $P$ on the line.

• Vector Equation: $\vec{r} = \vec{a} + \lambda\vec{b}$. Here, $a, b, c$ are the direction ratios of the line.
• Parametric Cartesian Equations: By equating coefficients of $\hat{i}, \hat{j}, \hat{k}$, we get $x = x_1 + \lambda a$, $y = y_1 + \lambda b$, $z = z_1 + \lambda c$.
• Symmetric Cartesian Equation: Eliminating $\lambda$ yields $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. If direction cosines $l, m, n$ are given, the equation is $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$.

Line Passing Through Two Given Points

• Vector Equation: For points with position vectors $\vec{a}$ and $\vec{b}$, the parallel vector is $(\vec{b} - \vec{a})$, giving $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
• Cartesian Equation: $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.

Angle Between Two Lines

Let two lines have direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$.

• Vector Form: If lines are $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$, the acute angle $\theta$ is $\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}$.
• Cartesian Cosine Form: $\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$.
• Direction Cosines Form: $\cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|$.
• Cartesian Sine Form: Using Lagrange's identity, $\sin\theta = \frac{\sqrt{(a_1b_2-a_2b_1)^2 + (b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2}}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$. Or $\sin\theta = \sqrt{(l_1m_2-l_2m_1)^2 + (m_1n_2-m_2n_1)^2 + (n_1l_2-n_2l_1)^2}$.

Conditions for Parallel and Perpendicular Lines

• Perpendicular ($\theta = 90^\circ$): $\vec{b}_1 \cdot \vec{b}_2 = 0 \implies a_1a_2 + b_1b_2 + c_1c_2 = 0$.
• Parallel ($\theta = 0^\circ$): Vectors are collinear $\implies \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. → [JEE TIP]

Shortest Distance Between Two Lines

By shortest distance, we mean the length of the line segment joining a point on one line to a point on the other such that the segment is the smallest. For skew lines, this segment is strictly perpendicular to both lines.

Distance Between Skew Lines

• Vector Form: Given $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$. The unit vector perpendicular to both is $\hat{n} = \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$. The shortest distance $d$ is the magnitude of the projection of $(\vec{a}_2 - \vec{a}_1)$ along $\hat{n}$. Formula: $d = \frac{|(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 \times \vec{b}_2|}$.
• Cartesian Form: For lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$. Formula: $d = \frac{\left| \begin{matrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{matrix} \right|}{\sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2}}$.

Distance Between Parallel Lines If $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ are parallel, they are coplanar. The shortest distance is the perpendicular distance from point $T(\vec{a}_2)$ to the line $l_1$. Formula: $d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}$.

Equations of a Plane (JEE Advanced Extension)

• General Cartesian Form: $Ax + By + Cz + D = 0$, where $A, B, C$ are the direction ratios of the normal.
• Normal Form: $\vec{r} \cdot \hat{n} = d$ (where $d \geq 0$ is the perpendicular distance from origin and $\hat{n}$ is the unit normal) or $lx + my + nz = p$.
• Point-Normal Form: A plane passing through position vector $\vec{a}$ with normal vector $\vec{n}$: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, which in Cartesian becomes $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$.
• Intercept Form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are intercepts on the axes.
• Passing Through Three Non-Collinear Points: $[\vec{r} - \vec{a}, \vec{b} - \vec{a}, \vec{c} - \vec{a}] = 0$.
• Family of Planes: Equation of a plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0$.

Line and Plane Interactions (JEE Advanced Extension)

• Angle Between Line and Plane: The angle $\phi$ between $\vec{r} = \vec{a} + \lambda\vec{b}$ and plane $\vec{r} \cdot \vec{n} = d$ is the complement of the angle between $\vec{b}$ and $\vec{n}$. Therefore, $\sin\phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.
• Line Parallel to Plane: $\vec{b} \cdot \vec{n} = 0$.
• Coplanarity of Two Lines: Lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ are coplanar if their shortest distance is zero $\implies (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$.
• Plane Containing Two Coplanar Lines: $(\vec{r} - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$.
• Distance of a Point from a Plane: The perpendicular distance of $P(x_1, y_1, z_1)$ from $Ax + By + Cz + D = 0$ is $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2+B^2+C^2}}$.
• Distance Between Parallel Planes: $d = \frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}}$ for planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$.

