Anand Classes presents detailed NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry Exercise 11.2, helping students learn how to find the distance between two points in 3D space and understand the application of distance formula in three-dimensional geometry. These step-by-step solutions are prepared as per the latest CBSE and NCERT syllabus, ensuring clarity, accuracy, and easy comprehension. Each problem is solved methodically to help students strengthen their spatial understanding and perform well in exams. Click the print button to download study material and notes.
NCERT Question.1 : Find the distance between the following pairs of points
(i) Points (2, 3, 5) and (4, 3, 1)
(ii) Points (–3, 7, 2) and (2, 4, –1)
(iii) Points (–1, 3, –4) and (1, –3, 4)
(iv) Points (2, –1, 3) and (–2, 1, 3)
Solution :
(i) Points (2, 3, 5) and (4, 3, 1)
Let $P(2, 3, 5)$ and $Q(4, 3, 1)$.
By using the distance formula:
$$
\text{Distance } PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting the values:
$$
x_1 = 2,\ y_1 = 3,\ z_1 = 5 \
x_2 = 4,\ y_2 = 3,\ z_2 = 1
$$
$$
PQ = \sqrt{(4 – 2)^2 + (3 – 3)^2 + (1 – 5)^2}
$$
$$
= \sqrt{(2)^2 + (0)^2 + (-4)^2}
$$
$$
= \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5}
$$
∴ The distance between the points is $2\sqrt{5}$ units.
(ii) Points (–3, 7, 2) and (2, 4, –1)
Let $P(-3, 7, 2)$ and $Q(2, 4, -1)$.
By the distance formula:
$$
PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting:
$$
x_1 = -3,\ y_1 = 7,\ z_1 = 2 \
x_2 = 2,\ y_2 = 4,\ z_2 = -1
$$
$$
PQ = \sqrt{(2 – (-3))^2 + (4 – 7)^2 + (-1 – 2)^2}
$$
$$
= \sqrt{(5)^2 + (-3)^2 + (-3)^2}
$$
$$
= \sqrt{25 + 9 + 9} = \sqrt{43}
$$
∴ The distance between the points is $\sqrt{43}$ units.
(iii) Points (–1, 3, –4) and (1, –3, 4)
Let $P(-1, 3, -4)$ and $Q(1, -3, 4)$.
By the distance formula:
$$
PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting:
$$
x_1 = -1,\ y_1 = 3,\ z_1 = -4 \
x_2 = 1,\ y_2 = -3,\ z_2 = 4
$$
$$
PQ = \sqrt{(1 – (-1))^2 + (-3 – 3)^2 + (4 – (-4))^2}
$$
$$
= \sqrt{(2)^2 + (-6)^2 + (8)^2}
$$
$$
= \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}
$$
∴ The distance between the points is $2\sqrt{26}$ units.
(iv) Points (2, –1, 3) and (–2, 1, 3)
Let $P(2, -1, 3)$ and $Q(-2, 1, 3)$.
By the distance formula:
$$
PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting:
$$
x_1 = 2,\ y_1 = -1,\ z_1 = 3 \
x_2 = -2,\ y_2 = 1,\ z_2 = 3
$$
$$
PQ = \sqrt{(-2 – 2)^2 + (1 – (-1))^2 + (3 – 3)^2}
$$
$$
= \sqrt{(-4)^2 + (2)^2 + (0)^2}
$$
$$
= \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}
$$
∴ The required distance is $2\sqrt{5}$ units.
Download detailed NCERT solutions by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.
Related topics: Distance formula in 3D geometry, coordinate geometry class 11, distance between two points formula, 3D coordinate system examples.
NCERT Question.2 : Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
Solution :
If three points are collinear, then they lie on the same straight line.
Step 1: Find the distance between each pair of points
Let
$P(-2, 3, 5)$, $Q(1, 2, 3)$ and $R(7, 0, -1)$.
