Introduction to Three-dimensional Geometry ISC Class 11

What Introduction to Three-dimensional Geometry is about

Weightage: Introduction to Three-dimensional Geometry belongs to the unit Coordinate Geometry, which CISCE weights at 20 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Axes, coordinate planes and octants

Take three mutually perpendicular lines XOX′, YOY′, ZOZ′ through O, arranged as a right-handed system: if the fingers of the right hand curl from the positive x-axis to the positive y-axis, the thumb points along the positive z-axis. The xy-plane contains the x- and y-axes (its equation is z = 0); similarly the yz-plane is x = 0 and the zx-plane is y = 0. The planes divide space into eight octants. A common numbering is I (+, +, +), II (−, +, +), III (−, −, +), IV (+, −, +), V (+, +, −), VI (−, +, −), VII (−, −, −), VIII (+, −, −), giving the signs of (x, y, z).

Worked example

State the octants (with the numbering above) of (−2, 3, −5) and (4, −1, 6).

Try it first, then show the solutionSolution

1. (−2, 3, −5): signs (−, +, −): octant VI
2. (4, −1, 6): signs (+, −, +): octant IV

Answer: VI; IV

Coordinates of a point and special positions

Through P draw planes parallel to the coordinate planes, meeting the axes at A, B, C. If OA = x, OB = y, OC = z (with signs), then P = (x, y, z). So x is the signed distance of P from the yz-plane, y from the zx-plane and z from the xy-plane. Points on the x-axis are (x, 0, 0), on the y-axis (0, y, 0), on the z-axis (0, 0, z); points in the xy-plane are (x, y, 0), and so on. The foot of the perpendicular from P to the xy-plane is (x, y, 0), and to the x-axis is (x, 0, 0). The distance of P from the x-axis is √(y² + z²).

Worked example

For P(3, −4, 5), find the feet of the perpendiculars to the xy-plane and to the x-axis, and the distances of P from the yz-plane and from the x-axis.

Try it first, then show the solutionSolution

1. Foot on the xy-plane: (3, −4, 0); on the x-axis: (3, 0, 0)
2. Distance from the yz-plane = |x| = 3
3. Distance from the x-axis = √(16 + 25) = √41

Answer: (3, −4, 0); (3, 0, 0); 3; √41

Distance formula

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). Draw the cuboid with PQ as a diagonal and edges parallel to the axes; its edges are |x₂ − x₁|, |y₂ − y₁|, |z₂ − z₁|. Applying Pythagoras twice: PQ = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²). In particular, the distance of P from the origin is √(x² + y² + z²). The order of the points does not matter because the differences are squared.

Worked example

Find the distance between A(1, −2, 3) and B(4, 2, 15).

Try it first, then show the solutionSolution

1. Differences 3, 4, 12
2. AB = √(9 + 16 + 144) = √169

Answer: 13

Using distances: collinearity and shapes

Three points A, B, C are collinear if the largest of AB, BC, CA equals the sum of the other two. A triangle is right-angled if the squares of two sides add to the square of the third, isosceles if two sides are equal, equilateral if all three are. For quadrilaterals, compare sides and diagonals (or midpoints of diagonals). The locus of a point satisfying a distance condition is found by writing the condition in x, y, z.

Worked example

Show that A(2, 1, −1), B(4, 3, 0) and C(8, 7, 2) are collinear.

Try it first, then show the solutionSolution

1. AB = √(4 + 4 + 1) = 3
2. BC = √(16 + 16 + 4) = 6; AC = √(36 + 36 + 9) = 9
3. AB + BC = 9 = AC

Answer: Collinear, with B between A and C

Section and midpoint formulae

The point R dividing the join of P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) internally in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n)). For external division replace + by − in both numerator and denominator. The midpoint (m = n) is ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2). The centroid of a triangle is the average of its three vertices.

Worked example

For A(2, −3, 4) and B(8, 0, 10), find the points dividing AB in the ratio 1 : 2 internally and externally, and the midpoint.

Try it first, then show the solutionSolution

1. Internal: ((8 + 4)/3, (0 − 6)/3, (10 + 8)/3) = (4, −2, 6)
2. External: ((8 − 4)/(1 − 2), (0 + 6)/(1 − 2), (10 − 8)/(1 − 2)) = (−4, −6, −2)
3. Midpoint: (5, −1.5, 7)

Answer: (4, −2, 6); (−4, −6, −2); (5, −3/2, 7)

Where a segment meets a coordinate plane

To find the ratio in which a coordinate plane divides the join of two points, let the ratio be k : 1 and write the general dividing point. On the yz-plane its x-coordinate is 0; on the zx-plane y = 0; on the xy-plane z = 0. Solve for k. A positive k means internal division; a negative k means external division (the plane meets the line extended beyond the segment). Substitute k back to get the point.

Worked example

Find the ratio in which the yz-plane divides the join of (−4, 5, 2) and (2, −1, 8), and the point of division.

