Three Dimensional Geometry
Master 3D coordinate geometry for JEE — direction cosines, equations of lines and planes, angles, distances, skew lines, and intersections with competitive-level shortcuts and tricks.
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01
Direction Cosines & Direction Ratios
Foundation of 3D geometry — representing directions in space using cosines and ratios. Essential building block for line and plane equations.
Direction Cosines & Ratios
Definition — Direction Cosines
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 its direction cosines (DCs). They satisfy $l^2 + m^2 + n^2 = 1$.
Direction Ratios
Any three numbers $a, b, c$ proportional to $l, m, n$ are direction ratios (DRs). If DRs are $(a, b, c)$, then DCs are:
$$l = \frac{a}{\sqrt{a^2+b^2+c^2}},\quad m = \frac{b}{\sqrt{a^2+b^2+c^2}},\quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$
JEE Shortcut — DRs of Line Joining Two Points
DRs of the line joining $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ are $(x_2-x_1,\; y_2-y_1,\; z_2-z_1)$.
Trick: For the angle between two lines with DRs $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$: $$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}$$
Trick: For the angle between two lines with DRs $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$: $$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}$$
Condition for Perpendicularity & Parallelism
Perpendicular: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Parallel: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
Parallel: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
📝 Example
Find the angle between lines with DRs $(2, 3, 6)$ and $(1, 2, 2)$.
$\cos\theta = \frac{|2(1)+3(2)+6(2)|}{\sqrt{4+9+36}\;\sqrt{1+4+4}} = \frac{|2+6+12|}{7 \cdot 3} = \frac{20}{21}$. So $\theta = \cos^{-1}\!\left(\frac{20}{21}\right)$.
02
Equations of Lines
Symmetric, vector, and parametric forms of straight lines in 3D — the backbone of JEE 3D geometry problems.
Forms of a Straight Line in 3D
Symmetric (Cartesian) Form
A line through $(x_1, y_1, z_1)$ with DRs $(a, b, c)$:
$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$
Vector Form
$\vec{r} = \vec{a} + \lambda\,\vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is the direction vector.
Parametric Form
$x = x_1 + at$, $y = y_1 + bt$, $z = z_1 + ct$, where $t \in \mathbb{R}$.
Line Through Two Points
Through $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$:
$$\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$$
JEE Trick: Any point on this line can be written as $\left(\frac{\lambda x_2 + x_1}{\lambda+1},\; \frac{\lambda y_2 + y_1}{\lambda+1},\; \frac{\lambda z_2 + z_1}{\lambda+1}\right)$ using section formula.
📝 Example
Write the equation of the line through $(1, -1, 2)$ and parallel to the line $\frac{x-3}{2} = \frac{y}{-1} = \frac{z+1}{3}$.
The direction ratios of the given line are $(2, -1, 3)$. The required line through $(1,-1,2)$ with same DRs: $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{3}$.
03
Equations of Planes
Multiple forms of plane equations — intercept, normal, general, and vector forms. High-weightage in JEE Advanced.
Forms of a Plane
General Form
$ax + by + cz + d = 0$, where $(a, b, c)$ is the normal to the plane.
Intercept Form
A plane cutting intercepts $a, b, c$ on the axes: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$.
Normal (Vector) Form
$\vec{r} \cdot \hat{n} = d$, where $\hat{n}$ is the unit normal and $d$ is the perpendicular distance from the origin.
Equivalently: $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, where $\vec{a}$ is a known point on the plane.
Equivalently: $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, where $\vec{a}$ is a known point on the plane.
Plane Through Three Non-Collinear Points
Through $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$:
$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$$
JEE Trick: The normal to this plane is $\vec{AB} \times \vec{AC}$.
Family of Planes
The plane through the intersection of planes $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0$.
JEE Shortcut: Use this family form and determine $\lambda$ from a given condition — saves a lot of computation.
JEE Shortcut: Use this family form and determine $\lambda$ from a given condition — saves a lot of computation.
📝 Example
Find the equation of the plane through $(1,1,1)$ and containing the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z}{1}$.
The line passes through $(1,2,0)$ with DR $(2,3,1)$. A vector from $(1,2,0)$ to $(1,1,1)$ is $(0,-1,1)$. Normal $= (2,3,1) \times (0,-1,1) = (3+1,\; 0-2,\; -2-0) = (4,-2,-2)$. Simplify to $(2,-1,-1)$.
Plane: $2(x-1) - 1(y-1) - 1(z-0) = 0 \Rightarrow 2x - y - z - 1 = 0$.
Plane: $2(x-1) - 1(y-1) - 1(z-0) = 0 \Rightarrow 2x - y - z - 1 = 0$.
