What is the section formula in 3D?

The coordinates of a point R that divides the line segment joining P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) in the ratio m:n internally are: R = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n)). For external division, replace n with −n.

What are the coordinates of the origin in 3D?

The origin in 3D space is O(0, 0, 0) — the point where all three coordinate axes intersect. The x-axis is the set of points (x, 0, 0), the y-axis is (0, y, 0), and the z-axis is (0, 0, z).

Introduction to Three Dimensional Geometry — NCERT Class 11 Maths

Q: What is the distance formula in three dimensions?

The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in 3D is d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²). This is a direct extension of the 2D distance formula, adding the z-coordinate difference.

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

This page provides complete, step-by-step NCERT solutions for every exercise in Chapter 11 — Introduction to Three Dimensional Geometry. Each solution applies the relevant 3D formula, substitutes the given values, and simplifies to arrive at the answer. Attempt each question on your own before consulting the solution.

Key Concepts (Quick Recap)

Coordinates in 3D: A point P is represented as (x, y, z)
Distance formula: d(P, Q) = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²)
Distance from origin: d = √(x² + y² + z²)
Section formula (internal, ratio m:n): R = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n))
Midpoint formula: M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
Centroid of triangle: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3)

Exercise 11.1

Q1. Locate the following points in 3D space: (1, 2, 3), (−1, 0, 4), (0, 0, 5).

Solution

(1, 2, 3): 1 unit along x-axis, 2 units along y-axis, 3 units along z-axis — in the first octant.
(−1, 0, 4): On the xz-plane (since y = 0), 1 unit in the negative x-direction and 4 units along z.
(0, 0, 5): On the z-axis itself, 5 units from the origin.

Q2. Name the octant in which the point (−5, 2, −7) lies.

Solution

x < 0, y > 0, z < 0. This is the octant where x and z are negative and y is positive. In the standard numbering, this is the second octant (or "x-negative, y-positive, z-negative" octant).

Q3. The x-coordinate of a point is positive and the y-coordinate is zero. What can you say about the z-coordinate to determine the octant?

Solution

If x > 0 and y = 0, the point lies on the xz-plane. If z > 0, the point is in the first octant region of the xz-plane. If z < 0, it lies in the fifth octant region of the xz-plane.

Q4. Name the octants in which the following points lie: (4, 2, −5), (−4, −2, 7), (4, −2, −5), (−4, 2, −5).

Solution

(4, 2, −5): x > 0, y > 0, z < 0 → Fifth octant
(−4, −2, 7): x < 0, y < 0, z > 0 → Third octant
(4, −2, −5): x > 0, y < 0, z < 0 → Sixth octant
(−4, 2, −5): x < 0, y > 0, z < 0 → Second octant

Exercise 11.2

Q1. Find the distance between the points P(1, −3, 4) and Q(−4, 1, −2).

Solution

Using d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²):

d = √((−4−1)² + (1−(−3))² + (−2−4)²)

d = √((−5)² + (4)² + (−6)²)

d = √(25 + 16 + 36)

d = √77

Distance = √77

Q2. Show that the points A(0, 7, 10), B(−1, 6, 6) and C(−4, 9, 6) form an isosceles right-angled triangle.

Solution

AB = √((−1−0)² + (6−7)² + (6−10)²) = √(1 + 1 + 16) = √18 = 3√2

BC = √((−4−(−1))² + (9−6)² + (6−6)²) = √(9 + 9 + 0) = √18 = 3√2

AC = √((−4−0)² + (9−7)² + (6−10)²) = √(16 + 4 + 16) = √36 = 6

AB = BC = 3√2 (isosceles) and AB² + BC² = 18 + 18 = 36 = AC² (right-angled at B).

The triangle is isosceles right-angled.

Q3. Show that the points A(1, 2, 3), B(−1, −1, −1) and C(3, 5, 7) are collinear.

Solution

AB = √((−1−1)² + (−1−2)² + (−1−3)²) = √(4 + 9 + 16) = √29

BC = √((3−(−1))² + (5−(−1))² + (7−(−1))²) = √(16 + 36 + 64) = √116 = 2√29

AC = √((3−1)² + (5−2)² + (7−3)²) = √(4 + 9 + 16) = √29

Wait — let me recheck: AB = √29, BC = 2√29, AC = √29.

AC ≠ AB + BC, so they are not collinear. Let me recompute AC.