Try it first, then show the solutionSolution

1. x-coordinate of the dividing point: (2k − 4)/(k + 1) = 0 ⇒ k = 2, i.e. ratio 2 : 1 internally
2. y = (−k + 5)/(k + 1) = 3/3 = 1; z = (8k + 2)/(k + 1) = 18/3 = 6

Answer: 2 : 1 internally; (0, 1, 6)

Maths in real life: Introduction to Three-dimensional Geometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Coordinates, Octants and the Distance Formula in Space
• Section Formula, Midpoint and Centroid in Space

Introduction to Three-dimensional Geometry formula sheet

Distance between two points

$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

From origin: √(x² + y² + z²).

Internal division

$\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)$

Ratio m : n from P towards Q.

External division

$\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n},\frac{m{z}_2-n{z}_1}{m-n}\right)$

m ≠ n.

Midpoint

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Special case m = n.

Centroid of a triangle

$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$

Divides each median 2 : 1.

Distance from an axis

$\text{from }x\text{-axis}=\sqrt{y^2+z^2}$

From the xy-plane: |z|.

Introduction to Three-dimensional Geometry: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Introduction to Three-dimensional Geometry

The question patterns to expect, and the method that works for each.

Tricks to remember Introduction to Three-dimensional Geometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Same formula, one more term

  Every 2-D formula (distance, section, midpoint, centroid) gains a z-part that looks exactly like the x- and y-parts.

  Use it when: Recalling formulas.

• Missing letter is zero

  On the xy-plane, z is missing, so z = 0; on the x-axis, only x survives.

  Use it when: Points on planes and axes.

• Distance from an axis skips that axis

  From the x-axis: √(y² + z²), leave out x.

  Use it when: Distances from axes.

• Negative k means outside

  k : 1 negative ⇒ external division.

  Use it when: Ratio of division by a plane.

• Octant count by signs

  Two choices of sign for each of three coordinates: 2³ = 8 octants.

  Use it when: Octants.

Introduction to Three-dimensional Geometry: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Introduction to Three-dimensional Geometry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Distance from an axis confused with distance from a plane

  Both involve 'dropping a perpendicular'.

  Wrong working

  Distance of (3, −4, 5) from the x-axis = 3.

  Right working

  From the x-axis: √(16 + 25) = √41; 3 is the distance from the yz-plane.

  ✓Do this: Axis: leave out that coordinate inside a square root; plane: take the missing coordinate's size.

  Usually costs: 1 mark.

• ✗Mistake: Dropping one root

  Students take only the positive square root.

  Wrong working

  (x − 2)² = 16 ⇒ x = 6.

  Right working

  x − 2 = ±4 ⇒ x = 6 or −2.

  ✓Do this: Write ± whenever you take a square root.

  Usually costs: 1 mark.

• ✗Mistake: Sign errors in external division

  Students keep + signs.

  Wrong working

  External 1 : 2 of (2, −3, 4), (8, 0, 10): (4, −2, 6).

  Right working

  (−4, −6, −2).

  ✓Do this: External: minus in numerator and denominator.

  Usually costs: 1-2 marks.

• ✗Mistake: Setting the wrong coordinate to zero

  The yz-plane is x = 0, not z = 0.

  Wrong working

  yz-plane divides AB: set z = 0.

  Right working

  yz-plane is x = 0.

  ✓Do this: The plane is named by the two coordinates that remain free; the third is zero.

  Usually costs: Full marks for the part.

• ✗Mistake: Proving collinearity by one equal pair

  Students compute only two distances.

  Wrong working

  AB = 3 and BC = 6, so collinear.

  Right working

  Also need AC = 9 = AB + BC.

  ✓Do this: Compute all three distances and check the sum.

  Usually costs: 1 mark.

Practice questions on Introduction to Three-dimensional Geometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The distance of (3, −4, 5) from the xy-plane is:
   (a) 3
   (b) 4
   (c) 5
   (d) √50

   Hint

   |z|.

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   With the octants numbered I (+, +, +), II (−, +, +), III (−, −, +), IV (+, −, +), the point (−1, −2, 3) lies in octant:
   (a) I
   (b) II
   (c) III
   (d) IV

   Hint

   Signs (−, −, +).

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Write the foot of the perpendicular from (4, −3, 7) to the yz-plane.

   Hint

   x = 0.

   Answer

   (0, −3, 7)

4. Q4 · 1 mark · Very short answerExam level

   Find the distance of (6, 2, −3) from the origin.

   Hint

   √(36 + 4 + 9).

   Answer

   7

5. Q5 · 2 marks · Short answerExam level

   Find the midpoint of the join of (2, 5, −1) and (−4, 3, 7).

   Hint

   Average each coordinate.

   Answer

   (−1, 4, 3)

6. Q6 · 2 marks · Short answerHarder exam level

   Find the distances of (3, −4, 12) from the x-axis and from the z-axis.

   Hint

   Leave out that axis's coordinate.

