MathsCBSE Class 11 · Introduction to Three Dimensional Geometry

Distance Formula in Three Dimensions: Collinearity, Shapes and Loci

Updated 17 October 2026 · Fact-checked

The distance between P(x1, y1, z1) and Q(x2, y2, z2) is √((x2 − x1)² + (y2 − y1)² + (z2 − z1)²). Three points are collinear if the largest of the three distances equals the sum of the other two. A triangle is right-angled if the squares of two sides add up to the square of the third. A locus is found by writing the distance condition for P(x, y, z) and simplifying: equidistance from two points gives a plane.

Apply Pythagoras on the base to get d², then again with dz to get PQ², so PQ² is the sum of the three squared differences.

Illustration drawn for QuestionBuddy.

Understand Distance Formula in Space: Collinearity, Shapes and Loci

The distance formula. For $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ ,
$ $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2} .$ $
It follows from applying Pythagoras twice inside a box with $P Q$ as space diagonal. Always compare squared distances to avoid roots until the end.

Collinearity. Points $A, B, C$ are collinear if one of $A B + B C = A C$ , $A B + A C = B C$ , $B C + C A = A B$ holds. Compute the three distances, identify the largest and check. (Do not use the section formula in space: it is formative only in 2026-27.)

Shapes.

Equilateral triangle: all three sides equal. Isosceles: two equal sides.
Right-angled triangle: $a^{2} + b^{2} = c^{2}$ for the three squared sides (the largest square is the sum of the other two).
Square: four equal sides and two equal diagonals. Rhombus: four equal sides. Rectangle: opposite sides equal and equal diagonals. Parallelogram: both pairs of opposite sides equal (but check the points are not collinear).

For a quadrilateral $A B C D$ , first list the sides $A B, B C, C D, D A$ , then the diagonals $A C, B D$ .

Loci. Take a general point $P (x, y, z)$ and translate the words into an equation of squared distances.

$P A = P B$ : the two squared distances are equal; the quadratic terms cancel, leaving a linear equation, a plane (the perpendicular bisector plane of $A B$ ).
$P A = k P B$ with $k \neq = 1$ : the quadratic terms stay, giving a sphere (it can be rewritten as $(x - h)^{2} + y^{2} + z^{2} = ρ^{2}$ after completing the square).
$P A^{2} + P B^{2} = constant$ : also a sphere.

**Equations such as $x^{2} + y^{2} + z^{2} = r^{2}$ ** describe all points at distance $r$ from the origin. Always check one or two points on your locus.

Distance Formula in Space: Collinearity, Shapes and Loci: formulas to remember

Distance: $P Q = \sqrt ((x 2 - x 1)^{2} + (y 2 - y 1)^{2} + (z 2 - z 1)^{2})$; Space diagonal of the box with P and Q as opposite corners.
Collinear points: $A B + B C = A C (w i t h A C t h e l a r g es t)$; Test all three distances.
Right angle: $A B^{2} + B C^{2} = A C^{2}$; Converse of Pythagoras theorem.
Equidistant locus: $P A^{2} = P B^{2}$; Gives a plane.
Distance from origin: $\sqrt (x^{2} + y^{2} + z^{2})$; Sphere of radius r: x² + y² + z² = r².

How to solve Distance Formula in Space: Collinearity, Shapes and Loci questions

Use these steps for proving collinearity or a shape and for finding loci.

1List the points and compute the squared distance between each relevant pair; keep the squares.
2For collinearity, check that the square root of the largest squared distance equals the sum of the other two roots.
3For a triangle, compare squares: equal squares mean equal sides; $a^{2} + b^{2} = c^{2}$ means a right angle.
4For a quadrilateral, compute all four sides and both diagonals before naming the shape.
5For a locus, take $P (x, y, z)$ , write the squared distances to each given point, equate as stated and expand.
6Simplify to a plane or a sphere (complete the square) and verify with a point.

