NCERT Plus Two (Class 12) Mathematics • Chapter 11

Three Dimensional Geometry

Direction cosines & ratios, 3D line equations (vector & Cartesian), angle between lines & skew/parallel shortest distance.

📄 Official NCERT Textbook PDF

Topic 11.2: Direction Cosines (l² + m² + n² = 1) & Direction Ratios

Imagine aiming a laser drone from the corner of a Thrissur classroom toward a ceiling fan! On a flat notebook page, a single slope m = tan θ tells you where a line points. Out in 3D space, a line tilts against ALL THREE axes (X, Y, Z) at angles α, β, γ! Taking their cosines gives the line's normalized 3D GPS compass (l, m, n) — its Direction Cosines!
DIRECTION COSINES (l, m, n)l = cos α, m = cos β, n = cos γ
Identity: l² + m² + n² = 1 always!
DIRECTION RATIOS (a, b, c)Any numbers proportional to (l, m, n):
l = ±ar, m = ±br, n = ±cr, r = √(a²+b²+c²)
TWO-POINT & AXESLine PQ has DRs (x₂−x₁, y₂−y₁, z₂−z₁).
X-axis DCs: (1,0,0); Y: (0,1,0); Z: (0,0,1)
🧠 Memory Trick: A directed arrow has ONE unit vector lî + mĵ + nk̂, while a straight line runs BOTH ways — so converting DRs (a,b,c) to DCs of a line ALWAYS carries ±!
💡 Simple Kid-Check: Think of Direction Ratios (a, b, c) as a rough recipe like "walk 2 steps forward, −1 step sideways, −2 steps down". Divide each step by the total straight-line distance r = √(2² + (−1)² + (−2)²) = 3 to shrink it to length 1 — boom, you get the Direction Cosines (2/3, −1/3, −2/3)!
  • 1
    From Angles α, β, γ: Compute l = cos α, m = cos β, n = cos γ (must obey cos²α + cos²β + cos²γ = 1).
  • 2
    From DRs (a, b, c): Find r = √(a² + b² + c²) and divide: (±a/r, ±b/r, ±c/r).
  • 3
    Collinearity Check: Three points A, B, C lie on one line iff DRs of AB and BC are proportional!
Key Relations:  l² + m² + n² = 1  |  sin²α + sin²β + sin²γ = 2  |  Equal angles to axes: l = m = n = ±1√3
🧒 Super-Easy NCERT Examples:
• Angles 90°, 60°, 30°: l = cos 90° = 0, m = cos 60° = ½, n = cos 30° = √3/2. Check: 0² + (½)² + (√3/2)² = 0 + ¼ + ¾ = 1 ✅
• Equal-Angle Line: l = m = n ⇒ l² + l² + l² = 1 ⇒ 3l² = 1 ⇒ l = m = n = ±1/√3!
Watch THREE 1-mark DHSE traps! (1) Writing l + m + n = 1 — WRONG! Squares add to 1: l² + m² + n² = 1! (2) Claiming a line makes 30°, 30°, 30° with axes — impossible because cos²30° + cos²30° + cos²30° = 9/4 ≠ 1! (3) Forgetting ± when asked for DCs of a line with DRs (2, −1, −2) — remember ±(2/3, −1/3, −2/3)!

