3D Geometry — Complete Guide with Solved Examples

3D Geometry — The Chapter That Separates JEE Toppers from the Rest

Three-dimensional geometry feels abstract until the moment it clicks. Once it does, you’ll realize this chapter is almost entirely formula-based — and that’s actually good news. With the right approach, you can score full marks on 3D geometry questions in JEE Main without breaking a sweat.

The chapter builds on two pillars: lines and planes in 3D space. Everything else — angles between them, distances, foot of perpendicular — follows from those two. The coordinate geometry you already know from Class 11 extends naturally into three dimensions; we’re just adding a $z$ -axis and working with vectors.

In JEE Main, 3D geometry typically contributes 1–2 questions per paper, often from lines in space or the distance between a point and a plane. In CBSE board exams, expect a 5-mark question guaranteed — usually involving the equation of a line/plane or proving that two lines are coplanar. This is a high-reward, low-variance topic. Learn the formulas cold, practice recognizing question types, and the marks come reliably.

Key Terms and Definitions

Direction Cosines (DCs): If a line makes angles $\alpha$ , $\beta$ , $\gamma$ with the positive $x$ -, $y$ -, $z$ -axes respectively, then $l = \cos\alpha$ , $m = \cos\beta$ , $n = \cos\gamma$ are the direction cosines.

The fundamental identity: $l^2 + m^2 + n^2 = 1$ always holds. If you’re given direction ratios, normalize them using this identity to get DCs.

Direction Ratios (DRs): Any three numbers $a, b, c$ proportional to $l, m, n$ . If DRs are $(a, b, c)$ , then:

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

Position Vector: The vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ locates a point $(x,y,z)$ from the origin. Every 3D geometry formula has both a vector form and a Cartesian form — always learn both.

Skew Lines: Two lines in 3D that neither intersect nor are parallel. This concept doesn’t exist in 2D, so many students get confused initially.

Core Formulas

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

If direction ratios are $(a, b, c)$ , then:

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

Vector form: $\vec{r} = \vec{a} + \lambda\vec{b}$

Cartesian form (Symmetric form):

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

where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are direction ratios.

Vector form: $\vec{r} \cdot \hat{n} = d$

General form: $ax + by + cz + d = 0$

Three-point form (through intercepts): $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$

Point $P(x_1, y_1, z_1)$ from plane $ax + by + cz + d = 0$ :

\text{Distance} = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2+b^2+c^2}}

Distance between skew lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ :

d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}

Angle between two lines with DRs $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ :

\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\cdot\sqrt{a_2^2+b_2^2+c_2^2}}

Angle between a line and a plane (with normal $(a, b, c)$ and line direction $(l, m, n)$ ):

\sin\theta = \frac{|al + bm + cn|}{\sqrt{a^2+b^2+c^2}\cdot\sqrt{l^2+m^2+n^2}}

Methods and Concepts

Lines in 3D Space

The equation of a line needs two things: a point on the line and a direction. That’s it. In vector form: $\vec{r} = \vec{a} + \lambda\vec{b}$ .

To find the equation passing through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ :

Direction vector = $\vec{AB} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$
Cartesian form: $\dfrac{x-x_1}{x_2-x_1} = \dfrac{y-y_1}{y_2-y_1} = \dfrac{z-z_1}{z_2-z_1}$

The denominators in the symmetric form ARE the direction ratios. When checking if a line is perpendicular to a plane, compare the line’s DRs with the plane’s normal vector — they should be proportional.

Planes in 3D Space

A plane is determined by either a point + normal vector, or three non-collinear points.

Normal to the plane is the vector $(a, b, c)$ in $ax + by + cz + d = 0$ . This is crucial — the normal to a plane is perpendicular to every line lying in that plane.

To find the plane through three points, set up the equation $ax + by + cz = 1$ and solve the $3\times3$ system. Alternatively, use the vector method: find two vectors in the plane, take their cross product for the normal, then use the point-normal form.

Coplanarity of Lines

Two lines are coplanar if and only if:

(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0

In Cartesian form, for lines $\dfrac{x-x_1}{l_1} = \dfrac{y-y_1}{m_1} = \dfrac{z-z_1}{n_1}$ and $\dfrac{x-x_2}{l_2} = \dfrac{y-y_2}{m_2} = \dfrac{z-z_2}{n_2}$ :

\begin{vmatrix} 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{vmatrix} = 0

Solved Examples

Example 1 — Easy (CBSE Level)

Find the direction cosines of the line joining points $A(2, -3, 4)$ and $B(4, 1, -2)$ .