04
Angles & Distances
Angle between line and plane, distance from a point to a plane, distance between parallel planes — JEE staples with elegant formulas.
Angle Between Line & Plane
Angle Between Two Planes
For planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$:
$$\cos\theta = \frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}$$
Angle Between a Line and a Plane
If a line has DRs $(a,b,c)$ and the plane has normal $(l,m,n)$:
$$\sin\theta = \frac{|al+bm+cn|}{\sqrt{a^2+b^2+c^2}\;\sqrt{l^2+m^2+n^2}}$$
JEE Trick: Note it is $\sin\theta$ (not $\cos\theta$) because the angle is measured from the plane, not the normal.
Distance from Point to Plane
Distance from $(x_1,y_1,z_1)$ to plane $ax+by+cz+d=0$:
$$D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$$
Distance Between Parallel Planes
For $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$: $D = \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$.
JEE Shortcut: First ensure both planes have the same normal coefficients before applying this formula.
JEE Shortcut: First ensure both planes have the same normal coefficients before applying this formula.
📝 Example
Find the distance from $(2, 3, -5)$ to the plane $x + 2y - 2z = 9$.
$D = \frac{|2 + 6 + 10 - 9|}{\sqrt{1+4+4}} = \frac{|9|}{3} = \boxed{3}$.
05
Skew Lines, Shortest Distance & Intersections
Skew lines, shortest distance between lines, and intersection of planes and lines — advanced 3D concepts frequently tested in JEE Advanced.
Skew Lines & Shortest Distance
Definition — Skew Lines
Two lines in 3D that are neither parallel nor intersecting are called skew lines. They lie in different planes.
Shortest Distance Between Two Skew Lines
For lines $\vec{r} = \vec{a}_1 + \lambda\,\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\,\vec{b}_2$:
$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$
JEE Trick: If $d = 0$, the lines are coplanar (either intersecting or parallel).
Cartesian Form — Shortest Distance
For lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$:
$$d = \frac{\left|\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}\right|}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}$$
Distance Between Parallel Lines
For parallel lines $\vec{r} = \vec{a}_1 + \lambda\,\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\,\vec{b}$:
$$d = \frac{|(\vec{a}_2-\vec{a}_1) \times \vec{b}|}{|\vec{b}|}$$
📝 Example
Find the shortest distance between $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.
$\vec{a}_2 - \vec{a}_1 = (1, 2, 2)$, $\vec{b}_1 = (2,3,4)$, $\vec{b}_2 = (3,4,5)$.
$\vec{b}_1 \times \vec{b}_2 = (15-16,\; 12-10,\; 8-9) = (-1, 2, -1)$, $|\vec{b}_1 \times \vec{b}_2| = \sqrt{1+4+1} = \sqrt{6}$.
$(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2) = -1+4-2 = 1$.
$d = \frac{|1|}{\sqrt{6}} = \boxed{\frac{1}{\sqrt{6}}}$.
$\vec{b}_1 \times \vec{b}_2 = (15-16,\; 12-10,\; 8-9) = (-1, 2, -1)$, $|\vec{b}_1 \times \vec{b}_2| = \sqrt{1+4+1} = \sqrt{6}$.
$(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2) = -1+4-2 = 1$.
$d = \frac{|1|}{\sqrt{6}} = \boxed{\frac{1}{\sqrt{6}}}$.
Intersection of Planes & Lines
Line of Intersection of Two Planes
The intersection of two non-parallel planes is a line. To find it:
1. Find a point common to both planes (set one variable to 0 and solve).
2. The direction of the line is $\vec{n}_1 \times \vec{n}_2$ (cross product of normals).
JEE Trick: The cross product $\vec{n}_1 \times \vec{n}_2$ directly gives the DRs of the line of intersection.
1. Find a point common to both planes (set one variable to 0 and solve).
2. The direction of the line is $\vec{n}_1 \times \vec{n}_2$ (cross product of normals).
JEE Trick: The cross product $\vec{n}_1 \times \vec{n}_2$ directly gives the DRs of the line of intersection.
Intersection of a Line and a Plane
Substitute the parametric form of the line into the plane equation, solve for $t$, then find the point.
Condition for line lying in plane: The line is parallel to the plane AND passes through a point on the plane.
Condition for line lying in plane: The line is parallel to the plane AND passes through a point on the plane.
📝 Example
Find the point of intersection of the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{1}$ and the plane $x+y+z=6$.
Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{1}=t$. Then $x=2t+1$, $y=3t-1$, $z=t$.
Substituting: $(2t+1)+(3t-1)+t = 6 \Rightarrow 6t = 6 \Rightarrow t=1$.
