NCERT Plus One (Class 11) Mathematics • Chapter 11

Introduction to Three Dimensional Geometry

Q: Name the octants in which the points (a) (−3, −1, 2) and (b) (2, −4, −7) lie.

• (a) (−, −) is Q3 and z > 0 → Octant III . • (b) (+, −) is Q4 and z < 0 (add 4) → Octant VIII .

Q: Find the coordinates of the foot of the perpendicular from P(4, −3, 5) on the ZX-plane and its distance from the ZX-plane.

• Foot on ZX-plane (set y = 0 ): (4, 0, 5) . • Perpendicular distance = |y| = |−3| = 3 units .

Q: Find the perpendicular distance of the point P(6, 7, 8) from (a) the XY-plane and (b) the Z-axis.

• (a) Distance from XY-plane = |z| = 8 units . • (b) Foot on Z-axis is (0, 0, 8) → Distance = √(6² + 7²) = √(36 + 49) = √85 units .

Q: Find the distance between the points P(2, −1, 3) and Q(−2, 1, 3).

PQ = √[(−2 − 2)² + (1 − (−1))² + (3 − 3)²] = √[(−4)² + 2² + 0²] = √(16 + 4) = √20 = 2√5 units .

Q: Show that the points P(−2, 3, 5), Q(1, 2, 3) and R(7, 0, −1) are collinear.

PQ = √[(1+2)² + (2−3)² + (3−5)²] = √(9+1+4) = √14 . QR = √[(7−1)² + (0−2)² + (−1−3)²] = √(36+4+16) = √56 = 2√14 . PR = √[(7+2)² + (0−3)² + (−1−5)²] = √(81+9+36) = √126 = 3√14 . Since PQ + QR = √14 + 2√14 = 3√14 = PR , the points are collinear .

Q: Verify that (0, 7, 10), (−1, 6, 6) and (−4, 9, 6) are the vertices of an isosceles right-angled triangle.

Let points be A, B, C: AB² = (−1)² + (−1)² + (−4)² = 1 + 1 + 16 = 18 BC² = (−3)² + 3² + 0² = 9 + 9 = 18 AC² = (−4)² + 2² + (−4)² = 16 + 4 + 16 = 36 . Since AB = BC = √18 (isosceles) AND AB² + BC² = 18 + 18 = 36 = AC² (right-angled at B), verified!

Q: Find the equation of the set of points which are equidistant from the points A(1, 2, 3) and B(3, 2, −1).

Let P(x, y, z) satisfy PA² = PB² : (x−1)² + (y−2)² + (z−3)² = (x−3)² + (y−2)² + (z+1)² Cancel (y−2)² : x²−2x+1 + z²−6z+9 = x²−6x+9 + z²+2z+1 4x − 8z = 0 ⇒ x − 2z = 0 .

Q: Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(−4, 0, 0) is equal to 10.

PA + PB = 10 ⇒ √[(x−4)²+y²+z²] = 10 − √[(x+4)²+y²+z²] . Squaring: (x−4)²+y²+z² = 100 + (x+4)²+y²+z² − 20√[(x+4)²+y²+z²] −16x − 100 = −20√[(x+4)²+y²+z²] ⇒ 4x + 25 = 5√[(x+4)²+y²+z²] . Squaring again: 16x² + 200x + 625 = 25(x² + 8x + 16 + y² + z²) → 9x² + 25y² + 25z² = 225 .

Three mutually perpendicular coordinate planes, eight 3D octants, coordinate signs, projections & 3D distance formula.

📄 Official NCERT Textbook PDF

Topic 11.2: Coordinate Axes & Mutually Perpendicular XY, YZ, ZX Planes

Look up at your room's ceiling fan! To tell a friend EXACTLY where the spinning blade tip hangs, measuring distance along the floor length (x) and floor width (y) isn't enough — you ALSO need the height above the floor (z)! Welcome to 3D Space, built from one floor and two corner walls meeting at right angles at the corner Origin O(0, 0, 0)!