(i) Distance $PQ$
By the distance formula:
$$
PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting values:
$$
x_1 = -2,\ y_1 = 3,\ z_1 = 5 \
x_2 = 1,\ y_2 = 2,\ z_2 = 3
$$
$$
PQ = \sqrt{(1 – (-2))^2 + (2 – 3)^2 + (3 – 5)^2}
$$
$$
= \sqrt{(3)^2 + (-1)^2 + (-2)^2}
$$
$$
= \sqrt{9 + 1 + 4} = \sqrt{14}
$$
Hence,
$$PQ = \sqrt{14}$$
(ii) Distance $QR$
By the distance formula:
$$
QR = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting:
$$
x_1 = 1,\ y_1 = 2,\ z_1 = 3 \
x_2 = 7,\ y_2 = 0,\ z_2 = -1
$$
$$
QR = \sqrt{(7 – 1)^2 + (0 – 2)^2 + (-1 – 3)^2}
$$
$$
= \sqrt{(6)^2 + (-2)^2 + (-4)^2}
$$
$$
= \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}
$$
Hence,
$$QR = 2\sqrt{14}$$
(iii) Distance $PR$
By the distance formula:
$$
PR = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
Substituting:
$$
x_1 = -2,\ y_1 = 3,\ z_1 = 5 \
x_2 = 7,\ y_2 = 0,\ z_2 = -1
$$
$$
PR = \sqrt{(7 – (-2))^2 + (0 – 3)^2 + (-1 – 5)^2}
$$
$$
= \sqrt{(9)^2 + (-3)^2 + (-6)^2}
$$
$$
= \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}
$$
Hence,
$$PR = 3\sqrt{14}$$
Step 2: Check for collinearity
We have:
$$PQ = \sqrt{14}, \quad QR = 2\sqrt{14}, \quad PR = 3\sqrt{14}$$
Now,
$$PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$$
∴ The three points P, Q, and R are collinear.
Conclusion:
The points $(-2, 3, 5)$, $(1, 2, 3)$, and $(7, 0, -1)$ lie on the same straight line, proving they are collinear.
Explore more with Anand Classes: Learn 3D geometry concepts like distance formula, section formula, and collinearity of points for Class 11 Maths, JEE Main, and NDA preparation.
NCERT Question.3 : Verify 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.
Solution:
(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.
Let us consider the points be
$P(0, 7, -10),\ Q(1, 6, -6)\ \text{and}\ R(4, 9, -6)$
If any two sides are equal, hence it will be an isosceles triangle.
So firstly let us calculate the distance of $PQ, QR$.
P ≡ $(0, 7, -10)$ and Q ≡ $(1, 6, -6)$
Now, by using the distance formula,
$$
\text{Length of distance } PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}
$$
So here,
$x_1 = 0,\ y_1 = 7,\ z_1 = -10$
$x_2 = 1,\ y_2 = 6,\ z_2 = -6$
$$
PQ = \sqrt{(1 – 0)^2 + (6 – 7)^2 + \big(-6 – (-10)\big)^2}
$$
$$
= \sqrt{(1)^2 + (-1)^2 + (4)^2}
$$
$$
= \sqrt{1 + 1 + 16} = \sqrt{18}
$$
Calculating $QR$
Q ≡ $(1, 6, -6)$ and R ≡ $(4, 9, -6)$
Using the distance formula,
$$
QR = \sqrt{(4 – 1)^2 + (9 – 6)^2 + \big(-6 – (-6)\big)^2}
$$
$$
= \sqrt{(3)^2 + (3)^2 + (0)^2}
$$
$$
= \sqrt{9 + 9 + 0} = \sqrt{18}
$$
Hence,
$$
\text{Length of } PQ = \text{Length of } QR \quad\text{i.e.}\quad \sqrt{18} = \sqrt{18}
$$
∴ Two sides are equal.
∴ $PQR$ is an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right-angled triangle.
$(0, 7, 10),\ (-1, 6, 6)\ \text{and}\ (-4, 9, 6)$ are the vertices of a right-angled triangle.
Let the points be
$P(0, 7, 10), Q(-1, 6, 6), R(-4, 9, 6)$
Firstly let us calculate the distances $PQ , QR$ and $PR$.
