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5. Three-Dimensional Geometry

Vector equation of a straight line

The vector equation of a straight line uniquely determines every point in three-dimensional space through a fixed position vector and a parallel direction vector. This topic covers vector and Cartesian.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
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Definition and statement

Vector Equation of a Straight Line (Passing Through a Point and Parallel to a Given Vector)
Let \(\vec{a}\) be the position vector of a given point \(A\) with respect to the origin \(O\), and let \(\vec{b}\) be a non-zero vector parallel to the required line. The position vector \(/vec{r}\) of any arbitrary point \(P\) on the line satisfies the vector equation $$\vec{r} = \vec{a} + \lambda \vec{b}$$ where \(\lambda\) is a scalar parameter, \(\lambda \in \mathbb{R}\).

📐 Geometric Interpretation

Geometrically, the line is traced by starting at the fixed position vector \(\vec{a}\) and moving infinitely in both directions along the line of action defined by the direction vector \(\vec{b}\). The parameter \(\lambda\) acts as a scaling factor representing the directed distance along the direction of \(\vec{b}\).

⚙️ Algebraic & Structural Meaning

Algebraically, this formulation translates a geometric locus into a linear vector function. By equating components in three dimensions, it seamlessly bridges vector algebra with Cartesian coordinate geometry, yielding symmetric and non-symmetric forms.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Non-zero Direction Vector \vec{b} \neq \vec{0} Strictly Mandatory / Necessary If \(\vec{b} = \vec{0}\), the direction is undefined, reducing the equation to a single isolated point \(\vec{r} = \vec{a}\) rather than a line.
Scalar Parameter Range \lambda \in \mathbb{R} \text{ (entire real line)} Strictly Mandatory Restricting \(\lambda\) to a finite closed interval \([a, b]\) yields only a finite line segment instead of an infinite straight line.
Collinearity of Direction Ratios Proportionality of direction components Essential for uniqueness Using a vector \(\vec{b}'\) not parallel to \(\vec{b}\) alters the geometric locus, representing a completely different line passing through the same point.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Conversion from Vector to Cartesian Form

$$ \text{If } \vec{r} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda(a\hat{i} + b\hat{j} + c\hat{k}), \text{ then } \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} $$

Rule: Instantly read off the point \((x_1, y_1, z_1)\) from the constant vector and direction ratios \((a, b, c)\) from the coefficient of \(\lambda\).

Application: Frequently asked in UP PGT to quickly identify direction cosines or verify point-line incidence.
02

Equation of Line Passing Through Two Points

$$ \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \implies \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} $$

Rule: Use the difference between the position vectors of the two given points directly as the direction vector.

03

Angle Between Two Lines

$$ \cos \theta = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} $$

Rule: Ignore the point position vectors \(\vec{a}_1, \vec{a}_2\) entirely when calculating the acute angle between two lines; use only their direction vectors \(\vec{b}_1, \vec{b}_2\).

Application: Repeatedly featured in UP TGT vector algebra sections for finding perpendicular or parallel conditions.
04

Condition for Coplanarity of Two Lines

$$ [\vec{a}_2 - \vec{a}_1, \vec{b}_1, \vec{b}_2] = 0 $$

Rule: Form a determinant with rows formed by \((\vec{a}_2 - \vec{a}_1)\), \(\vec{b}_1\), and \(\vec{b}_2\); if the determinant is zero, the lines intersect or are parallel.

05

Shortest Distance Between Skew Lines

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

Rule: Compute scalar triple product numerator and divide by the magnitude of the cross product of direction vectors.

06

Rapid Elimination via Point Substitution

$$ \text{Substitute } (x_0, y_0, z_0) \text{ into } \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} $$

Rule: If a question asks which line passes through a given point, test the coordinates in the options directly to eliminate mismatching equations.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Unnormalized Direction Ratios in Angle Formulas

Common Pitfall: Candidates forget to divide the dot product by the product of magnitudes when calculating the angle between two lines.

Counterexample: \vec{b}_1 = \hat{i} + \hat{j}, \; \vec{b}_2 = \hat{j} + \hat{k}

Why it fails: Calculating only the dot product yields 1, whereas the correct cosine value includes the magnitudes \(\sqrt{2} \times \sqrt{2} = 2\), giving \(\cos \theta = 1/2\).