3 AXESX'OX, Y'OY, Z'OZ meet at 90° at Origin O(0,0,0)

3 COORDINATE PLANESXY-plane: z = 0 | YZ-plane: x = 0
ZX-plane: y = 0

POINTS ON AXESX-axis: (x, 0, 0) | Y-axis: (0, y, 0)
Z-axis: (0, 0, z)

🧠 Memory Trick: Every plane sets the MISSING letter to ZERO! So XY-plane has z = 0, YZ-plane has x = 0, and ZX-plane has y = 0!

💡 Simple Kid-Check: Stand facing the front corner of a room: let the floor be the XY-plane, the left wall be the ZX-plane, and the back wall be the YZ-plane. Anything resting flat on the floor has height z = 0! Anything glued to the back wall (YZ-plane) has forward distance x = 0!

Location	Zero Coordinate(s)	General Point	Perpendicular Axis
XY-plane	`z = 0`	`(x, y, 0)`	Z-axis
YZ-plane	`x = 0`	`(0, y, z)`	X-axis
ZX-plane	`y = 0`	`(x, 0, z)`	Y-axis
X-axis / Y-axis / Z-axis	Two zeros!	`(x,0,0)`, `(0,y,0)`, `(0,0,z)`	Intersection of two planes

Classic 1-mark Kerala HSE trap: "What is the equation of the YZ-plane?" Students sometimes write y = 0, z = 0 — WRONG! That is the X-axis! The YZ-plane spans everywhere along y and z while keeping x = 0! Two zero coordinates make an axis; ONE zero coordinate makes a plane!

Topic 11.2: The Eight Octants & Quick Coordinate Sign Lookup

Think of a 2-storey house where each floor has 4 rooms meeting at a central pillar! The three coordinate planes slice all of 3D space into 8 rooms called Octants (Roman numerals I to VIII): 4 upstairs rooms where z > 0, and 4 matching basement rooms directly underneath where z < 0!

UPSTAIRS (z > 0)I: (+,+,+) | II: (−,+,+)
III: (−,−,+) | IV: (+,−,+)

BASEMENT (z < 0)V: (+,+,−) | VI: (−,+,−)
VII: (−,−,−) | VIII: (+,−,−)

INSTANT TRICKFind 2D Quadrant from (x, y)!
If z > 0 keep Q1–Q4; if z < 0 add +4!

🧠 Memory Trick: Check (x, y) just like Class 10 Quadrants 1, 2, 3, 4! Positive z → Octants I, II, III, IV. Negative z → just ADD 4 to get Octants V, VI, VII, VIII!

💡 Simple Kid-Check: Don't memorize 24 signs! Take any point like (−3, 1, −2). Step 1: Ignore z and look at (−3, 1) → negative x, positive y is Quadrant 2! Step 2: Now look at z = −2 (negative = basement!). So add 4: 2 + 4 = 6 → Octant VI! Done in 3 seconds!

Octant	I	II	III	IV	V	VI	VII	VIII
x	+	−	−	+	+	−	−	+
y	+	+	−	−	+	+	−	−
z	+	+	+	+	−	−	−	−
NCERT Name	`XOYZ`	`X'OYZ`	`X'OY'Z`	`XOY'Z`	`XOYZ'`	`X'OYZ'`	`X'OY'Z'`	`XOY'Z'`

Watch out for two traps! (1) Points with ANY zero coordinate — like (2, 0, −4) or (0, 3, 5) — lie ON a coordinate plane or axis, NOT inside any octant! (2) Always read the Roman numerals in order: I–IV cycle counterclockwise in XY with z > 0, and V–VIII repeat the EXACT same XY cycle with z < 0!

Topic 11.3: Locating Coordinates (x, y, z) & Projections onto Axes/Planes

To reach a mango hanging at P(3, 4, 5) in your garden from the corner peg O(0, 0, 0), walk 3 steps forward along X to L, turn right 4 steps parallel to Y to reach shadow M(3, 4, 0) on the ground, then climb 5 steps straight up to P! Dropping shadows like M onto walls and axes gives Projections & Perpendicular Distances!