Distance Formula in Space: Collinearity, Shapes and Loci in real life

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.

Distance between two points in space

Suppose two cable-car stations are at $A (- 1, 2, 5)$ and $B (1, 5, 11)$ in a coordinate system measured in hundreds of metres.

How long is the straight cable between them in metres?

Try it first, then show the solutionSolution

$A B^{2} = (1 - (- 1))^{2} + (5 - 2)^{2} + (11 - 5)^{2} = 4 + 9 + 36 = 49$ .
$A B = 7$ units $= 700$ m.

Answer: 700 m.

Memory hook: Difference each coordinate, square, add, root.

Sphere of coverage

Suppose a Wi-Fi router stands at the origin and a repeater at $(3, 0, 0)$ . A technician marks the boundary surface of all points that are exactly twice as far from the router as from the repeater.

Find the equation of this boundary surface.

Try it first, then show the solutionSolution

Let $P (x, y, z)$ . Condition $O P = 2 P R$ gives $x^{2} + y^{2} + z^{2} = 4 [(x - 3)^{2} + y^{2} + z^{2}]$ .
Expand: $3 x^{2} + 3 y^{2} + 3 z^{2} - 24 x + 36 = 0$ , so $x^{2} + y^{2} + z^{2} - 8 x + 12 = 0$ .
Complete the square: $(x - 4)^{2} + y^{2} + z^{2} = 4$ , a sphere of centre $(4, 0, 0)$ and radius 2.

Answer: The sphere $(x - 4)^{2} + y^{2} + z^{2} = 4$ .

Memory hook: Equal distances give a plane; a fixed ratio other than 1 gives a sphere.

Tricks to remember Distance Formula in Space: Collinearity, Shapes and Loci

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

Keep the squares
Compare squared distances; take roots only for the final answer.
Use it when: Shape and collinearity proofs.
Largest equals the sum
For collinear points the longest distance is the sum of the other two.
Use it when: Collinearity.
Equal distance kills the squares
In PA = PB the x², y², z² terms cancel, leaving a plane.
Use it when: Locus questions.
Ratio keeps the squares
If PA = k PB with k not 1, the quadratic terms survive: a sphere.
Use it when: Apollonius-type loci.

Distance Formula in Space: Collinearity, Shapes and Loci: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Plain distance · 2 marks · Very short answerBoard level

Find the distance between $P (- 1, 2, 5)$ and $Q (1, 5, 11)$ .

Try it first, then show the solutionSolution

$P Q^{2} = (1 + 1)^{2} + (5 - 2)^{2} + (11 - 5)^{2} = 4 + 9 + 36$ . (1 mark)
$P Q = 49 = 7$ . (1 mark)

Answer: 7 units

Example 2: Collinearity · 3 marks · Short answerBoard level

Show that $A (1, 2, - 1)$ , $B (4, 6, 3)$ and $C (7, 10, 7)$ are collinear.

Try it first, then show the solutionSolution

$A B^{2} = 9 + 16 + 16 = 41$ , so $A B = 41$ . (1 mark)
$B C^{2} = 9 + 16 + 16 = 41$ , so $B C = 41$ . (½ mark)
$A C^{2} = 36 + 64 + 64 = 164$ , so $A C = 241$ . (1 mark)
$A B + B C = 241 = A C$ , so the points are collinear (and $B$ is the midpoint of $A C$ ). (½ mark)

Answer: Collinear

Example 3: A right-angled triangle · 3 marks · Short answerHarder board level

Show that $A (1, - 1, 2)$ , $B (3, 2, 8)$ and $C (4, 1, 0)$ form a right-angled triangle, and find its area.

