Chapter 11 of 14

Updated 6 August 2026

NCERT Solutions

Introduction to Three Dimensional Geometry — NCERT Solutions

Q: What are the important topics in Introduction to Three Dimensional Geometry for CBSE Class 11 Mathematics?

Key topics in Introduction to Three Dimensional Geometry include Coordinate Axes, Coordinate Planes, and Octants, Coordinates of a Point in Space, Distance Between Two Points in Space, Worked Examples and Important Results. Study these first, then practise questions on each for Class 11 exams.

Q: Are these NCERT Solutions for Introduction to Three Dimensional Geometry free?

The first 7 of the 13 solutions on this page are open to read. The other 6 are free with a Super Tutor account — signing up is free and needs no card.

Q: How should I revise Introduction to Three Dimensional Geometry for Class 11 exams?

Learn the core ideas first, then work through the 162 practice questions on Introduction to Three Dimensional Geometry. Revise definitions regularly and use flashcards for quick recall before the exam.

CBSE · Class 11 · Mathematics

NCERT Solutions for Introduction to Three Dimensional Geometry, CBSE Class 11 Mathematics: 13 textbook questions solved step by step.

162 questions48 flashcards2 formulas & key relations5 concepts

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13 Questions Solved · 3 Sections

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Exercise 11.1

1A point is on the x-axis. What are its y-coordinate and z-coordinates?Show solution

Given: A point lies on the x-axis.

Any point on the x-axis is of the form $(x, 0, 0)$ .

Therefore, the y-coordinate and z-coordinate of the point are both $\mathbf{0}$ .

2A point is in the XZ-plane. What can you say about its y-coordinate?Show solution

Given: A point lies in the XZ-plane.

Any point in the XZ-plane has its y-coordinate equal to zero, because the XZ-plane is defined as the set of all points where $y = 0$ .

Therefore, the y-coordinate of the point is $\mathbf{0}$ .

3Name the octants in which the following points lie: (1,2,3), (4,−2,3), (4,−2,−5), (4,2,−5), (−4,2,−5), (−4,2,5), (−3,−1,6), (−2,−4,−7).Show solution

The sign convention for octants is:

Octant	x	y	z
I	+	+	+
II	−	+	+
III	−	−	+
IV	+	−	+
V	+	+	−
VI	−	+	−
VII	−	−	−
VIII	+	−	−

$(1, 2, 3)$ : $x>0,\ y>0,\ z>0$ → Octant I
$(4, -2, 3)$ : $x>0,\ y<0,\ z>0$ → Octant IV
$(4, -2, -5)$ : $x>0,\ y<0,\ z<0$ → Octant VIII
$(4, 2, -5)$ : $x>0,\ y>0,\ z<0$ → Octant V
$(-4, 2, -5)$ : $x<0,\ y>0,\ z<0$ → Octant VI
$(-4, 2, 5)$ : $x<0,\ y>0,\ z>0$ → Octant II
$(-3, -1, 6)$ : $x<0,\ y<0,\ z>0$ → Octant III
$(-2, -4, -7)$ : $x<0,\ y<0,\ z<0$ → Octant VII

4Fill in the blanks:
(i) The x-axis and y-axis taken together determine a plane known as ___
(ii) The coordinates of points in the XY-plane are of the form ___
(iii) Coordinate planes divide the space into ___ octants.Show solution

(i) The x-axis and y-axis taken together determine a plane known as the XY-plane (also called the xy-plane).

(ii) The coordinates of points in the XY-plane are of the form $\mathbf{(x,\ y,\ 0)}$ , since the z-coordinate is zero for every point in the XY-plane.

(iii) Coordinate planes divide the space into eight octants.