Conditions & Limitations

• Parallel Shortest Distance Condition: The formula $d = |\vec{b} \times (\vec{a}_2 - \vec{a}_1)| / |\vec{b}|$ CANNOT be used if $\vec{b}_1$ and $\vec{b}_2$ are merely proportional but not factored to be identical in the equation. You must make the parallel vector $\vec{b}$ uniform in both line equations before substituting into the formula.
• Skew vs. Parallel: Always evaluate $\vec{b}_1 \times \vec{b}_2$ first. If it yields $\vec{0}$, the lines are parallel. You must strictly use the parallel lines distance formula, as the skew formula will yield an undefined $0/0$ form.
• Unique Direction Cosines Limitations: Direction ratios $a, b, c$ can be any real numbers, but direction cosines $l, m, n$ are strictly bound by $[-1, 1]$ and their sum of squares must be $1$.

Previous Year JEE Topics

• Shortest Distance Between Skew Lines: Almost guaranteed 1 question in JEE Main and highly featured in mixed geometry problems in JEE Advanced.
• Coplanarity of Lines & Intersection Points: Setting determinant $= 0$ to find unknown parameters (like $\lambda$ or $k$).
• Foot of Perpendicular / Image of a Point: Highly tested as it combines plane equations and line equations.
• Family of Planes: Using $P_1 + \lambda P_2 = 0$ to find a plane satisfying a third condition (e.g., perpendicular to a given line).

Standard Derivations & Step-by-Step Problem Solving

Step-by-Step Problem Solving: Finding Foot of Perpendicular from Point $P(\vec{\alpha})$ to Line $\vec{r} = \vec{a} + \lambda\vec{b}$

1. Assume the foot of the perpendicular $F$ has position vector $\vec{f} = \vec{a} + \lambda_0\vec{b}$ for some specific $\lambda_0$.
2. Write the direction vector of the perpendicular line segment $PF$ as $\vec{f} - \vec{\alpha} = (\vec{a} - \vec{\alpha}) + \lambda_0\vec{b}$.
3. Since $PF$ is perpendicular to the given line, their dot product must be zero: $[(\vec{a} - \vec{\alpha}) + \lambda_0\vec{b}] \cdot \vec{b} = 0$.
4. Solve this linear equation for $\lambda_0$: $\lambda_0 = \frac{(\vec{\alpha} - \vec{a}) \cdot \vec{b}}{|\vec{b}|^2}$.
5. Substitute $\lambda_0$ back into $\vec{f}$ to get the exact coordinates.
6. The perpendicular distance is $|\vec{f} - \vec{\alpha}|$, and the image of $P$ across the line is $2\vec{f} - \vec{\alpha}$. → [JEE TIP]

Memory Aids & JEE Traps

• JEE TIPIn the Cartesian angle formula $\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\dots}$, students often assume $a_1^2+b_1^2+c_1^2 = 1$. This is ONLY true if $a_1, b_1, c_1$ are direction cosines. If they are direction ratios, you must divide by their magnitudes.
• JEE TIPWhen given two lines to find the distance between, the first step is checking proportionality $a_1/a_2 = b_1/b_2 = c_1/c_2$. Applying the skew determinant formula to parallel lines yields an indeterminate form $0/0$.
• JEE TIPWhen making equations symmetric, constants must be properly moved. Example: $x = 3, y = 4z$ translates to $\frac{x-3}{0} = \frac{y-0}{4} = \frac{z-0}{1}$. Never ignore the zero denominator; it means the line is perpendicular to the $x$-axis.

Top 10 JEE MCQ Traps

• [JEE TIP] Trap 11 - The Fractional Direction Ratio:

  • Misconception: Direction ratios ($a, b, c$) of a line must always be expressed strictly as integers.
  • Correct Understanding: Direction ratios can be any real numbers, including fractions, decimals, or irrationals (e.g., $\sqrt{2}, 1, -\pi$). Their only requirement is to remain directly proportional to the true direction cosines ($l, m, n$) of the line.

• [JEE TIP] Trap 12 - The Non-Unity Variable Coefficient:

  • Misconception: In a Cartesian line expression like $\frac{2x-1}{3} = \frac{y-2}{4}$, the direction ratios can be directly read off from the denominators as $(3, 4)$.
  • Correct Understanding: To correctly determine direction ratios, the coefficients of all variables ($x, y, z$) in the numerators must strictly be $+1$. Factoring out the $2$ from the first term rewrites it as $\frac{x - 0.5}{1.5}$. Consequently, the true direction ratios are $(1.5, 4)$, or symmetrically $(3, 8)$ when scaled.