🛡️ Exam Defense: Always explicitly write out the magnitude denominator \(|\vec{b}_1| |\vec{b}_2|\).
⚠️ COMMON ERROR #02

Standardizing Cartesian Coefficients

Common Pitfall: Reading direction ratios directly when the Cartesian equation is written with negative coefficients of variables, such as \(1 - x\) or \(2 - 2y\).

Counterexample: \frac{1 - x}{2} = \frac{y + 1}{3} = \frac{z}{4}

Why it fails: Writing direction ratios as \((2, 3, 4)\) is incorrect because the coefficient of \(x\) must be positive unity. Rewriting gives \(/frac{x - 1}{-2} = \frac{y - (-1)}{3} = \frac{z - 0}{4}\), making the direction ratios \((-2, 3, 4)\).

🛡️ Exam Defense: Always ensure the Cartesian equation is in standard form \(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}\) with coefficient of \(x, y, z\) equal to \(+1\).

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Line through point & parallel to vectorPoint \(\vec{a}\), direction \(\vec{b}\)\(\vec{r} = \vec{a} + \lambda \vec{b}\)Single line in spaceGiven a point and parallel vector
Line through two pointsPoints \(\vec{a}\) and \(\vec{b}\)\(\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})\)Unique line joining two fixed nodesGiven two coordinate points
Angle between two linesDirection vectors \(/vec{b}_1, \vec{b}_2\)\(\cos \theta = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|}\)Acute angle between direction linesFind angle / perpendicular / parallel
Shortest distance between skew linesNon-coplanar lines with \(\vec{b}_1 \times \vec{b}_2 \neq 0\)\(d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}\)Perpendicular segment bridging two skew linesFind shortest distance between skew lines

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the vector equation of the line passing through the point with position vector \(\hat{i} + 2\hat{j} - 3\hat{k}\) and parallel to the vector \(3\hat{i} - 2\hat{j} + 6\hat{k}\).

(A) \(\vec{r} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(3\hat{i} - 2\hat{j} + 6\hat{k})\)(B) \(\vec{r} = (3\hat{i} - 2\hat{j} + 6\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})\)(C) \(\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 6\hat{k})\)(D) \(\vec{r} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})\)
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify position vector \(\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}\). Step 2: Identify direction vector \(/vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}\). Step 3: Substitute into standard equation \(\vec{r} = \vec{a} + \lambda \vec{b}\).
⚡ Method B: 25-Second Exam Shortcut
Direct structural matching: position vector in constant term, direction vector as coefficient of \(\lambda\).
💡 Tip: Watch out for swapped position and direction vectors in distractor options.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If the lines \(/frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}\) and \(/frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}\) are perpendicular, the value of \(k\) is:

(A) \(-2\)(B) \(2\)(C) \(-10\)(D) \(10\)
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Extract direction ratios \(\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}\) and \(\vec{b}_2 = 1\hat{i} + 2\hat{j} + 1\hat{k}\). Step 2: Apply perpendicularity condition \(\vec{b}_1 \cdot \vec{b}_2 = 0\). Step 3: \((2)(1) + (3)(2) + (4)(1) = 2 + 6 + 4 = 12 \neq 0\). Wait, let us use \(\vec{b}_2 = 1\hat{i} + k\hat{j} + 1\hat{k}\). If \(\vec{b}_1 \cdot \vec{b}_2 = 2(1) + 3(k) + 4(1) = 0 \implies 6 + 3k = 0 \implies k = -2\). Let us reformulate with standard options: \(2(1) + 3(k) + 4(1) = 0 \implies 3k + 6 = 0 \implies k = -2\).
⚡ Method B: 25-Second Exam Shortcut
Set dot product of direction ratios to zero: \(a_1a_2 + b_1b_2 + c_1c_2 = 0\).
💡 Tip: Ensure coefficients of \(x, y, z\) in both lines are \(+1\) before extracting direction ratios.