AC = √((3−1)² + (5−2)² + (7−3)²) = √(4 + 9 + 16) = √29.

But AB + BC = √29 + 2√29 = 3√29 ≠ √29.

Actually, AC should be the full distance from A to C. Let me recheck: AB = √29, AC = √29... that means A and C are the same distance from B, but that would make B the midpoint.

Check: Midpoint of AC = ((1+3)/2, (2+5)/2, (3+7)/2) = (2, 3.5, 5) ≠ B(−1, −1, −1).

So they are not collinear. Let me verify by direction ratios:

AB direction ratios: (−2, −3, −4)

AC direction ratios: (2, 3, 4)

These are proportional (AC = −1 × AB), so A, B, C are collinear since AB and AC have proportional direction ratios.

AB + BC = √29 + 2√29 = 3√29, and AC should be 3√29... Let me recompute AC.

AC = √(4 + 9 + 16) = √29. Hmm, same as AB. But direction ratios show they are collinear.

Actually, since AC ratios = (2, 3, 4) and AB ratios = (−2, −3, −4), B lies on the extension of AC beyond A. So AB + BC should equal AC... but B is between A and C only if direction is same.

Since AB ratios are (−2, −3, −4) and AC ratios are (2, 3, 4), B is on the opposite side. So BC = AB + AC = √29 + √29... Let me recompute BC.

BC = √((3+1)² + (5+1)² + (7+1)²) = √(16 + 36 + 64) = √116 = 2√29.

AB = √29, AC = √29, BC = 2√29 = AB + AC. Collinear. B is between A and C on the line, but on the opposite side of A. Actually BC = AB + AC means B is not between A and C. The collinearity is confirmed by proportional direction ratios.

Proved. Direction ratios AB = (−2, −3, −4) and AC = (2, 3, 4) are proportional, so the points are collinear.

Q4. Find the distance of the point (2, 3, −5) from the origin.

Solution

d = √(2² + 3² + (−5)²) = √(4 + 9 + 25) = √38

Distance = √38

Q5. Verify whether the points (5, −2, 3), (6, 4, 5) and (8, 2, 1) form a right-angled triangle.

Solution

Let A(5, −2, 3), B(6, 4, 5), C(8, 2, 1).

AB² = 1 + 36 + 4 = 41

BC² = 4 + 4 + 16 = 24

AC² = 9 + 16 + 4 = 29

Check: AB² = 41, BC² + AC² = 24 + 29 = 53 ≠ 41.

AB² + BC² = 41 + 24 = 65, AC² + BC² = 29 + 24 = 53, AB² + AC² = 41 + 29 = 70.

None sum to give the third. The points do not form a right-angled triangle.

Exercise 11.3

Q1. Find the coordinates of the point which divides the line segment joining (1, −2, 3) and (3, 4, −5) in the ratio 2 : 3 internally.

Solution

Using section formula: R = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n))

m = 2, n = 3.

x = (2×3 + 3×1)/(2+3) = (6 + 3)/5 = 9/5

y = (2×4 + 3×(−2))/5 = (8 − 6)/5 = 2/5

z = (2×(−5) + 3×3)/5 = (−10 + 9)/5 = −1/5

R = (9/5, 2/5, −1/5)

Q2. Find the coordinates of the point which divides the line segment joining (−2, 3, 5) and (1, −4, 3) in the ratio 2 : 3 externally.

Solution

For external division in ratio m:n, replace n with −n:

R = ((2×1 + (−3)×(−2))/(2−3), (2×(−4) + (−3)×3)/(2−3), (2×3 + (−3)×5)/(2−3))

x = (2 + 6)/(−1) = −8

y = (−8 − 9)/(−1) = 17

z = (6 − 15)/(−1) = 9

R = (−8, 17, 9)

Q3. Find the coordinates of the midpoint of the line segment joining (2, −3, 5) and (4, 7, −9).

Solution

M = ((2+4)/2, (−3+7)/2, (5−9)/2) = (6/2, 4/2, −4/2) = (3, 2, −2)

Q4. Find the ratio in which the line segment joining (4, 2, −3) and (2, −3, 5) is divided by the yz-plane.

Solution

On the yz-plane, x = 0. Let the ratio be k : 1.

0 = (k×2 + 1×4)/(k + 1)

2k + 4 = 0 → k = −2

The ratio is −2 : 1, meaning 2 : 1 externally on the side of (4, 2, −3).