Direction ratios of $\vec{AB}$ : $(4-2, 1-(-3), -2-4) = (2, 4, -6)$ .

Magnitude: $\sqrt{4 + 16 + 36} = \sqrt{56} = 2\sqrt{14}$

Direction cosines:

l = \frac{2}{2\sqrt{14}} = \frac{1}{\sqrt{14}}, \quad m = \frac{4}{2\sqrt{14}} = \frac{2}{\sqrt{14}}, \quad n = \frac{-6}{2\sqrt{14}} = \frac{-3}{\sqrt{14}}

Verify: $\dfrac{1}{14} + \dfrac{4}{14} + \dfrac{9}{14} = 1$ ✓

Example 2 — Medium (JEE Main Level)

Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ .

Rewrite as $x + 2y - 2z - 9 = 0$ . Here $a=1$ , $b=2$ , $c=-2$ , $d=-9$ , and point is $(2,3,-5)$ .

\text{Distance} = \frac{|1(2) + 2(3) + (-2)(-5) - 9|}{\sqrt{1+4+4}} = \frac{|2+6+10-9|}{3} = \frac{|9|}{3} = 3

The distance is 3 units.

This exact formula appears in almost every JEE Main paper. The most common slip is forgetting the absolute value in the numerator or computing $\sqrt{a^2+b^2+c^2}$ incorrectly. Always check: does your denominator equal the magnitude of the normal vector?

Example 3 — Hard (JEE Main 2024 Shift 1 Type)

Find the shortest distance between the lines:

L_1: \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} + \hat{k})

L_2: \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \mu(2\hat{i} - \hat{j} + 3\hat{k})

Step 1: Identify components.

$\vec{a}_1 = 2\hat{i} - \hat{j} + 3\hat{k}$ , $\vec{b}_1 = \hat{i} + 2\hat{j} + \hat{k}$
$\vec{a}_2 = \hat{i} + \hat{j} - \hat{k}$ , $\vec{b}_2 = 2\hat{i} - \hat{j} + 3\hat{k}$

Step 2: $\vec{a}_2 - \vec{a}_1 = -\hat{i} + 2\hat{j} - 4\hat{k}$

Step 3: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&1\\2&-1&3\end{vmatrix}$

$= \hat{i}(6+1) - \hat{j}(3-2) + \hat{k}(-1-4) = 7\hat{i} - \hat{j} - 5\hat{k}$

Step 4: $|\vec{b}_1 \times \vec{b}_2| = \sqrt{49+1+25} = \sqrt{75} = 5\sqrt{3}$

Step 5: $(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2) = (-1)(7) + (2)(-1) + (-4)(-5) = -7-2+20 = 11$

Shortest distance $= \dfrac{|11|}{5\sqrt{3}} = \dfrac{11}{5\sqrt{3}} = \dfrac{11\sqrt{3}}{15}$

Exam-Specific Tips

CBSE Board Exams (Class 12)

CBSE typically asks 3D geometry in two places: a 3-mark question (equation of a line/plane, or angle between two lines) and a 5-mark question (foot of perpendicular, distance from a plane, or proving coplanarity). In the 5-mark question, show all steps — each intermediate result carries partial marks in CBSE marking scheme. Writing just the final answer gets you at most 1 mark.

The vector form and Cartesian form are equally acceptable in CBSE. Use whichever you’re more comfortable with — but be consistent within one solution.

JEE Main Strategy

3D geometry in JEE Main is almost always calculation-heavy but concept-light. The most frequently tested subtopics (based on papers from 2020–2024):

Distance of a point from a plane (~40% of 3D questions)
Shortest distance between skew lines (~25%)
Equation of a plane through given conditions (~20%)
Angle between line and plane (~15%)

Time yourself: a clean 3D geometry problem should take under 4 minutes. If you’re going beyond that, you’ve likely made an arithmetic error — restart the calculation rather than continuing.

Common Mistakes to Avoid

Mistake 1: Confusing direction ratios with direction cosines. DRs are any proportional set — $(2, 4, -6)$ and $(1, 2, -3)$ are both valid DRs for the same line. DCs are the normalized version where $l^2+m^2+n^2=1$ . Never use raw DRs in the angle formula without normalizing — it gives wrong answers.