Topic 11.3.1: Vector & Cartesian Equations of a Line in Space

How do you guide a drone along a straight zip-line above Munnar's tea hills? You only need TWO ingredients: one anchor point A(x₁, y₁, z₁) on the line (position vector a) and one compass arrow b = aî + bĵ + ck̂ parallel to the line! Slide forward or backward by any number λ, and r = a + λb visits EVERY point on the line!
VECTOR EQUATIONr = a + λb  (λ ∈ ℝ)
a = fixed point, b = parallel direction vector
CARTESIAN EQUATIONx − x₁a = y − y₁b = z − z₁c
Point (x₁,y₁,z₁), DRs (a, b, c)
ZERO DR CASEIf b = 0 (say parallel to XZ-plane), just equate numerator to zero: y = y₁ (valid notation: (x−x₁)/a = (y−y₁)/0 = (z−z₁)/c)
🧠 Memory Trick: In Cartesian form, TOP (with flipped signs!) = Anchor Point and BOTTOM = Direction Arrow b!
💡 Simple Kid-Check: Picture standing at bus stop a and walking along straight road direction b. Take λ = 1 step forward → you land at a + b. Take λ = −2 steps backward → a − 2b. That's literally all r = a + λb says!
  • 1
    Write Vector Form: Put point (x₁,y₁,z₁) into a = x₁î + y₁ĵ + z₁k̂ and parallel vector b = aî + bĵ + ck̂.
  • 2
    Derive Cartesian Form: Let r = xî + yĵ + zk̂. Equate î, ĵ, k̂: x = x₁ + λa, y = y₁ + λb, z = z₁ + λc. Eliminate λ!
  • 3
    General Point on a Line: Set the Cartesian fractions equal to λ to get (x₁ + aλ, y₁ + bλ, z₁ + cλ) — super handy for finding intersections!
🧒 NCERT Example 6: Find vector & Cartesian equations of the line through (5, 2, −4) parallel to 3î + 2ĵ − 8k̂.
• a = 5î + 2ĵ − 4k̂,  b = 3î + 2ĵ − 8k̂.
• Vector Form: r = (5î + 2ĵ − 4k̂) + λ(3î + 2ĵ − 8k̂).
• Cartesian Form: x − 53 = y − 22 = z + 4−8.
Watch the SIGN FLIP in Cartesian form! In x + 32 = y − 54 = z + 62, the passing point is (−3, 5, −6), NOT (3, −5, 6)! Always rewrite x + 3 mentally as x − (−3)! Also never confuse a (point) with b (the vector multiplied by λ)!

Topic 11.3: Converting Vector ↔ Symmetric Cartesian Forms (& Two-Point Line)

Sometimes exam papers hand you a line in clean vector form, and other times in sneaky disguised Cartesian clothes like (2x − 1)/4 = (1 − y)/3 = 2z! Before reading the point or direction ratios, you MUST tailor the equation so x, y, z stand completely naked with coefficient +1!
CLEAN-UP RULEForce numerator coefficients of x, y, z to be +1! Divide both top & bottom of each fraction!
TWO-POINT LINEThrough A(a) & B(b):
r = a + λ(b − a) (DRs = B − A!)
DISGUISED EXAMPLES1 − x3 = x − 1−3  |  2x − 64 = x − 32
🧠 Memory Trick: "+1x, +1y, +1z FIRST, read denominators LAST!" Never read (a, b, c) until every numerator starts cleanly with x − ..., y − ..., z − ...!
💡 Simple Kid-Check: Imagine converting currency before adding prices! If one fraction says (7y − 14) / 2p, divide top and bottom by 7 to reveal (y − 2) / (2p/7). Only NOW is the denominator the true direction ratio!
  • 1
    Vector → Cartesian: Read a = (x₁, y₁, z₁) and b = (a, b, c) directly, then write (x−x₁)/a = (y−y₁)/b = (z−z₁)/c.
  • 2
    Cartesian → Vector: Normalize numerators to +1x, +1y, +1z. Read point a (flipped top signs) and direction b (cleaned bottoms), then write r = a + λb.
  • 3
    Non-Fraction Form like 6x − 2 = 3y + 1 = 2z − 2: Divide all three expressions by LCM/coefficients (divide by 6 here) to get (x − 1/3)/1 = (y + 1/3)/2 = (z − 1)/3!
🧒 Board Exam Converter Walkthrough:
Convert x + 32 = 1 − y5 = 2z − 46 into Vector Form.
• Fix 1st term: x − (−3)2  |  Fix 2nd (× −1 top/bottom): y − 1−5  |  Fix 3rd (÷ 2 top/bottom): z − 23.
• Anchor a = −3î + ĵ + 2k̂, Direction b = 2î − 5ĵ + 3k̂.
• Vector Form: r = (−3î + ĵ + 2k̂) + λ(2î − 5ĵ + 3k̂)!
THE #1 Kerala DHSE 2-mark trap! In (1 − x)/3 = (7y − 14)/2p = (z − 3)/2, reading DRs blindly as (3, 2p, 2) ruins the entire problem! You MUST pull −1 out of (1 − x) and 7 out of (7y − 14) so the true DRs are (−3, 2p/7, 2)! Also note: Two-Point line through A, B isn't a new formula — just set b = B − A!