FEET ON PLANESOn XY: (x, y, 0) | On YZ: (0, y, z)
On ZX: (x, 0, z) (zero out missing letter!)

DISTANCE FROM PLANESFrom YZ-plane: |x|
From ZX-plane: |y| | From XY-plane: |z|

DISTANCE FROM AXESFrom X-axis: √(y² + z²)
From Y-axis: √(x² + z²) | Z-axis: √(x² + y²)

🧠 Memory Trick: Projection keeps the named letters and zeroes the rest! Distance from a PLANE uses 1 missing letter (|x| from YZ-plane); distance from an AXIS uses Pythagoras of the OTHER 2 letters!

💡 Simple Kid-Check: Shine a flashlight straight down on point P(x, y, z)! Its shadow on the XY floor is (x, y, 0), and its height above the floor is |z|. Its shadow on the X-axis bar is (x, 0, 0) — so its slanted rope distance to the X-axis uses BOTH sideways y and vertical z: √(y² + z²)!

🪜 Quick Walkthrough for P(3, −4, 12):
• Foot of perpendicular on XY-plane = (3, −4, 0); distance from XY-plane = |12| = 12.
• Foot of perpendicular on Y-axis = (0, −4, 0); distance from Y-axis = √(3² + 12²) = √153.
• Distance from Z-axis = √(3² + (−4)²) = √(9 + 16) = 5!

THE #1 Kerala HSE 1-mark trap in Chapter 11: "Find the distance of P(3, 4, 5) from the X-axis." Students rush and answer 3! WRONG — 3 is the distance from the YZ-plane! Distance from the X-axis is the distance between (3, 4, 5) and (3, 0, 0), which is √(4² + 5²) = √41! Always read carefully: plane vs axis!

Topic 11.4: Distance Formula in 3D Space

Want to measure the tight diagonal wire stretched from one corner of a rectangular room P(x₁, y₁, z₁) to the opposite upper corner Q(x₂, y₂, z₂)? First use Pythagoras across the floor (Δx² + Δy²), then use Pythagoras a second time upward along height Δz — giving the sleek 3D Distance Formula!

3D DISTANCE FORMULAPQ = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]

FROM ORIGIN O(0,0,0)OP = √(x² + y² + z²)
(just add + z² to 2D Pythagoras!)

ORDER FREEDOM(x₂ − x₁)² = (x₁ − x₂)² — squaring wipes out signs, so PQ = QP!

🧠 Memory Trick: 3D Distance is literally 2D Distance with ONE extra room height term + (z₂ − z₁)² tucked under the same square-root roof!

💡 Simple Kid-Check: Inside any shoebox of length a, width b, and height c, the floor diagonal squared is a² + b², and the long 3D space diagonal squared is (a² + b²) + c²! That's all the 3D Distance Formula is: square the step in x, square the step in y, square the step in z, add them, and take the square root!

Distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]

📏 Easy Worked Example: Find distance between P(1, −3, 4) and Q(−4, 1, 2).
• Δx = −4 − 1 = −5 → (−5)² = 25
• Δy = 1 − (−3) = +4 → 4² = 16
• Δz = 2 − 4 = −2 → (−2)² = 4
• PQ = √(25 + 16 + 4) = √45 = 3√5 units!

Watch two silly sign blunders that lose marks: (1) Subtracting negative numbers: 1 − (−3) = 4, NOT −2! Always wrap negative coordinates in brackets when subtracting! (2) Squaring a negative number like (−5)² = +25, NEVER −25! Also remember to simplify surds cleanly at the end: √45 = 3√5, √50 = 5√2.