• [JEE TIP] Trap 13 - The Non-Parallel Intersection Fallacy:

  • Misconception: Just like in 2D geometry, any two straight lines in a 3D coordinate system that are not parallel must eventually intersect at a common point.
  • Correct Understanding: In three dimensions, non-parallel lines do not have to intersect. They can exist as skew lines, which are non-parallel, non-intersecting lines that lie in parallel but distinct planes.

• [JEE TIP] Trap 14 - The 2D-Parallel Distance Copycat:

  • Misconception: The shortest distance between two parallel lines in 3D space can be calculated using the simple scalar constant difference formula $\frac{|D_1 - D_2|}{\sqrt{A^2+B^2}}$ used for 2D parallel lines.
  • Correct Understanding: The 2D constant-difference formula applies to parallel planes in 3D, not lines. For two parallel lines in space sharing a direction vector $\vec{b}$, you must use the cross-product vector projection formula: $d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}$.

• [JEE TIP] Trap 15 - The Cross Product Distance Confusion:

  • Misconception: Computing the vector cross product $\vec{b}_1 \times \vec{b}_2$ of the direction vectors in the skew lines formula yields the shortest distance directly.
  • Correct Understanding: The cross product $\vec{b}_1 \times \vec{b}_2$ only provides the spatial direction of the shortest distance vector (since it is perpendicular to both lines). To find the actual scalar distance, you must project the position connector vector $(\vec{a}_2 - \vec{a}_1)$ onto this normal direction via a dot product.

• [JEE TIP] Trap 16 - The Parallel Line-Plane Orthogonality Condition:

  • Misconception: If a straight line runs perfectly parallel to a flat plane, the dot product of the line's direction vector $\vec{b}$ and the plane's normal vector $\vec{n}$ must be equal to $1$.
  • Correct Understanding: If a line is parallel to a plane, the line is geometrically perpendicular to the plane's normal vector. Therefore, their vector dot product must strictly equal zero ($\vec{b} \cdot \vec{n} = 0$).

• [JEE TIP] Trap 17 - The Ratio Squaring Blindspot:

  • Misconception: Squaring and adding the components of any direction ratio vector allows you to apply the unit identity $a^2 + b^2 + c^2 = 1$.
  • Correct Understanding: This mathematical constraint applies exclusively to direction cosines. For general direction ratios, squaring and adding yields a non-unity scaling factor: $a^2 + b^2 + c^2 = k^2$, where $k$ is the absolute magnitude of the chosen vector components.

• [JEE TIP] Trap 18 - The Plane Image Coordinate Detour:

  • Misconception: When asked to find the coordinates of the image of a point reflected in a plane, calculating the scalar perpendicular distance is sufficient.
  • Correct Understanding: The distance formula only provides a length, not a spatial position. To directly evaluate the true coordinates of the reflected image, you must use the parametric coordinate shift formula: $\frac{x_2-x_1}{A} = \frac{y_2-y_1}{B} = \frac{z_2-z_1}{C} = -2\frac{Ax_1 + By_1 + Cz_1 + D}{A^2 + B^2 + C^2}$.

• [JEE TIP] Trap 19 - The Vector Origin Assignment:

  • Misconception: In the standard vector line equation $\vec{r} = \vec{a} + \lambda \vec{b}$, the fixed vector $\vec{a}$ must always represent the position vector of the line's origin coordinate.
  • Correct Understanding: The vector $\vec{a}$ is not a unique fixed point; it represents the position vector of any arbitrary point that lies on that line. The line equation remains perfectly identical if you substitute $\vec{a}$ with any other coordinate vector situated along the line's path.

• [JEE TIP] Trap 20 - The Zero-Denominator Equation Illusion:

  • Misconception: A Cartesian line equation containing a zero in its denominator, such as $\frac{x-2}{0} = \frac{y-1}{3} = \frac{z+4}{5}$, is mathematically invalid and represents an undefined line.
  • Correct Understanding: In 3D geometry, this expression is a highly valid parametric shorthand. The zero denominator indicates that the line has no directional component along that specific axis, meaning the $x$-coordinate is completely constant ($x=2$). It tells you that the line lies within a plane that runs perfectly parallel to the YZ-plane.