Topic 11.4: Angle Between Two Lines & Perpendicular / Parallel Tests

Picture two airplanes flying along straight skyways in 3D space — even if one flies above the other without crossing, we find the angle between their compass arrows b₁ and b₂! Just pull out the Dot Product from Chapter 10: completely ignore the starting points a₁, a₂ and compute the acute angle between b₁ and b₂!
ACUTE ANGLE (VECTOR)cos θ = | b₁ · b₂|b₁| |b₂| |
Absolute value gives acute 0° ≤ θ ≤ 90°!
CARTESIAN & DC FORMcos θ = |a₁a₂ + b₁b₂ + c₁c₂√(a₁²+b₁²+c₁²)√(a₂²+b₂²+c₂²)|
Using DCs: cos θ = |l₁l₂ + m₁m₂ + n₁n₂|
PERPENDICULAR & PARALLEL⊥ (90°): a₁a₂ + b₁b₂ + c₁c₂ = 0
∥ (0°): a₁a₂ = b₁b₂ = c₁c₂
🧠 Memory Trick: Angle ONLY cares about b₁ and b₂! Starting points a₁ and a₂ NEVER enter the angle formula!
💡 Simple Kid-Check: Two lines crossing in space actually form two pairs of angles — one acute (say 60°) and one obtuse (120°). Wrapping the dot product in modulus bars |...| guarantees you always get the neat acute angle θ ≤ 90°! And when cos θ = 0, the lines stand at 90° (perpendicular)!
  • 1
    Extract Directions: From vector equations pick b₁, b₂ (with λ, μ); from Cartesian pick cleaned denominators (a₁,b₁,c₁), (a₂,b₂,c₂).
  • 2
    Dot Product & Lengths: Compute |b₁ · b₂| and magnitudes |b₁|, |b₂|.
  • 3
    Find θ or Test ⊥: For perpendicularity, just verify a₁a₂ + b₁b₂ + c₁c₂ = 0 (no square roots needed!).
🧒 NCERT Example 7 & Classic Perpendicular PYQ:
• Example 7: b₁ = 3î + 2ĵ + 6k̂ (|b₁| = √(9+4+36) = 7), b₂ = î + 2ĵ + 2k̂ (|b₂| = √(1+4+4) = 3).
  cos θ = |3(1)+2(2)+6(2)| / (7 × 3) = 19/21 ⇒ θ = cos⁻¹(19/21).
• Find p (Ex 11.2 Q10): 1−x3 = 7y−142p = z−32 ⊥ 7−7x3p = y−51 = 6−z5.
  Clean numerators → DRs₁ = (−3, 2p/7, 2), DRs₂ = (−3p/7, 1, −5). Set dot = 0: 9p/7 + 2p/7 − 10 = 0 ⇒ 11p/7 = 10 ⇒ p = 70/11!
Two instant-loss mistakes in Board Exams: (1) Taking the dot product of a₁ and a₂ instead of b₁ and b₂ — anchor points do NOT set direction! (2) Forgetting to flip signs in (1 − x)/3 or (6 − z)/5 before checking a₁a₂ + b₁b₂ + c₁c₂ = 0!