Try it first, then show the solutionSolution

$A B^{2} = 4 + 9 + 36 = 49$ , $A C^{2} = 9 + 4 + 4 = 17$ , $B C^{2} = 1 + 1 + 64 = 66$ . (1 mark)
$A B^{2} + A C^{2} = 49 + 17 = 66 = B C^{2}$ , so the angle at $A$ is a right angle. (1 mark)
Area $= \frac{1}{2} \cdot A B \cdot A C = \frac{1}{2} \cdot 7 \cdot 17 = \frac{7 17}{2} \approx 14.43$ square units. (1 mark)

Answer: Right angle at $A$ ; area $\frac{7 17}{2}$

Example 4: Locus: a plane · 3 marks · Short answerHarder board level

Find the locus of a point $P$ equidistant from $A (2, 0, - 1)$ and $B (4, 2, 3)$ .

Try it first, then show the solutionSolution

$P A^{2} = (x - 2)^{2} + y^{2} + (z + 1)^{2}$ and $P B^{2} = (x - 4)^{2} + (y - 2)^{2} + (z - 3)^{2}$ . (1 mark)
Equate and expand; the $x^{2}, y^{2}, z^{2}$ terms cancel: $- 4 x + 2 z + 5 = - 8 x - 4 y - 6 z + 29$ . (1 mark)
Simplify: $4 x + 4 y + 8 z = 24$ , so $x + y + 2 z = 6$ . The midpoint $(3, 1, 1)$ satisfies it: $3 + 1 + 2 = 6$ . (1 mark)

Answer: The plane $x + y + 2 z = 6$

Example 5: Locus: a sphere · 5 marks · Long answerHOTS / competency

Find the locus of a point whose distance from the origin is twice its distance from $A (3, 0, 0)$ , and identify it.

Try it first, then show the solutionSolution

Condition $O P = 2 P A$ , so $O P^{2} = 4 P A^{2}$ . (1 mark)
$x^{2} + y^{2} + z^{2} = 4 [(x - 3)^{2} + y^{2} + z^{2}]$ . (1 mark)
Expand: $3 x^{2} + 3 y^{2} + 3 z^{2} - 24 x + 36 = 0$ , i.e. $x^{2} + y^{2} + z^{2} - 8 x + 12 = 0$ . (1 mark)
Complete the square: $(x - 4)^{2} + y^{2} + z^{2} = 4$ . (1 mark)
A sphere with centre $(4, 0, 0)$ and radius 2. Check: $(2, 0, 0)$ is at distance 2 from $O$ and 1 from $A$ ; $(6, 0, 0)$ at 6 and 3. (1 mark)

Answer: The sphere $(x - 4)^{2} + y^{2} + z^{2} = 4$

Where students lose marks in Distance Formula in Space: Collinearity, Shapes and Loci

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

Mistake: Forgetting the third coordinate
The 2D formula is used by habit.
Wrong working
$P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
Right working
Add $(z_{2} - z_{1})^{2}$ inside the root.
Do this: Write three brackets every time.
Usually costs: The answer.
Mistake: Adding squares instead of roots for collinearity
Squares must not be added to test the sum of lengths.
Wrong working
$41 + 41 = 82 \neq = 164$ , so not collinear.
Right working
$41 + 41 = 241 = 164$ , so collinear.
Do this: Convert to lengths before adding.
Usually costs: The conclusion.
Mistake: Naming a quadrilateral from sides only
Equal sides alone do not prove a square.
Wrong working
"All four sides are 3, hence a square."
Right working
All sides equal gives a rhombus; equal diagonals as well give a square.
Do this: Always compute both diagonals.
Usually costs: 1 to 2 marks.
Mistake: Leaving a sphere locus as a raw equation
The question asks to identify the locus.
Wrong working
Stops at $x^{2} + y^{2} + z^{2} - 8 x + 12 = 0$ .
Right working
Complete the square and state centre $(4, 0, 0)$ and radius 2.
Do this: Finish with the name of the surface.
Usually costs: The last mark.

Exam tips

Always show the three squared differences inside the root; the method mark depends on them.
For collinearity name the longest distance and show the sum, not just the numbers.
For quadrilaterals, list four sides and both diagonals in a small table.
For loci end with a verification point and the name of the surface (plane or sphere).
The section formula is formative only in 2026-27; use distances for midpoint-type checks.