Point: $(3, 2, 1)$.
Substituting: $(2t+1)+(3t-1)+t = 6 \Rightarrow 6t = 6 \Rightarrow t=1$.
Point: $(3, 2, 1)$.
★ Key Takeaways
- Direction cosines satisfy $l^2+m^2+n^2=1$ — use this to normalize direction ratios quickly.
- For angle between line and plane, use $\sin\theta$ (not $\cos\theta$) — a common JEE mistake.
- The shortest distance formula uses scalar triple product in the numerator — if it equals zero, the lines are coplanar.
- Use the family of planes $P_1 + \lambda P_2 = 0$ to bypass lengthy simultaneous equations.
- The cross product of two plane normals gives the DRs of their line of intersection — saves significant computation.
✎ Practice Problems
Problem 1
Find the direction cosines of the line joining $(1, 2, 3)$ and $(4, 6, 6)$.
Show Solution ▼
DRs $= (3, 4, 3)$. Magnitude $= \sqrt{9+16+9} = \sqrt{34}$. DCs $= \left(\frac{3}{\sqrt{34}},\; \frac{4}{\sqrt{34}},\; \frac{3}{\sqrt{34}}\right)$.
Problem 2
Find the equation of the plane through $(1,1,1)$, $(1,-1,1)$, and $(-7,3,-5)$.
Show Solution ▼
$\vec{AB} = (0,-2,0)$, $\vec{AC} = (-8,2,-6)$. Normal $= \vec{AB}\times\vec{AC} = (12,0,−16)$, simplified $(3,0,-4)$. Plane: $3(x-1)-4(z-1)=0 \Rightarrow 3x - 4z + 1 = 0$.
Problem 3
Find the image of the point $(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$.
Show Solution ▼
Foot of perpendicular: parametric line through $(1,3,4)$ with DRs $(2,-1,1)$: $(1+2t, 3-t, 4+t)$. Substituting in plane: $2(1+2t)-(3-t)+(4+t)+3=0 \Rightarrow 6t+6=0 \Rightarrow t=-1$. Foot $= (-1,4,3)$. Image $= (2(-1)-1, 2(4)-3, 2(3)-4) = (-3, 5, 2)$.
Problem 4
Show that the lines $\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{0}$ and $\frac{x-4}{2}=\frac{y-1}{0}=\frac{z}{1}$ are skew, and find the shortest distance.
Show Solution ▼
$\vec{b}_1 \times \vec{b}_2 = (3,-1,0)\times(2,0,1) = (-1,-3,2)$. $\vec{a}_2-\vec{a}_1=(3,-1,-3)$. Scalar triple product $= (-1)(3)+(-3)(-1)+(2)(-3) = -3+3-6=-6 \neq 0$, so skew. $d = \frac{|-6|}{\sqrt{1+9+4}} = \frac{6}{\sqrt{14}}$.
Problem 5
Find the angle between the line $\frac{x+1}{1}=\frac{y}{2}=\frac{z-3}{-2}$ and the plane $x+y+4=0$.
Show Solution ▼
Line DRs $(1,2,-2)$, plane normal $(1,1,0)$. $\sin\theta = \frac{|1+2+0|}{3 \cdot \sqrt{2}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$. So $\theta = \frac{\pi}{4}$ or $45°$.
🎯 Interactive MCQs
1. If the direction cosines of a line are $\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)$, then $c$ equals:
A $1$
B $2$
C $\sqrt{3}$
D $3$
2. The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is:
A $1$
B $\frac{1}{\sqrt{3}}$
C $7$
D $\frac{1}{7}$
3. The plane containing the line $\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z}{0}$ and parallel to the line $\frac{x}{0}=\frac{y}{1}=\frac{z-1}{1}$ passes through the point:
A $(0, 0, 0)$
B $(0, 6, 3)$
C $(1, 2, 0)$
D $(2, 3, 1)$
4. The shortest distance between the lines $\frac{x}{1}=\frac{y}{-1}=\frac{z}{1}$ and $\frac{x+1}{-1}=\frac{y-1}{2}=\frac{z}{1}$ is:
A $0$
B $\frac{1}{\sqrt{6}}$
C $\frac{1}{\sqrt{14}}$
D $\frac{3}{\sqrt{14}}$
5. The equation of the plane through the intersection of $x+y+z=6$ and $2x+3y+4z+5=0$, and passing through $(1,1,1)$, is:
A $20x + 23y + 26z = 69$
B $20x + 23y + 26z = 46$
C $2x + 3y + 4z = 9$
D $20x + 23y + 26z = 50$