Topic 11.5.1: Shortest Distance Between Two Skew Lines

Picture a metro viaduct in Kochi running straight East-West 10 m up in the air, while a highway straight underneath runs North-South on the ground! Those two straight lines are neither parallel nor intersecting — they live in completely different planes! In 3D geometry, we call them Skew Lines, and their Shortest Distance is the unique bridge segment standing perpendicular to BOTH lines!
SKEW LINES MEANINGLines in 3D space that are neither parallel nor intersecting (non-coplanar!).
VECTOR SHORTEST DISTANCEd = | (b₁ × b₂) · (a₂ − a₁)|b₁ × b₂| |
INTERSECTION TESTTwo non-parallel lines intersect iff d = 0, i.e., (b₁ × b₂) · (a₂ − a₁) = 0!
🧠 Memory Trick: Cross the directions (b₁ × b₂) to get the shared elevator shaft, then project the connector (a₂ − a₁) onto it!
💡 Simple Kid-Check: How do you measure the vertical height between two skewed bridges? Find a unit vector pointing straight perpendicular to BOTH bridges: n̂ = (b₁ × b₂) / |b₁ × b₂|! Then connect ANY point a₁ on bridge 1 to ANY point a₂ on bridge 2, and take the shadow (dot projection) of (a₂ − a₁) along n̂!
  • 1
    Find Connector: Compute a₂ − a₁.
  • 2
    Find Mutual Perpendicular: Compute cross product b₁ × b₂ and its length |b₁ × b₂|.
  • 3
    Project & Absolute Value: Compute |(b₁ × b₂) · (a₂ − a₁)| / |b₁ × b₂|.
Cartesian Determinant Form:  Numerator = |det [x₂−x₁, y₂−y₁, z₂−z₁ ; a₁, b₁, c₁ ; a₂, b₂, c₂]|, Denominator = √[(b₁c₂−b₂c₁)² + (c₁a₂−c₂a₁)² + (a₁b₂−a₂b₁)²]. (Tip: converting Cartesian → Vector is much faster!)
🧒 NCERT Example 9 (Guaranteed 4-Mark Question!):
l₁: r = (î + ĵ) + λ(2î − ĵ + k̂),  l₂: r = (2î + ĵ − k̂) + μ(3î − 5ĵ + 2k̂).
1. a₂ − a₁ = (2−1)î + (1−1)ĵ + (−1−0)k̂ = î − k̂.
2. b₁ × b₂ = |î ĵ k̂; 2 −1 1; 3 −5 2| = 3î − ĵ − 7k̂, so |b₁ × b₂| = √(9 + 1 + 49) = √59.
3. Dot product: (3î − ĵ − 7k̂) · (î − k̂) = 3(1) + (−1)(0) + (−7)(−1) = 10.
4. Shortest Distance: d = 10√59 units!
TWO classic 4-mark pitfalls! (1) Dropping the outer absolute value bars — if the dot product comes out −10, distance MUST still be written positive as +10/√59! (2) Plugging parallel lines into the skew formula — when b₁ and b₂ are parallel, b₁ × b₂ = 0, giving an undefined 0/0! Always check if b₁ ∥ b₂ FIRST!