Q5. Find the ratio in which the line segment joining (2, 4, 5) and (3, 5, −4) is divided by the xy-plane.

Solution

On the xy-plane, z = 0. Let the ratio be k : 1.

0 = (k×(−4) + 1×5)/(k + 1)

−4k + 5 = 0 → k = 5/4

The ratio is 5 : 4 internally.

Q6. Find the coordinates of the point which divides the line segment joining (1, −1, 2) and (3, 3, 4) in the ratio 1 : 3.

Solution

R = ((1×3 + 3×1)/4, (1×3 + 3×(−1))/4, (1×4 + 3×2)/4)

= (6/4, 0/4, 10/4)

= (3/2, 0, 5/2)

Q7. The centroid of a triangle with vertices (2, 3, −4), (−3, 2, 1) and (5, −4, 3) is:

Solution

G = ((2−3+5)/3, (3+2−4)/3, (−4+1+3)/3) = (4/3, 1/3, 0/3) = (4/3, 1/3, 0)

Q8. A point R with x-coordinate 4 lies on the line segment joining P(2, −3, 4) and Q(8, 0, 10). Find the coordinates of R.

Solution

Let R divide PQ in ratio k : 1.

4 = (k×8 + 1×2)/(k + 1)

4k + 4 = 8k + 2 → 4k − 2 = 0 → k = 1/2

Ratio = 1 : 2.

y = (1×0 + 2×(−3))/3 = −6/3 = −2

z = (1×10 + 2×4)/3 = 18/3 = 6

R = (4, −2, 6)

Q9. Find the point on the y-axis which is equidistant from A(3, 1, 2) and B(5, 5, 2).

Solution

Point on y-axis: P(0, y, 0).

PA² = 9 + (y−1)² + 4 = (y−1)² + 13

PB² = 25 + (y−5)² + 4 = (y−5)² + 29

PA² = PB² → (y−1)² + 13 = (y−5)² + 29

y² − 2y + 1 + 13 = y² − 10y + 25 + 29

−2y + 14 = −10y + 54

8y = 40 → y = 5

P = (0, 5, 0)

Miscellaneous Exercise

Q1. Three vertices of a parallelogram are A(3, −1, 2), B(1, 2, −4) and C(−1, 1, 2). Find the fourth vertex D.

Solution

The diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD.

Midpoint of AC = ((3−1)/2, (−1+1)/2, (2+2)/2) = (1, 0, 2)

Let D = (x, y, z). Midpoint of BD = ((1+x)/2, (2+y)/2, (−4+z)/2) = (1, 0, 2)

1+x = 2 → x = 1; 2+y = 0 → y = −2; −4+z = 4 → z = 8.

D = (1, −2, 8)

Q2. Find the lengths of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0).

Solution

Median from A: midpoint of BC = (3, 2, 0). Length = √(9 + 4 + 36) = 7

Median from B: midpoint of AC = (3, 0, 3). Length = √(9 + 16 + 9) = √34 = √34

Median from C: midpoint of AB = (0, 2, 3). Length = √(36 + 4 + 9) = √49 = 7

Q3. If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(−4, 3b, −10) and R(8, 14, 2c), find the values of a, b, c.

Solution

Centroid = ((2a − 4 + 8)/3, (2 + 3b + 14)/3, (6 − 10 + 2c)/3) = (0, 0, 0)

2a + 4 = 0 → a = −2

3b + 16 = 0 → b = −16/3

2c − 4 = 0 → c = 2

Q4. Find the coordinates of the points which trisect the line segment joining (4, −1, −2) and (−2, 3, 4).

Solution

First trisection point (ratio 1:2):

P₁ = ((1×(−2) + 2×4)/3, (1×3 + 2×(−1))/3, (1×4 + 2×(−2))/3)

= (6/3, 1/3, 0/3) = (2, 1/3, 0)

Second trisection point (ratio 2:1):

P₂ = ((2×(−2) + 1×4)/3, (2×3 + 1×(−1))/3, (2×4 + 1×(−2))/3)

= (0/3, 5/3, 6/3) = (0, 5/3, 2)

Internal Links

[Chapter 11 Guide — Introduction to Three Dimensional Geometry](/school/cbse/class/11/maths/cbse-introduction-to-3d-geometry)
[All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
[Class 11 Maths Subject Guide](/school/cbse/class/11/maths)

Frequently Asked Questions

The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in 3D is d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²). This is a direct extension of the 2D distance formula, adding the z-coordinate difference.