Exercise 11.2

1Find the distance between the following pairs of points:
(i) (2,3,5) and (4,3,1)
(ii) (−3,7,2) and (2,4,−1)
(iii) (−1,3,−4) and (1,−3,4)
(iv) (2,−1,3) and (−2,1,3).Show solution

Formula used: Distance between $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is
$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

(i) $P(2,3,5)$ and $Q(4,3,1)$ :
$PQ = \sqrt{(4-2)^2+(3-3)^2+(1-5)^2} = \sqrt{4+0+16} = \sqrt{20} = 2\sqrt{5}$

(ii) $P(-3,7,2)$ and $Q(2,4,-1)$ :
$PQ = \sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2} = \sqrt{25+9+9} = \sqrt{43}$

(iii) $P(-1,3,-4)$ and $Q(1,-3,4)$ :
$PQ = \sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2} = \sqrt{4+36+64} = \sqrt{104} = 2\sqrt{26}$

(iv) $P(2,-1,3)$ and $Q(-2,1,3)$ :
$PQ = \sqrt{(-2-2)^2+(1-(-1))^2+(3-3)^2} = \sqrt{16+4+0} = \sqrt{20} = 2\sqrt{5}$

2Show that the points (−2,3,5), (1,2,3) and (7,0,−1) are collinear.Show solution

Let $A(-2,3,5)$ , $B(1,2,3)$ , $C(7,0,-1)$ .

Using the distance formula:
$AB = \sqrt{(1+2)^2+(2-3)^2+(3-5)^2} = \sqrt{9+1+4} = \sqrt{14}$

$BC = \sqrt{(7-1)^2+(0-2)^2+(-1-3)^2} = \sqrt{36+4+16} = \sqrt{56} = 2\sqrt{14}$

$AC = \sqrt{(7+2)^2+(0-3)^2+(-1-5)^2} = \sqrt{81+9+36} = \sqrt{126} = 3\sqrt{14}$

Now, $AB + BC = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = AC$ .

Since $AB + BC = AC$ , the three points are collinear.

3Verify the following:
(i) (0,7,−10), (1,6,−6) and (4,9,−6) are the vertices of an isosceles triangle.
(ii) (0,7,10), (−1,6,6) and (−4,9,6) are the vertices of a right angled triangle.
(iii) (−1,2,1),(1,−2,5),(4,−7,8) and (2,−3,4) are the vertices of a parallelogram.Show solution

(i) Let $A(0,7,-10)$ , $B(1,6,-6)$ , $C(4,9,-6)$ .

$AB = \sqrt{(1-0)^2+(6-7)^2+(-6+10)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$

$BC = \sqrt{(4-1)^2+(9-6)^2+(-6+6)^2} = \sqrt{9+9+0} = \sqrt{18} = 3\sqrt{2}$

$CA = \sqrt{(0-4)^2+(7-9)^2+(-10+6)^2} = \sqrt{16+4+16} = \sqrt{36} = 6$

Since $AB = BC = 3\sqrt{2}$ and $CA = 6$ , two sides are equal.

Hence, the triangle is isosceles. ✓

(ii) Let $A(0,7,10)$ , $B(-1,6,6)$ , $C(-4,9,6)$ .

$AB^2 = (-1-0)^2+(6-7)^2+(6-10)^2 = 1+1+16 = 18$

$BC^2 = (-4+1)^2+(9-6)^2+(6-6)^2 = 9+9+0 = 18$

$CA^2 = (0+4)^2+(7-9)^2+(10-6)^2 = 16+4+16 = 36$

Check: $AB^2 + BC^2 = 18 + 18 = 36 = CA^2$ .

Since $AB^2 + BC^2 = CA^2$ , by the converse of Pythagoras' theorem, the triangle is a right angled triangle (right angle at B). ✓

(iii) Let $A(-1,2,1)$ , $B(1,-2,5)$ , $C(4,-7,8)$ , $D(2,-3,4)$ .

For a parallelogram, opposite sides must be equal: $AB = DC$ and $BC = AD$ .