Topic 11.5.2: Distance Between Parallel Lines in 3D Space

Now imagine two straight railway tracks running side-by-side in Palakkad (b₁ = b₂ = b)! Since both tracks share the SAME direction b, b × b = 0 — so the skew-line formula breaks into 0/0! Instead, connect a₁ to a₂ to form a neat parallelogram with b: just divide Area |b × (a₂ − a₁)| by Base |b| to get the perpendicular height d!
PARALLEL CONDITIONr = a₁ + λb & r = a₂ + μb (or b₂ = kb₁ — watch for scaled vectors!)
PARALLEL DISTANCE FORMULAd = |b × (a₂ − a₁)||b|
(Cross product on top, NOT dot product!)
SKEW vs PARALLEL RULEParallel: |b × (a₂−a₁)| / |b|
Skew: |(b₁×b₂)·(a₂−a₁)| / |b₁×b₂|
🧠 Memory Trick: FIRST glance at b₁ & b₂ saves 4 marks! Proportional bs? Use the CROSS formula (Area ÷ Base). Different directions? Use the DOT skew formula!
💡 Simple Kid-Check: Remember Parallelogram Area = Base × Height! Make vector b your base (length |b|) and vector (a₂ − a₁) your slanted side. The parallelogram's area is |b × (a₂ − a₁)|, so the perpendicular height between the parallel tracks is simply Height = Area ÷ Base!
Line Pair TypeDirection TestFormula UsedIf Result = 0
Parallel Linesb₂ = kb₁|b × (a₂−a₁)| / |b|Coincident (same line!)
Skew / Intersectingb₁ × b₂ ≠ 0|(b₁×b₂)·(a₂−a₁)| / |b₁×b₂|Intersecting lines!
🧒 NCERT Example 10 (Classic Parallel Tracks):
l₁: r = (î + 2ĵ − 4k̂) + λ(2î + 3ĵ + 6k̂),  l₂: r = (3î + 3ĵ − 5k̂) + μ(2î + 3ĵ + 6k̂).
1. Notice b₁ = b₂ = b = 2î + 3ĵ + 6k̂ → Parallel!  |b| = √(4 + 9 + 36) = 7.
2. a₂ − a₁ = (3−1)î + (3−2)ĵ + (−5−(−4))k̂ = 2î + ĵ − k̂.
3. b × (a₂ − a₁) = |î ĵ k̂; 2 3 6; 2 1 −1| = −9î + 14ĵ − 4k̂.
4. Magnitude: √(81 + 196 + 16) = √293 ⇒ d = √2937 units!
THE sneakiest 4-mark DHSE trick! Suppose line 1 has b₁ = 2î + 3ĵ + 6k̂ and line 2 writes b₂ = 4î + 6ĵ + 12k̂! Because 4 = 2×2, 6 = 2×3, 12 = 2×6, they ARE parallel (just scaled by 2)! Students who jump straight into the skew formula get b₁ × b₂ = 0 and panic at 0/0! Absorb the 2 into μ so both lines use the same b, then apply the parallel formula!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March / Model PYQ • 1 Marks): If a line has direction ratios 2, −1, −2, determine its direction cosines.

Given DRs: a = 2, b = −1, c = −2.

r = √(a² + b² + c²) = √(2² + (−1)² + (−2)²) = √(4 + 1 + 4) = √9 = 3.

Direction cosines of the line are ±(23, −13, −23).

Q2 (Kerala HSE Say / Model PYQ • 1 Marks): Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and is parallel to the line x + 33 = y − 45 = z + 86.

A parallel line shares the exact same direction ratios (3, 5, 6).

Passing through (x₁, y₁, z₁) = (−2, 4, −5), the required Cartesian equation is:

x + 23 = y − 45 = z + 56.

Q3 (Kerala HSE PYQ • 2 Marks): The Cartesian equation of a line is x − 53 = y + 47 = z − 62. Write its vector form.

Rewrite as x − 53 = y − (−4)7 = z − 62.

Point: a = 5î − 4ĵ + 6k̂, Direction: b = 3î + 7ĵ + 2k̂.

Vector Equation: r = (5î − 4ĵ + 6k̂) + λ(3î + 7ĵ + 2k̂).

Q4 (NCERT / Kerala HSE PYQ • 2 Marks): Show that the three points A(2, 3, −4), B(1, −2, 3), and C(3, 8, −11) are collinear.

Direction ratios of AB: (1−2, −2−3, 3−(−4)) = (−1, −5, 7).

Direction ratios of BC: (3−1, 8−(−2), −11−3) = (2, 10, −14).

Notice (2, 10, −14) = −2 × (−1, −5, 7), so DRs are proportional and B is a common point.

Hence A, B, C are collinear.