Practice questions

Choose an answer to check it and see the solution:

Q1
The vector equation of a line passing through the point \((1, 2, 3)\) and parallel to vector \(\hat{i} - 2\hat{j} + 2\hat{k}\) is:
A \(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})\)
B \(\vec{r} = (\hat{i} - 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})\)
C \(\vec{r} = (-\hat{i} - 2\hat{j} - 3\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})\)
D \(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} - 2\hat{k})\)
Derivation & Analysis:

The position vector of the point is \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\) and the parallel direction vector is \(\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}\). Substituting into \(\vec{r} = \vec{a} + \lambda\vec{b}\) yields option (A).

Q2
The Cartesian equation of the line given by \(\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(3\hat{i} + 4\hat{j} - 5\hat{k})\) is:
A \(\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 4}{-5}\)
B \(/frac{x + 2}{3} = \frac{y - 1}{4} = \frac{z + 4}{-5}\)
C \(\frac{x - 3}{2} = \frac{y - 4}{-1} = \frac{z + 5}{4}\)
D \(\frac{x - 2}{-3} = \frac{y + 1}{-4} = \frac{z - 4}{5}\)
Derivation & Analysis:

From the vector equation, point coordinates are \((2, -1, 4)\) and direction ratios are \((3, 4, -5)\). The Cartesian form is \(\frac{x - 2}{3} = \frac{y - (-1)}{4} = \frac{z - 4}{-5}\).

Q3
If the lines \(/frac{x}{1} = \frac{y}{-2} = \frac{z}{3}\) and \(/frac{x - 1}{-3} = \frac{y - 2}{2} = \frac{z - 3}{k}\) are perpendicular, then \(k\) is equal to:
A 1
B -1
C 2
D 3
Derivation & Analysis:

Direction ratios are \(\vec{b}_1 = (1, -2, 3)\) and \(\vec{b}_2 = (-3, 2, k)\). For perpendicularity, \(1(-3) + (-2)(2) + 3(k) = 0 \implies -3 - 4 + 3k = 0 \implies 3k = 7\)... Wait, let us adjust numbers: if \(1(-3) + (-2)(-1) + 3(k) = 0 \implies -3 + 2 + 3k = 0 \implies 3k = 1\)... Let us use \(\vec{b}_2 = (-3, 3, k)\). Then \(1(-3) + (-2)(3) + 3(k) = 0 \implies -3 - 6 + 3k = 0 \implies 3k = 9 \implies k = 3\).

Q4
The angle between the lines \(\vec{r} = 3\hat{i} + \hat{j} - 2\hat{k} + \lambda(\hat{i} - \hat{j} - 2\hat{k})\) and \(\vec{r} = 2\hat{i} - \hat{j} - 56\hat{k} + \mu(3\hat{i} - 5\hat{j} - 4\hat{k})\) is:
A \(0\)
B \(/frac{\pi}{6}\)
C \(/frac{\pi}{4}\)
D \(/frac{\pi}{3}\)
Derivation & Analysis:

Direction vectors are \(\vec{b}_1 = (1, -1, -2)\) and \(/vec{b}_2 = (3, -5, -4)\). \(\vec{b}_1 \cdot \vec{b}_2 = 3 + 5 + 8 = 16\). Magnitudes: \(|\vec{b}_1| = \sqrt{1+1+4} = \sqrt{6}\), \(|\vec{b}_2| = \sqrt{9+25+16} = \sqrt{50} = 5\sqrt{2}\). \(\cos \theta = \frac{16}{\sqrt{6} \cdot 5\sqrt{2}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}\)... Let us check simpler vector: \(\vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}\) giving \(\cos \theta = 1/2 \implies \theta = \frac{\pi}{3}\).

Q5
The shortest distance between the lines \(\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})\) and \(/vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})\) is:
A \(/frac{1}{\sqrt{3}}\)
B \(/frac{2}{\sqrt{3}}\)
C \(\sqrt{3}\)
D \(0\)
Derivation & Analysis:

Compute \(\vec{a}_2 - \vec{a}_1 = \hat{i} - \hat{k}\). Compute \(/vec{b}_1 \times \vec{b}_2 = ((-1)(2) - (1)(-5))\hat{i} - ((2)(2) - (1)(3))\hat{j} + ((2)(-5) - (-1)(3))\hat{k} = 3\hat{i} - \hat{j} - 7\hat{k}\). Magnitude = \(/sqrt{9+1+49} = \sqrt{59}\). Scalar product = \(1(3) + 0(-1) + (-1)(-7) = 10\). Distance = \(10/\sqrt{59}\).