$AB = \sqrt{(1+1)^2+(-2-2)^2+(5-1)^2} = \sqrt{4+16+16} = \sqrt{36} = 6$

$DC = \sqrt{(4-2)^2+(-7+3)^2+(8-4)^2} = \sqrt{4+16+16} = \sqrt{36} = 6$

$BC = \sqrt{(4-1)^2+(-7+2)^2+(8-5)^2} = \sqrt{9+25+9} = \sqrt{43}$

$AD = \sqrt{(2+1)^2+(-3-2)^2+(4-1)^2} = \sqrt{9+25+9} = \sqrt{43}$

Since $AB = DC$ and $BC = AD$ , the quadrilateral is a parallelogram. ✓

4Find the equation of the set of points which are equidistant from the points (1,2,3) and (3,2,−1).

Let $P(x,y,z)$ be any point equidistant from $A(1,2,3)$ and $B(3,2,-1)$ .

Given condition: $PA = PB$ , i.e., $PA^2 = PB^2$ .

$PA^2 = (x-1)^2+(y-2)^2+(z-3)^2$

$PB^2 = (x-3)^2+(y-2)^2+(z+1)^2$

Setting $PA^2 = PB^2$ :
$(x-1)^2+(y-2)^2+(z-3)^2 = (x-3)^2+(y-2)^2+(z+1)^2$

$(x-1)^2+(z-3)^2 = (x-3)^2+(z+1)^2$

$x^2-2x+1+z^2-6z+9 = x^2-6x+9+z^2+2z+1$

$-2x-6z+10 = -6x+2z+10$

$4x - 8z = 0$

$\boxed{x - 2z = 0}$

This is the required equation of the set of points.

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5Find 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.

Let $P(x,y,z)$ be any point such that $PA + PB = 10$ .

$PA = \sqrt{(x-4)^2+y^2+z^2}, \quad PB = \sqrt{(x+4)^2+y^2+z^2}$

Given: $PA + PB = 10$ , so $PA = 10 - PB$ .

Squaring both sides:
$PA^2 = 100 - 20\cdot PB + PB^2$

$(x-4)^2+y^2+z^2 = 100 - 20\cdot PB + (x+4)^2+y^2+z^2$

$x^2-8x+16 = 100 - 20\cdot PB + x^2+8x+16$

$-8x = 100 - 20\cdot PB + 8x$

$20\cdot PB = 100 + 16x$

$PB = \frac{100+16x}{20} = 5 + \frac{4x}{5}$

Squaring again:
$PB^2 = \left(5+\frac{4x}{5}\right)^2$

$(x+4)^2+y^2+z^2 = 25 + 8x + \frac{16x^2}{25}$

$x^2+8x+16+y^2+z^2 = 25+8x+\frac{16x^2}{25}$

$x^2+16+y^2+z^2 = 25+\frac{16x^2}{25}$

Multiplying throughout by 25:
$25x^2+400+25y^2+25z^2 = 625+16x^2$

$9x^2+25y^2+25z^2 = 225$

Dividing by 225:
$\boxed{\frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{9} = 1}$

This is the required equation of the set of points P.

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Miscellaneous Exercise on Chapter 11

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

Given: $A(3,-1,2)$ , $B(1,2,-4)$ , $C(-1,1,2)$ . Let $D(x,y,z)$ be the fourth vertex.

Concept: In a parallelogram, the diagonals bisect each other. So the midpoint of diagonal $AC$ = midpoint of diagonal $BD$ .

Midpoint of $AC$ :
$\left(\frac{3+(-1)}{2},\ \frac{-1+1}{2},\ \frac{2+2}{2}\right) = (1,\ 0,\ 2)$

Midpoint of $BD$ :
$\left(\frac{1+x}{2},\ \frac{2+y}{2},\ \frac{-4+z}{2}\right)$

Setting them equal:
$\frac{1+x}{2} = 1 \Rightarrow x = 1$
$\frac{2+y}{2} = 0 \Rightarrow y = -2$
$\frac{-4+z}{2} = 2 \Rightarrow z = 8$

The coordinates of the fourth vertex $D$ are $\mathbf{(1,\ -2,\ 8)}$ .