Calculating $PQ$
P ≡ $(0, 7, 10)$ and Q ≡ $(-1, 6, 6)$
$$
PQ = \sqrt{(-1 – 0)^2 + (6 – 7)^2 + (6 – 10)^2}
$$
$$
= \sqrt{(-1)^2 + (-1)^2 + (-4)^2}
$$
$$
= \sqrt{1 + 1 + 16} = \sqrt{18}
$$
Calculating $QR$
Q ≡ $(-1, 6, 6)$ and R ≡ $(-4, 9, 6)$
$$
QR = \sqrt{(-4 – (-1))^2 + (9 – 6)^2 + (6 – 6)^2}
$$
$$
= \sqrt{(-3)^2 + (3)^2 + (0)^2}
$$
$$
= \sqrt{9 + 9 + 0} = \sqrt{18}
$$
Calculating $PR$
P ≡ $(0, 7, 10)$ and R ≡ $(-4, 9, 6)$
$$
PR = \sqrt{(-4 – 0)^2 + (9 – 7)^2 + (6 – 10)^2}
$$
$$
= \sqrt{(-4)^2 + (2)^2 + (-4)^2}
$$
$$
= \sqrt{16 + 4 + 16} = \sqrt{36} = 6
$$
Now,
$$
PQ^2 + QR^2 = 18 + 18 = 36 = PR^2
$$
By the converse of Pythagoras’ theorem, the triangle is right-angled at $Q$.
∴ The given vertices $P, Q, R$ are the vertices of a right-angled triangle at $Q$.
(iii) $( -1, 2, 1 ),\ (1, -2, 5),\ (4, -7, 8)$ and $(2, -3, 4)$ are the vertices of a parallelogram.
$(-1, 2, 1),\ (1, -2, 5),\ (4, -7, 8)\ \text{and}\ (2, -3, 4)$ are the vertices of a parallelogram.
Let the points be:
$A(-1, 2, 1),\ B(1, -2, 5),\ C(4, -7, 8)\, \ D(2, -3, 4)$
$ABCD$ can be vertices of a parallelogram only if opposite sides are equal, i.e. $AB = CD$ and $BC = AD$.
Firstly let us calculate the distances.
A ≡ $(-1, 2, 1)$ and B ≡ $(1, -2, 5)$
$$
AB = \sqrt{(1 – (-1))^2 + (-2 – 2)^2 + (5 – 1)^2}
$$
$$
= \sqrt{(2)^2 + (-4)^2 + (4)^2}
$$
$$
= \sqrt{4 + 16 + 16} = \sqrt{36} = 6
$$
B ≡ $(1, -2, 5)$ and C ≡ $(4, -7, 8)$
$$
BC = \sqrt{(4 – 1)^2 + (-7 – (-2))^2 + (8 – 5)^2}
$$
$$
= \sqrt{(3)^2 + (-5)^2 + (3)^2}
$$
$$
= \sqrt{9 + 25 + 9} = \sqrt{43}
$$
C ≡ $(4, -7, 8)$ and D ≡ $(2, -3, 4)$
$$
CD = \sqrt{(2 – 4)^2 + (-3 – (-7))^2 + (4 – 8)^2}
$$
$$
= \sqrt{(-2)^2 + (4)^2 + (-4)^2}
$$
$$
= \sqrt{4 + 16 + 16} = \sqrt{36} = 6
$$
D ≡ $(2, -3, 4)$ and A ≡ $(-1, 2, 1)$
$$
DA = \sqrt{(-1 – 2)^2 + (2 – (-3))^2 + (1 – 4)^2}
$$
$$
= \sqrt{(-3)^2 + (5)^2 + (-3)^2}
$$
$$
= \sqrt{9 + 25 + 9} = \sqrt{43}
$$
Since $AB = CD$ and $BC = DA$, both pairs of opposite sides are equal.
Check: Are the diagonals equal (is the parallelogram actually a rectangle)?
We check the diagonals (AC) and (BD) for the points
$$
A(-1,2,1),\quad B(1,-2,5),\quad C(4,-7,8),\quad D(2,-3,4).
$$
Diagonal (AC)
Differences:
$$
\Delta x = 4 – (-1) = 5,\quad
\Delta y = -7 – 2 = -9,\quad
\Delta z = 8 – 1 = 7.
$$
Diagonal Length:
$$
AC = \sqrt{5^2 + (-9)^2 + 7^2} = \sqrt{25 + 81 + 49} = \sqrt{155}.
$$
Diagonal (BD)
Differences:
$
\Delta x = 2 – 1 = 1,\quad
\Delta y = -3 – (-2) = -1,\quad
\Delta z = 4 – 5 = -1.
$
Diagonal Length:
$$
BD = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}.
$$
Conclusion
$$
AC = \sqrt{155}\neq \sqrt{3} = BD.
$$
Since the diagonals are not equal, the parallelogram is not a rectangle.