Topic 11.4: Proving Collinearity & Checking Right/Isosceles Triangles in 3D

Imagine three birds perched on electric poles A, B, and C in mid-air! How do you know whether they sit along one tight straight wire (Collinear) or form the corners of a triangle — maybe even a Right-Angled or Isosceles triangle — without ever climbing up with a protractor? Just measure the three lengths AB, BC, and AC!

COLLINEAR TESTSum of two smaller lengths = largest length!
E.g., AB + BC = AC → Straight line!

RIGHT TRIANGLECheck squares: s₁² + s₂² = s₃²
(Largest side s₃ is hypotenuse!)

ISOSCELES / PARALLELOGRAMIsosceles: any 2 sides equal.
Parallelogram: AB = CD, BC = DA & diagonals unequal (not rectangle).

🧠 Memory Trick: Collinearity adds lengths with square roots (√14 + 2√14 = 3√14)! Right-angle adds SQUARED lengths without roots (18 + 18 = 36)!

💡 Simple Kid-Check: Triangle inequality says the shortcut AC is ALWAYS shorter than going via B (AB + BC > AC) — UNLESS point B sits right on the straight wire between A and C, where AB + BC = AC!

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Collinearity Proof: Always simplify surds first! Example: P(−2, 3, 5), Q(1, 2, 3), R(7, 0, −1) → PQ = √14, QR = √56 = 2√14, PR = √126 = 3√14. Since PQ + QR = √14 + 2√14 = 3√14 = PR, they are collinear!
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Right / Isosceles Triangle: For A(0,7,10), B(−1,6,6), C(−4,9,6): AB² = 18, BC² = 18, AC² = 36. Two equal sides (AB = BC) AND AB² + BC² = AC² → Isosceles Right-Angled Triangle (right angle at B)!

NEVER claim points are NOT collinear just because √14 + √56 looks different from √126! You MUST factor out perfect squares (√56 = √(4×14) = 2√14, √126 = √(9×14) = 3√14) before adding! Also in 3-mark parallelogram proofs (AB = CD, BC = DA), if the question says "not a rectangle", you MUST also compute diagonals AC ≠ BD!

Topic 11.4: Equidistant Locus Equations in 3D Coordinates

Suppose two radio beacons sit at A(1, 2, 3) and B(3, 2, −1), and a drone flies so it ALWAYS stays equally far from both beacons (PA = PB). Where in 3D space can the drone fly? Squaring PA² = PB² magically cancels out x², y², z² and leaves a flat slicing wall — a Plane Equation!

EQUIDISTANT RULESet PA = PB ⇒ PA² = PB²
(Squaring kills square roots instantly!)

MAGIC CANCELLATIONx², y², z² cancel on BOTH sides,
leaving clean linear ax + by + cz + d = 0!

SPECIAL AXES / SUMSPoint on Z-axis: take P(0, 0, z)!
Sum PA + PB = 2a: square twice!

🧠 Memory Trick: Whenever an exam says "equidistant from A and B", IMMEDIATELY start with PA² = PB² (don't waste time writing giant square roots that you will immediately square away!).

💡 Simple Kid-Check: If two points A and B sit in space, all points equidistant from both form the flat "middle mirror wall" (perpendicular bisector plane) cutting halfway between A and B! And if the question restricts P to the Y-axis, just plug in P(0, y, 0) and solve for y!

🎯 NCERT Classic (Ex 11.2 Q4): Find the equation of the set of points equidistant from A(1, 2, 3) and B(3, 2, −1).
Let P(x, y, z) satisfy PA² = PB²:
(x − 1)² + (y − 2)² + (z − 3)² = (x − 3)² + (y − 2)² + (z + 1)²
Cancel (y − 2)² and expand:
(x² − 2x + 1) + (z² − 6z + 9) = (x² − 6x + 9) + (z² + 2z + 1)
Cancel x², z² and constants 10 → −2x − 6z = −6x + 2z ⇒ 4x − 8z = 0 ⇒ x − 2z = 0!