Mistake 2: Wrong formula for angle between a line and a plane. The angle between a line and a plane uses $\sin\theta$ (not $\cos\theta$ ), because you’re measuring from the plane (not from the normal). Students who memorize only the line-line formula (which uses $\cos\theta$ ) substitute wrongly here.

Mistake 3: Forgetting the absolute value in distance formulas. The distance formula always gives a non-negative result. The expression $ax_1 + by_1 + cz_1 + d$ can be negative if the point is on the opposite side of the plane — always take the modulus.

Mistake 4: Assuming two non-parallel lines must intersect. In 3D, two lines can be neither parallel nor intersecting — that’s what skew lines are. Before finding the intersection of two lines, first verify they are coplanar using the determinant condition.

Mistake 5: Not converting to the correct form before applying formulas. If the plane is given as $2x - y + 3z = 7$ and you need to use $ax + by + cz + d = 0$ , write it as $2x - y + 3z - 7 = 0$ . Getting the sign of $d$ wrong changes your distance calculation entirely.

Practice Questions

Q1 (CBSE, 3 marks): Find the angle between the lines whose direction ratios are $(1, 1, 2)$ and $(\sqrt{3}-1, -\sqrt{3}-1, 4)$ .

\cos\theta = \frac{1 \cdot(\sqrt{3}-1) + 1\cdot(-\sqrt{3}-1) + 2\cdot4}{\sqrt{6}\cdot\sqrt{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2+16}}

Numerator: $\sqrt{3}-1 - \sqrt{3} - 1 + 8 = 6$

Denominator: $\sqrt{6} \cdot \sqrt{4-2\sqrt{3} + 4+2\sqrt{3}+16} = \sqrt{6}\cdot\sqrt{24} = \sqrt{144} = 12$

$\cos\theta = \dfrac{6}{12} = \dfrac{1}{2}$ , so $\theta = 60°$ .

Q2 (CBSE, 5 marks): Find the foot of the perpendicular from the point $(3, -1, 11)$ to the line $\dfrac{x}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$ .

Let the foot of perpendicular be $F = (2\lambda, 3\lambda+2, 4\lambda+3)$ for some $\lambda$ .

Direction of the line: $(2, 3, 4)$ .

Vector from $F$ to the point: $(3-2\lambda, -3-3\lambda, 8-4\lambda)$ .

For perpendicularity, dot product with $(2,3,4) = 0$ :

2(3-2\lambda) + 3(-3-3\lambda) + 4(8-4\lambda) = 0

6 - 4\lambda - 9 - 9\lambda + 32 - 16\lambda = 0

29 - 29\lambda = 0 \implies \lambda = 1

Foot $F = (2, 5, 7)$ .

Q3 (JEE Main level): Find the distance between the parallel planes $2x - y + 3z - 4 = 0$ and $6x - 3y + 9z + 13 = 0$ .

Rewrite the second plane by dividing by 3: $2x - y + 3z + \frac{13}{3} = 0$ .

Distance between parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is $\dfrac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$ .

Here $d_1 = -4$ , $d_2 = \frac{13}{3}$ .

d = \frac{\left|-4 - \frac{13}{3}\right|}{\sqrt{4+1+9}} = \frac{\left|\frac{-25}{3}\right|}{\sqrt{14}} = \frac{25}{3\sqrt{14}} = \frac{25\sqrt{14}}{42}

Q4 (JEE Main level): Show that the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$ and $\dfrac{x-4}{5} = \dfrac{y-1}{2} = \dfrac{z}{1}$ intersect. Find their point of intersection.

Any point on $L_1$ : $(2\lambda+1, 3\lambda+2, 4\lambda+3)$

Any point on $L_2$ : $(5\mu+4, 2\mu+1, \mu)$

Equating $z$ -coordinates: $4\lambda+3 = \mu$ … (i)

Equating $x$ : $2\lambda+1 = 5\mu+4 \implies 2\lambda = 5\mu+3$ … (ii)

Equating $y$ : $3\lambda+2 = 2\mu+1 \implies 3\lambda = 2\mu-1$ … (iii)

From (i): $\mu = 4\lambda+3$ . Substitute in (ii): $2\lambda = 5(4\lambda+3)+3 = 20\lambda+18$ , so $-18\lambda = 18$ , $\lambda = -1$ .