Q5 (NCERT Ex 11.2 / HSE PYQ • 3 Marks): Find the value of p so that the lines 1 − x3 = 7y − 142p = z − 32 and 7 − 7x3p = y − 51 = 6 − z5 are at right angles.

Step 1 (Normalize numerators to +1x, +1y, +1z):

Line 1: x − 1−3 = y − 22p/7 = z − 32 → DRs₁ = (−3, 2p/7, 2).

Line 2: x − 1−3p/7 = y − 51 = z − 6−5 → DRs₂ = (−3p/7, 1, −5).

Step 2 (Apply a₁a₂ + b₁b₂ + c₁c₂ = 0):

(−3)(−3p/7) + (2p/7)(1) + (2)(−5) = 0 ⇒ 9p7 + 2p7 = 10 ⇒ 11p7 = 10 ⇒ p = 7011.

Q6 (Kerala HSE March PYQ • 3 Marks): Find the angle between the pair of lines x − 22 = y − 15 = z + 3−3 and x + 2−1 = y − 48 = z − 54.

DRs of first line: (2, 5, −3); DRs of second line: (−1, 8, 4).

a₁a₂ + b₁b₂ + c₁c₂ = 2(−1) + 5(8) + (−3)(4) = −2 + 40 − 12 = 26.

Magnitudes: √(4 + 25 + 9) = √38 and √(1 + 64 + 16) = √81 = 9.

cos θ = 269√38 ⇒ θ = cos⁻¹(269√38).

Q7 (Kerala HSE March PYQ (High Yield) • 4 Marks): Find the shortest distance between the lines r = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r = (2î − ĵ − k̂) + μ(2î + ĵ + 2k̂).

Here a₁ = î + 2ĵ + k̂, b₁ = î − ĵ + k̂; a₂ = 2î − ĵ − k̂, b₂ = 2î + ĵ + 2k̂.

1. a₂ − a₁ = î − 3ĵ − 2k̂.

2. b₁ × b₂ = |î ĵ k̂; 1 −1 1; 2 1 2| = −3î + 0ĵ + 3k̂, so |b₁ × b₂| = √(9 + 9) = 3√2.

3. (b₁ × b₂) · (a₂ − a₁) = (−3)(1) + 0(−3) + 3(−2) = −9.

4. d = |−93√2| = 3√2 = 3√22 units.

Q8 (Kerala HSE Model / NCERT PYQ • 4 Marks): Find the shortest distance between the lines x + 17 = y + 1−6 = z + 11 and x − 31 = y − 5−2 = z − 71.

Convert to vector quantities: a₁ = (−1, −1, −1), b₁ = (7, −6, 1); a₂ = (3, 5, 7), b₂ = (1, −2, 1).

1. a₂ − a₁ = (4, 6, 8).

2. b₁ × b₂ = |î ĵ k̂; 7 −6 1; 1 −2 1| = −4î − 6ĵ − 8k̂, |b₁ × b₂| = √(16+36+64) = √116 = 2√29.

3. (b₁ × b₂) · (a₂ − a₁) = (−4)(4) + (−6)(6) + (−8)(8) = −16 − 36 − 64 = −116.

4. d = |−1162√29| = 58√29 = 2√29 units!

Q9 (NCERT / Kerala HSE PYQ • 4 Marks): Find the distance between the parallel lines r = (î + 2ĵ − 4k̂) + λ(2î + 3ĵ + 6k̂) and r = (3î + 3ĵ − 5k̂) + μ(2î + 3ĵ + 6k̂).

Both lines share direction vector b = 2î + 3ĵ + 6k̂, so they are parallel with |b| = √(4 + 9 + 36) = 7.

a₂ − a₁ = 2î + ĵ − k̂.

b × (a₂ − a₁) = |î ĵ k̂; 2 3 6; 2 1 −1| = −9î + 14ĵ − 4k̂.

d = |b × (a₂ − a₁)||b| = √(81 + 196 + 16)7 = √2937 units.