∴ $ABCD$ is a parallelogram.
Useful practice: distance formula in three dimensions, detecting isosceles and right triangles using side lengths, and checking parallelogram conditions using opposite side equality — great for Class 11 coordinate geometry and competitive exam revision.
NCERT Question.4 : Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1)
Solution:
Let $A(1,2,3)$ & $B(3,2,-1)$.
Let point $P$ be $(x,y,z)$.
Since it is given that point $P(x,y,z)$ is equal distance from point $A(1,2,3)$ & $B(3,2,-1)$ i.e. $PA=PB$.
$P\equiv (x,y,z)$ and $A\equiv(1,2,3)$.
Now, by using the distance formula,
$$
PA=\sqrt{(1-x)^2+(2-y)^2+(3-z)^2}.
$$
$P\equiv (x,y,z)$ and $B\equiv(3,2,-1)$.
Now, by using the distance formula,
$$
PB=\sqrt{(3-x)^2+(2-y)^2+(-1-z)^2}.
$$
Since $PA=PB$, square both sides to get $PA^2=PB^2$:
$$
(1-x)^2+(2-y)^2+(3-z)^2=(3-x)^2+(2-y)^2+(-1-z)^2.
$$
Expand and simplify (terms $(2-y)^2$ cancel on both sides):
$$
(1 – 2x + x^2) + (9 – 6z + z^2) = (9 – 6x + x^2) + (1 + 2z + z^2).
$$
Collect like terms:
$$
-2x -6z +10 = -6x +2z +10.
$$
Bring all terms to one side:
$$
4x -8z = 0
$$
Divide by 4:
$$
x – 2z = 0.
$$
∴ The required equation is
$$
\boxed{x – 2z = 0.}
$$
NCERT Question.5 : 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 $10$.
Solution:
Let $A(4,0,0)$ & $B(-4,0,0)$.
Let the coordinates of point $P$ be $(x,y,z)$.
Calculating $PA$:
$P\equiv(x,y,z)$ and $A\equiv(4,0,0)$.
By the distance formula,
$$
PA=\sqrt{(4-x)^2+(0-y)^2+(0-z)^2}=\sqrt{(4-x)^2+y^2+z^2}.
$$
Calculating $PB$:
$P\equiv(x,y,z)$ and $B\equiv(-4,0,0)$.
By the distance formula,
$$
PB=\sqrt{(-4-x)^2+(0-y)^2+(0-z)^2}=\sqrt{( -4-x)^2+y^2+z^2}.
$$
Now it is given that:
$$
PA + PB = 10.
$$
Let $PA=10-PB$. Square both sides:
$$
PA^2=(10-PB)^2=100+PB^2-20PB.
$$
Substitute $PA^2=(4-x)^2+y^2+z^2$ and $PB^2=(-4-x)^2+y^2+z^2$:
$$
(4-x)^2+y^2+z^2 = 100 + (-4-x)^2+y^2+z^2 – 20PB.
$$
Cancel $y^2+z^2$ from both sides and expand:
$$
(16 + x^2 -8x) = 100 + (16 + x^2 +8x) – 20PB.
$$
Simplify:
$$
20PB = 16x + 100.
$$
So
$$
5PB = 4x + 25.
$$
Square again:
$$
25PB^2 = (4x+25)^2 = 16x^2 + 200x + 625.
$$
But $PB^2 = (-4-x)^2 + y^2 + z^2 = x^2 + 8x +16 + y^2 + z^2$. Substitute:
$$
25\big(x^2 + 8x + 16 + y^2 + z^2\big) = 16x^2 + 200x + 625.
$$
Expand left side:
$$
25x^2 + 200x + 400 + 25y^2 + 25z^2 = 16x^2 + 200x + 625.
$$
Bring all terms to one side:
$$
9x^2 + 25y^2 + 25z^2 -225 = 0.
$$
Thus the required equation is
$$
\boxed{9x^2 + 25y^2 + 25z^2 -225 = 0.}
$$
(Equivalently, dividing by $225$ gives the standard ellipsoid form $$\frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{9}=1$$ with foci at $(\pm4,0,0)$ and semi-major axis $a=5$.)
Locus is an ellipsoid with foci at $(4,0,0)$ and $(-4,0,0)$; principal semi-axes $a=5$ and $b=3$.