Practice questions on Distance Formula in Space: Collinearity, Shapes and Loci

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

Q1 · 1 mark · MCQBoard level
The distance of the point $(3, - 4, 12)$ from the origin is:
(a) 11
(b) 12
(c) 13
(d) 19
Hint
Use $x^{2} + y^{2} + z^{2}$ .
Answer
(c). $9 + 16 + 144 = 13$ .
Q2 · 1 mark · MCQHarder board level
The number of points on the $x$ -axis at a distance of 5 units from the point $(2, 3, 4)$ is:
(a) 0
(b) 1
(c) 2
(d) infinitely many
Hint
Take the point $(t, 0, 0)$ and form $(t - 2)^{2} + 9 + 16 = 25$ .
Answer
(b). $(t - 2)^{2} = 0$ , so only $t = 2$ : the point $(2, 0, 0)$ .
Q3 · 3 marks · Short answerHarder board level
Find the point on the $z$ -axis which is equidistant from $A (2, 3, 5)$ and $B (4, - 1, 1)$ .
Hint
Let the point be $(0, 0, z)$ and equate $P A^{2} = P B^{2}$ .
Answer
$4 + 9 + (z - 5)^{2} = 16 + 1 + (z - 1)^{2}$ , so $38 - 10 z = 18 - 2 z$ , giving $z = \frac{5}{2}$ . Point $(0, 0, \frac{5}{2})$ (both squared distances are $19.25$ ).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The points $(1, 2, 3)$ , $(2, 4, 6)$ and $(3, 6, 9)$ are collinear.
Reason (R): The distances are $A B = B C = 14$ and $A C = 214$ , so $A B + B C = A C$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compute the three distances.
Answer
(a). $A B^{2} = 1 + 4 + 9 = 14$ , $B C^{2} = 1 + 4 + 9 = 14$ , $A C^{2} = 4 + 16 + 36 = 56 = 4 \times 14$ .
Q5 · 4 marks · Short answerHOTS / competency
Show that $A (1, 1, 1)$ , $B (3, 3, 2)$ , $C (5, 2, 0)$ and $D (3, 0, - 1)$ are the vertices of a square.
Hint
Find the four sides and both diagonals.
Answer
$A B^{2} = 4 + 4 + 1 = 9$ , $B C^{2} = 4 + 1 + 4 = 9$ , $C D^{2} = 4 + 4 + 1 = 9$ , $D A^{2} = 4 + 1 + 4 = 9$ : all sides equal 3. $A C^{2} = 16 + 1 + 1 = 18$ , $B D^{2} = 0 + 9 + 9 = 18$ : equal diagonals. A rhombus with equal diagonals is a square (also $9 + 9 = 18$ gives a right angle at each vertex).

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.What is the distance between two points in space?Show answer
√((x2 − x1)² + (y2 − y1)² + (z2 − z1)²).
2.How do you test collinearity?Show answer
The longest of the three distances equals the sum of the other two.
3.What does PA = PB give?Show answer
A plane (the perpendicular bisector plane of AB).
4.What does PA = 2 PB give?Show answer
A sphere.
5.What do equal diagonals and equal sides give?Show answer
A square.

Distance Formula in Space: Collinearity, Shapes and Loci: frequently asked questions

Why is the formula the same as in the plane with one extra term?

Because applying Pythagoras twice in a box adds the square of the third difference.

Can three points in space form a triangle?

Yes, unless they are collinear. Check that no distance equals the sum of the other two.

Is a locus always a plane or a sphere here?

In this chapter, equal distances from two points gives a plane and a fixed ratio or a fixed sum of squares gives a sphere.

Is the section formula needed?

The chapter page marks the section formula in 3D as formative only in 2026-27, so distance-based reasoning is enough.