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

First check direction vectors $\vec{b}_1$ and $\vec{b}_2$. If $\vec{b}_1 = k\vec{b}_2$, the lines are parallel or coincident. If not parallel, compute scalar triple product $[\vec{a}_2 - \vec{a}_1, \vec{b}_1, \vec{b}_2]$. If this determinant value equals $0$, the lines intersect at a point; if non-zero, the lines are skew.

What is the fastest strategy to solve Foot of Perpendicular questions in JEE Main?

Always convert the line equation into parametric form $Q(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$. Express vector $\vec{PQ}$ in terms of $\lambda$ and immediately set the dot product $\vec{PQ} \cdot \vec{d} = 0$, where $\vec{d} = \langle a, b, c \rangle$. Solving this linear equation in $\lambda$ gives $Q$ directly without lengthy formula substitution.

Three Dimensional Geometry Notes, Formulas, Traps & PYQs | ExamiFly

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Key Formulas & High-Yield Equations

Direction Cosines Fundamental Relation

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

Valid strictly when

l = \cos\alpha

m = \cos\beta

n = \cos\gamma

represent direction cosines where

\alpha, \beta, \gamma \in [0, \pi]

are angles made with positive

x, y, z

axes respectively.

Angle Between Two Lines

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

Applies strictly when

\langle a_1, b_1, c_1 \rangle

and

\langle a_2, b_2, c_2 \rangle

are direction ratios of non-zero directional vectors

\vec{v}_1

and

\vec{v}_2

Vector Form of a Straight Line

$\vec{r} = \vec{a} + \lambda \vec{b}$

Valid for a line passing through a fixed point with position vector

\vec{a}

and parallel to a non-zero direction vector

\vec{b}

, with scalar parameter

\lambda \in \mathbb{R}

Shortest Distance Between Skew Lines

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

Valid strictly for non-parallel, non-intersecting lines where

|\vec{b}_1 \times \vec{b}_2| \neq 0

. If cross product vanishes, parallel line distance formula must be used.

Shortest Distance Between Parallel Lines

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

Applies strictly when lines share identical or collinear direction vectors

\vec{b}_1 = k\vec{b}_2 = \vec{b}

Solved High-Yield Sample Questions

Practice Mode

Question #1JEE Main 2024

Find the shortest distance between the two lines given by $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Extract position vectors $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and direction vectors $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ , $\vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$ . Thus $\vec{a}_2 - \vec{a}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$ .

Step 2

Calculate cross product $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$ . Its magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$ .

Step 3

Evaluate dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-1) + (2)(2) + (2)(-1) = 1$ . The shortest distance is $d = \frac{1}{\sqrt{6}}$ .

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Question #2JEE Main 2023

Let $P(1, 2, 3)$ be a point and line $L$ be $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ . If $Q$ is the foot of the perpendicular from $P$ onto line $L$ , find the distance $PQ$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write parametric coordinates of any arbitrary point $Q$ on line $L$ as $Q(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$ for parameter $\lambda \in \mathbb{R}$ .

Step 2

Determine vector $\vec{PQ} = (3\lambda + 5)\hat{i} + (2\lambda + 5)\hat{j} + (-2\lambda + 4)\hat{k}$ . Apply orthogonality condition $\vec{PQ} \cdot \vec{d} = 0$ with direction vector $\vec{d} = 3\hat{i} + 2\hat{j} - 2\hat{k}$ : $3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0 \implies 17\lambda + 17 = 0 \implies \lambda = -1$ .

Step 3

Substitute $\lambda = -1$ to get $Q(3, 5, 9)$ . Calculate distance $PQ = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .

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Mastering Three Dimensional Geometry: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Direction Cosines Fundamental Relation

Angle Between Two Lines

Vector Form of a Straight Line

Shortest Distance Between Skew Lines

Shortest Distance Between Parallel Lines

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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