3D Geometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Coordinate System Basics: In 3D Cartesian system, a point is represented as $P(x, y, z)$. The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

Section Formula: For internal division in ratio $m:n$: $$(x, y, z) = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$$

Direction Ratios and Cosines: For line joining $(x_1,y_1,z_1)$ to $(x_2,y_2,z_2)$: direction ratios are $(a,b,c) = (x_2-x_1, y_2-y_1, z_2-z_1)$. Direction cosines are: $$l = \frac{a}{\sqrt{a^2+b^2+c^2}},\ m = \frac{b}{\sqrt{a^2+b^2+c^2}},\ n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$ with $l^2 + m^2 + n^2 = 1$.

⚡ JEE Tip: For a line with direction ratios $(a,b,c)$, any scalar multiple $(ka, kb, kc)$ represents the same direction. Direction cosines are unique (up to sign).

⚡ Common Mistake: Remember that lines in 3D can be skew (neither parallel nor intersecting). Always check if lines actually intersect before solving intersection problems.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Line Equations:

• Vector form: $\vec{r} = \vec{a} + \lambda \vec{b}$ where $\vec{a}$ is position vector of a point on line and $\vec{b}$ is direction vector
• Cartesian form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ where $(a,b,c)$ are direction ratios
• Symmetric form: Same as Cartesian, written as single equation with parameter $t$

Plane Equations:

• General form: $ax + by + cz + d = 0$ where $\vec{n} = (a,b,c)$ is the normal vector
• Point-normal form: $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$
• Vector form: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$

Angle Formulas:

• Angle between two lines (with direction vectors $\vec{b}_1, \vec{b}_2$): $$\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|}$$

• Angle between two planes (with normals $\vec{n}_1, \vec{n}_2$): $$\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$$

• Angle between line and plane (line direction $\vec{b}$, plane normal $\vec{n}$): $$\sin\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}$$

Distance Formulas:

• Distance from point $(x_1,y_1,z_1)$ to plane $ax + by + cz + d = 0$: $$D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2+b^2+c^2}}$$

• Distance from point $\vec{p}$ to line $\vec{r} = \vec{a} + \lambda\vec{b}$: $$D = \frac{|\vec{b} \times (\vec{p} - \vec{a})|}{|\vec{b}|}$$

Worked Examples:

Example 1: Find the angle between lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{1}$ and $\frac{x+1}{1} = \frac{y}{2} = \frac{z-2}{3}$.

Direction vectors: $\vec{b}_1 = (2,3,1)$ and $\vec{b}_2 = (1,2,3)$. $\vec{b}_1 \cdot \vec{b}_2 = 2(1) + 3(2) + 1(3) = 2 + 6 + 3 = 11$. $|\vec{b}_1| = \sqrt{4+9+1} = \sqrt{14}$. $|\vec{b}_2| = \sqrt{1+4+9} = \sqrt{14}$. $\cos\theta = \frac{11}{14}$, so $\theta = \cos^{-1}\left(\frac{11}{14}\right) \approx 38.2°$.

Example 2: Find the distance from point $(1, 2, 3)$ to plane $2x + y - 2z + 5 = 0$.

$D = \frac{|2(1) + 1(2) - 2(3) + 5|}{\sqrt{4+1+4}} = \frac{|2 + 2 - 6 + 5|}{\sqrt{9}} = \frac{|3|}{3} = 1$.

Example 3 (JEE 2022): Find the equation of plane passing through $(1, 2, 3)$ and perpendicular to line joining $(3, 4, 5)$ and $(5, 6, 7)$.

Direction of line: $\vec{b} = (2, 2, 2)$ or $(1, 1, 1)$. This is the normal to our required plane. Plane: $1(x-1) + 1(y-2) + 1(z-3) = 0$ → $x + y + z - 6 = 0$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Skew Lines and Shortest Distance:

For two skew lines $\vec{r}_1 = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r}_2 = \vec{a}_2 + \mu\vec{b}_2$, the shortest distance is: $$D = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$

Coplanarity Condition:

Lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ are coplanar iff: $$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$$

Sphere:

General equation: $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$. Centre: $(-u, -v, -w)$, Radius: $\sqrt{u^2+v^2+w^2-d}$ (real if $u^2+v^2+w^2 > d$).

Intersection of Line and Sphere:

For line $\vec{r} = \vec{a} + \lambda\vec{b}$ and sphere $|\vec{r} - \vec{c}| = R$: Substitute: $|\vec{a} + \lambda\vec{b} - \vec{c}|^2 = R^2$. This gives $\lambda^2|\vec{b}|^2 + 2\lambda\vec{b} \cdot (\vec{a} - \vec{c}) + |\vec{a} - \vec{c}|^2 - R^2 = 0$. Solve quadratic in $\lambda$. If discriminant $< 0$, line misses sphere.

Advanced Problems:

Problem (JEE Advanced 2020): Find the locus of a point $P$ such that the sum of squares of its distances from planes $x + y + z = 3$ and $x + 2y + 2z = 4$ equals $10$.

Solution: Let $P(x,y,z)$. Distance to first plane: $d_1 = \frac{|x+y+z-3|}{\sqrt{3}}$. Distance to second: $d_2 = \frac{|x+2y+2z-4|}{\sqrt{9} = 3}$. Condition: $d_1^2 + d_2^2 = 10$. So $\frac{(x+y+z-3)^2}{3} + \frac{(x+2y+2z-4)^2}{9} = 10$. Multiplying by 9: $3(x+y+z-3)^2 + (x+2y+2z-4)^2 = 90$. Expand and simplify to get locus equation.

Problem: Find the equation of sphere passing through circle $x^2+y^2+z^2=9$ in plane $z=2$ and point $(0,0,0)$.

Plane $z=2$ gives circle of intersection: $x^2+y^2+4=9$ → $x^2+y^2=5$. Sphere general form: $x^2+y^2+z^2+2ux+2vy+2wz+d=0$. Passing through $(0,0,0)$: $d = 0$. Passing through circle means passing through three points on circle, e.g., $(\sqrt{5},0,2), (-\sqrt{5},0,2), (0,\sqrt{5},2)$. From $(\sqrt{5},0,2)$: $5+4+2u\sqrt{5}+2w(2) = 0$ → $2u\sqrt{5}+2w(2) = -9$. Similar equations give system to solve for $u, v, w$.

JEE Advanced Patterns (2018–2024):

• Shortest distance between skew lines is a recurring topic
• Sphere intersection with planes/lines tested nearly every year
• Combined problems with conics and 3D are common in Paper 2
• Conditions for perpendicularity/coplanarity appear frequently
• Projection-related questions have appeared in 2021, 2023

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