Then $\mu = -1$ . Check (iii): $3(-1) = 2(-1)-1 = -3$ ✓ Lines intersect.

Point: $(2(-1)+1, 3(-1)+2, 4(-1)+3) = (-1, -1, -1)$ .

Q5 (JEE Main level): Find the equation of the plane passing through the points $(1,0,-1)$ , $(3,2,2)$ and parallel to the line $\dfrac{x-1}{1} = \dfrac{y-1}{-2} = \dfrac{z-2}{3}$ .

Vectors in the plane: $\vec{v}_1 = (3-1, 2-0, 2-(-1)) = (2,2,3)$ and direction of the given line $\vec{v}_2 = (1,-2,3)$ .

Normal to the plane: $\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&2&3\\1&-2&3\end{vmatrix}$

$= \hat{i}(6+6) - \hat{j}(6-3) + \hat{k}(-4-2) = 12\hat{i} - 3\hat{j} - 6\hat{k}$

Simplify: $(4, -1, -2)$ . Using point $(1,0,-1)$ :

$4(x-1) - 1(y-0) - 2(z+1) = 0$

$4x - y - 2z - 6 = 0$

Q6 (JEE Main — Direct Formula): Find the angle between the line $\dfrac{x+1}{2} = \dfrac{y}{3} = \dfrac{z-3}{6}$ and the plane $10x + 2y - 11z = 3$ .

Line direction: $(2, 3, 6)$ . Plane normal: $(10, 2, -11)$ .

\sin\theta = \frac{|2(10) + 3(2) + 6(-11)|}{\sqrt{4+9+36}\cdot\sqrt{100+4+121}} = \frac{|20+6-66|}{\sqrt{49}\cdot\sqrt{225}} = \frac{40}{7\cdot15} = \frac{40}{105} = \frac{8}{21}

$\theta = \sin^{-1}\left(\dfrac{8}{21}\right)$

FAQs

What is the difference between direction ratios and direction cosines?

Direction ratios are any three numbers proportional to the actual cosines — they tell you the direction but aren’t normalized. Direction cosines satisfy $l^2+m^2+n^2=1$ . You convert by dividing each DR by $\sqrt{a^2+b^2+c^2}$ . Use DRs for setting up equations; use DCs when you need actual angles.

How do I find the equation of a plane if three points are given?

Set up the plane as $ax+by+cz=1$ and substitute the three points to get three equations in $a$ , $b$ , $c$ . Solve the system. Alternatively, find two vectors in the plane (difference of point pairs), take their cross product for the normal, then write the plane equation using any one of the three points.

When do I use $\cos\theta$ vs $\sin\theta$ in angle formulas?

Line-to-line angle uses $\cos\theta$ . Line-to-plane angle uses $\sin\theta$ (since the line makes angle $\theta$ with the plane, it makes $90°-\theta$ with the normal). Plane-to-plane angle (the dihedral angle) uses $\cos\theta$ between their normals.

How do I check if two lines in 3D are parallel, intersecting, or skew?

First check if direction vectors are parallel (proportional DRs). If yes, lines are parallel. If not, check coplanarity using the determinant condition. If coplanar and not parallel, they intersect. If neither parallel nor coplanar, they’re skew.

What’s the fastest way to find the foot of a perpendicular from a point to a plane?

If the point is $P(x_0,y_0,z_0)$ and the plane is $ax+by+cz+d=0$ , the foot is:

\left(x_0 - \frac{a(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2},\ y_0 - \frac{b(\ldots)}{a^2+b^2+c^2},\ z_0 - \frac{c(\ldots)}{a^2+b^2+c^2}\right)

Is 3D geometry important for JEE Advanced?

Yes, but JEE Advanced pushes harder — expect problems combining 3D geometry with vectors, and parametric line problems where you optimize a distance. The concepts are the same; the algebra gets denser. Practice finding the image of a point in a plane and locus problems.

How many marks does 3D geometry carry in CBSE Class 12?

Typically 8 marks: one 3-mark question and one 5-mark question. Some years it appears in internal choices. Given the predictability and formula-based nature, treat this as easy full marks in your board prep.

Can direction cosines be negative?

Yes. Direction cosines can be negative because the angles $\alpha$ , $\beta$ , $\gamma$ can be obtuse. The constraint is only $l^2+m^2+n^2=1$ — individual values range from $-1$ to $1$ . A line has two sets of direction cosines (opposite directions), differing only in sign.