   Answer

   x-axis: √(16 + 144) = 4√10; z-axis: √(9 + 16) = 5

7. Q7 · 2 marks · Short answerHarder exam level

   Find the point on the y-axis equidistant from (3, 1, 2) and (5, 5, 2).

   Hint

   Let it be (0, y, 0).

   Answer

   9 + (y − 1)² + 4 = 25 + (y − 5)² + 4 ⇒ 8y = 40 ⇒ y = 5: (0, 5, 0)

8. Q8 · 2 marks · Short answerHarder exam level

   Find the centroid of the triangle with vertices (1, −1, 2), (4, 3, 0) and (−2, 1, 7).

   Hint

   Average.

   Answer

   (1, 1, 3)

9. Q9 · 3 marks · Short answerHarder exam level

   The centroid of the triangle with vertices (a, 1, 3), (−2, b, −5) and (4, 7, c) is the origin. Find a, b, c.

   Hint

   Each coordinate sum is 0.

   Answer

   a − 2 + 4 = 0 ⇒ a = −2; 1 + b + 7 = 0 ⇒ b = −8; 3 − 5 + c = 0 ⇒ c = 2

10. Q10 · 2 marks · Short answerHarder exam level

    Find the point dividing the join of (−1, 2, 4) and (5, −4, 1) externally in the ratio 2 : 1.

    Hint

    Denominator m − n = 1.

    Answer

    ((10 + 1)/1, (−8 − 2)/1, (2 − 4)/1) = (11, −10, −2)

11. Q11 · 3 marks · Short answerHarder exam level

    Show that (−1, 4, 2), (1, 3, 5) and (5, 1, 11) are collinear.

    Hint

    Find all three distances.

    Answer

    AB = √(4 + 1 + 9) = √14, BC = √(16 + 4 + 36) = 2√14, AC = √(36 + 9 + 81) = 3√14 = AB + BC, so collinear.

12. Q12 · 3 marks · Short answerHarder exam level

    Find the ratio in which the xy-plane divides the join of (1, 2, −3) and (4, −1, 6), and the point of division.

    Hint

    Set z = 0.

    Answer

    (6k − 3)/(k + 1) = 0 ⇒ k = 1/2, i.e. 1 : 2 internally. Point ((4k + 1)/(k + 1), (−k + 2)/(k + 1), 0) = (2, 1, 0).

13. Q13 · 5 marks · Long answerHarder exam level

    A triangle has vertices A(1, 0, 2), B(3, 4, 6), C(5, −2, 4).
    (a) Find the length of the median from A. [2]
    (b) Find the centroid G. [1]
    (c) Show that G divides this median in the ratio 2 : 1. [2]

    Hint

    Midpoint of BC is D(4, 1, 5).

    Answer

    (a) AD = √(9 + 1 + 9) = √19. (b) G = (3, 2/3, 4). (c) The point dividing AD in 2 : 1 is ((8 + 1)/3, (2 + 0)/3, (10 + 2)/3) = (3, 2/3, 4) = G.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the equation of the set of points P for which the sum of the squares of the distances from (1, 0, 0) and (−1, 0, 0) is 10. Describe the set.

    Hint

    Expand both squared distances.

    Answer

    (x − 1)² + y² + z² + (x + 1)² + y² + z² = 10 ⇒ 2x² + 2y² + 2z² + 2 = 10 ⇒ x² + y² + z² = 4: all points 2 units from the origin (a sphere of radius 2).

15. Q15 · 3 marks · Short answerHOTS / competency

    A(1, 1, 1), B(1, 4, 5) and C(5, 4, 1) form a triangle. Show that it is isosceles and find the length of the bisector of angle A from A to BC.

    Hint

    The bisector of the vertex angle of an isosceles triangle meets the base at its midpoint.

    Answer

    AB = √(0 + 9 + 16) = 5, AC = √(16 + 9 + 0) = 5, so isosceles. The bisector meets BC at its midpoint (3, 4, 3); length √(4 + 9 + 4) = √17.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Equation of the xy-plane?Show answer

   z = 0.

2. 2.General point on the z-axis?Show answer

   (0, 0, z).

3. 3.How many octants are there?Show answer

   Eight.

4. 4.Distance between two points in space?Show answer

   √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²).

5. 5.Distance of (x, y, z) from the x-axis?Show answer

   √(y² + z²).

6. 6.Internal section formula?Show answer

   ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n)).

7. 7.Centroid of a triangle?Show answer

   Average of the three vertices.

8. 8.Condition for the yz-plane in a division problem?Show answer

   x-coordinate of the dividing point is 0.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name the octant of any point from its signs?
• ☐Can you write the foot of the perpendicular from a point to each coordinate plane?
• ☐Can you find the distance of a point from each axis?
• ☐Can you find an unknown coordinate from a given distance (both values)?
• ☐Can you prove three points collinear using distances?
• ☐Can you divide a segment internally and externally?
• ☐Can you find where a segment crosses a coordinate plane?
• ☐Can you find the fourth vertex of a parallelogram?

Introduction to Three-dimensional Geometry: frequently asked questions