Watch out for NCERT Ex 11.2 Q5: "PA + PB = 10" (SUM of distances, NOT PA = PB!). Students mistakenly write PA² + PB² = 100 — WRONG! Since (a + b)² ≠ a² + b², you MUST isolate one root first: PA = 10 − PB, square both sides, simplify, isolate the remaining root, and square a SECOND time to get 9x² + 25y² + 25z² = 225!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Name the octants in which the points (a) (−3, −1, 2) and (b) (2, −4, −7) lie.

• (a) (−, −) is Q3 and z > 0 → Octant III.
• (b) (+, −) is Q4 and z < 0 (add 4) → Octant VIII.

Q2 (Kerala HSE Say • 1 Marks): Find the coordinates of the foot of the perpendicular from P(4, −3, 5) on the ZX-plane and its distance from the ZX-plane.

• Foot on ZX-plane (set y = 0): (4, 0, 5).
• Perpendicular distance = |y| = |−3| = 3 units.

Q3 (Kerala HSE March • 2 Marks): Find the perpendicular distance of the point P(6, 7, 8) from (a) the XY-plane and (b) the Z-axis.

• (a) Distance from XY-plane = |z| = 8 units.
• (b) Foot on Z-axis is (0, 0, 8) → Distance = √(6² + 7²) = √(36 + 49) = √85 units.

Q4 (Kerala HSE March • 2 Marks): Find the distance between the points P(2, −1, 3) and Q(−2, 1, 3).

PQ = √[(−2 − 2)² + (1 − (−1))² + (3 − 3)²]
= √[(−4)² + 2² + 0²] = √(16 + 4) = √20 = 2√5 units.

Q5 (Kerala HSE Say • 3 Marks): Show that the points P(−2, 3, 5), Q(1, 2, 3) and R(7, 0, −1) are collinear.

PQ = √[(1+2)² + (2−3)² + (3−5)²] = √(9+1+4) = √14.
QR = √[(7−1)² + (0−2)² + (−1−3)²] = √(36+4+16) = √56 = 2√14.
PR = √[(7+2)² + (0−3)² + (−1−5)²] = √(81+9+36) = √126 = 3√14.
Since PQ + QR = √14 + 2√14 = 3√14 = PR, the points are collinear.

Q6 (Kerala HSE March • 3 Marks): Verify that (0, 7, 10), (−1, 6, 6) and (−4, 9, 6) are the vertices of an isosceles right-angled triangle.

Let points be A, B, C:
AB² = (−1)² + (−1)² + (−4)² = 1 + 1 + 16 = 18
BC² = (−3)² + 3² + 0² = 9 + 9 = 18
AC² = (−4)² + 2² + (−4)² = 16 + 4 + 16 = 36.
Since AB = BC = √18 (isosceles) AND AB² + BC² = 18 + 18 = 36 = AC² (right-angled at B), verified!

Q7 (NCERT / DHSE Model • 3 Marks): Find the equation of the set of points which are equidistant from the points A(1, 2, 3) and B(3, 2, −1).

Let P(x, y, z) satisfy PA² = PB²:
(x−1)² + (y−2)² + (z−3)² = (x−3)² + (y−2)² + (z+1)²
Cancel (y−2)²: x²−2x+1 + z²−6z+9 = x²−6x+9 + z²+2z+1
4x − 8z = 0 ⇒ x − 2z = 0.

Q8 (NCERT / DHSE Model • 4 Marks): Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(−4, 0, 0) is equal to 10.

PA + PB = 10 ⇒ √[(x−4)²+y²+z²] = 10 − √[(x+4)²+y²+z²].
Squaring: (x−4)²+y²+z² = 100 + (x+4)²+y²+z² − 20√[(x+4)²+y²+z²]
−16x − 100 = −20√[(x+4)²+y²+z²] ⇒ 4x + 25 = 5√[(x+4)²+y²+z²].
Squaring again: 16x² + 200x + 625 = 25(x² + 8x + 16 + y² + z²)
→ 9x² + 25y² + 25z² = 225.