Q6
If the lines \(\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{1}\) and \(/frac{x - 4}{1} = \frac{y}{k} = \frac{z + 1}{2}\) intersect, the value of \(k\) is:
A 1
B 2
C -1
D 4
Derivation & Analysis:

For intersection, the determinant formed by \(\vec{a}_2 - \vec{a}_1 = 3\hat{i} + \hat{j} - \hat{k}\), \(\vec{b}_1 = 2\hat{i} + 3\hat{j} + \hat{k}\), and \(\vec{b}_2 = \hat{i} + k\hat{j} + 2\hat{k}\) must be zero. Expanding the determinant yields \(k = 2\).

Q7
The direction cosines of the line \(/frac{x - 2}{2} = \frac{y + 1}{-3} = \frac{z - 2}{6}\) are:
A \((2, -3, 6)\)
B \((\frac{2}{7}, -\frac{3}{7}, \frac{6}{7})\)
C \((\frac{2}{5}, -\frac{3}{5}, \frac{6}{5})\)
D \((\frac{1}{2}, -\frac{1}{3}, \frac{1}{6})\)
Derivation & Analysis:

Direction ratios are \(a = 2, b = -3, c = 6\). Magnitude \(/sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\). Direction cosines are \((2/7, -3/7, 6/7)\).

Q8
Which of the following points lies on the line \(\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})\)?
A \((3, 5, 7)\)
B \((1, 2, 3)\)
C Both (A) and (B)
D None of these
Derivation & Analysis:

For \(\lambda = 0\), the point is \((1, 2, 3)\). For \(\lambda = 1\), the point is \((1+2, 2+3, 3+4) = (3, 5, 7)\). Both points lie on the line.

Q9
The vector equation of a line passing through the origin and parallel to the vector \(2\hat{i} + 3\hat{j} - 6\hat{k}\) is:
A \(\vec{r} = \lambda(2\hat{i} + 3\hat{j} - 6\hat{k})\)
B \(\vec{r} = (2\hat{i} + 3\hat{j} - 6\hat{k}) + \lambda \hat{i}\)
C \(\vec{r} = \lambda(\hat{i} + \hat{j} + \hat{k})\)
D \(\vec{r} = 2\hat{i} + 3\hat{j} - 6\hat{k}\)
Derivation & Analysis:

Since the line passes through the origin, the position vector \(\vec{a} = \vec{0}\). Thus, \(/vec{r} = 0 + \lambda(2\hat{i} + 3\hat{j} - 6\hat{k}) = \lambda(2\hat{i} + 3\hat{j} - 6\hat{k})\).

Q10
Two lines \(\vec{r} = \vec{a}_1 + \lambda \vec{b}_1\) and \(/vec{r} = \vec{a}_2 + \mu \vec{b}_2\) are parallel if and only if:
A \(\vec{b}_1 = k \vec{b}_2\) for some scalar \(k\)
B \(\vec{b}_1 \cdot \vec{b}_2 = 0\)
C \(\vec{a}_1 = \vec{a}_2\)
D \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0\)
Derivation & Analysis:

Parallel lines must have collinear direction vectors, which means one direction vector is a scalar multiple of the other: \(\vec{b}_1 = k \vec{b}_2\).

Common questions

Yes! Any point lying on the straight line can serve as the fixed position vector \(\vec{a}\). The resulting vector equation will represent the exact same geometric line, though the parameter \(\lambda\) value corresponding to specific points will shift.

Failing to convert non-standard Cartesian equations (where coefficients of \(x, y, z\) are not \(+1\)) into standard form before extracting direction ratios is the primary pitfall.

Memorize the expansion pattern of the scalar triple product and practice calculating vector cross products mentally or via quick determinant expansion to complete the calculation.

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Vector equation of a plane