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2Find the lengths of the medians of the triangle with vertices A(0,0,6), B(0,4,0) and C(6,0,0).

Given: $A(0,0,6)$ , $B(0,4,0)$ , $C(6,0,0)$ .

Let $D$ , $E$ , $F$ be the midpoints of $BC$ , $CA$ , $AB$ respectively.

Midpoint D of BC:
$D = \left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right) = (3,2,0)$

Median AD:
$AD = \sqrt{(3-0)^2+(2-0)^2+(0-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$

Midpoint E of CA:
$E = \left(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right) = (3,0,3)$

Median BE:
$BE = \sqrt{(3-0)^2+(0-4)^2+(3-0)^2} = \sqrt{9+16+9} = \sqrt{34}$

Midpoint F of AB:
$F = \left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right) = (0,2,3)$

Median CF:
$CF = \sqrt{(0-6)^2+(2-0)^2+(3-0)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$

The lengths of the medians are $AD = 7$ , $BE = \sqrt{34}$ , $CF = 7$ .

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

Given: Origin $O(0,0,0)$ is the centroid of $\triangle PQR$ with $P(2a,2,6)$ , $Q(-4,3b,-10)$ , $R(8,14,2c)$ .

Formula: Centroid $= \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3},\ \dfrac{z_1+z_2+z_3}{3}\right)$

Setting each coordinate of the centroid equal to 0:

x-coordinate:
$\frac{2a+(-4)+8}{3} = 0 \Rightarrow 2a+4 = 0 \Rightarrow a = -2$

y-coordinate:
$\frac{2+3b+14}{3} = 0 \Rightarrow 3b+16 = 0 \Rightarrow b = -\frac{16}{3}$

z-coordinate:
$\frac{6+(-10)+2c}{3} = 0 \Rightarrow 2c-4 = 0 \Rightarrow c = 2$

Therefore, $a = -2$ , $b = -\dfrac{16}{3}$ , $c = 2$ .

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4If A and B be the points (3,4,5) and (−1,3,−7), respectively, find the equation of the set of points P such that PA² + PB² = k², where k is a constant.

Let $P(x,y,z)$ be any point satisfying the given condition.

$PA^2 = (x-3)^2+(y-4)^2+(z-5)^2$
$= x^2-6x+9+y^2-8y+16+z^2-10z+25$

$PB^2 = (x+1)^2+(y-3)^2+(z+7)^2$
$= x^2+2x+1+y^2-6y+9+z^2+14z+49$

Adding:
$PA^2+PB^2 = 2x^2+2y^2+2z^2-4x-14y+4z+109$

Given $PA^2+PB^2 = k^2$ :
$2x^2+2y^2+2z^2-4x-14y+4z+109 = k^2$

$\boxed{2x^2+2y^2+2z^2-4x-14y+4z = k^2-109}$

This is the required equation of the set of points P.

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Frequently Asked Questions

What are the important topics in Introduction to Three Dimensional Geometry for CBSE Class 11 Mathematics?

Key topics in Introduction to Three Dimensional Geometry include Coordinate Axes, Coordinate Planes, and Octants, Coordinates of a Point in Space, Distance Between Two Points in Space, Worked Examples and Important Results. Study these first, then practise questions on each for Class 11 exams.

Are these NCERT Solutions for Introduction to Three Dimensional Geometry free?

The first 7 of the 13 solutions on this page are open to read. The other 6 are free with a Super Tutor account — signing up is free and needs no card.

How should I revise Introduction to Three Dimensional Geometry for Class 11 exams?

Learn the core ideas first, then work through the 162 practice questions on Introduction to Three Dimensional Geometry. Revise definitions regularly and use flashcards for quick recall before the exam.

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Exercise 11.1

Exercise 11.2

Miscellaneous Exercise on Chapter 11

Frequently Asked Questions

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Practice Quiz

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