NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

Question:1

Locate the following points:
(i) (1, – 1, 3), (ii) (– 1, 2, 4)
(iii) (– 2, – 4, –7) (iv) (– 4, 2, – 5).

Answer:

Question:2

Name the octant in which each of the following points lies.
(i) (1, 2, 3), (ii) (4, – 2, 3), (iii) (4, –2, –5) (iv) (4, 2, –5)
(v) (– 4, 2, 5) (vi) (–3, –1, 6) (vii) (2, – 4, – 7) (viii) (– 4, 2, – 5).

Answer:

The octant in which a point lies is determined by the signs of the coordinates of the point.

Thus,

(1,2,3) → first quadrant

(4,-2,3) → Fourth octant

(4,-2,-5) → eighth octant

(4,2,-5) → fifth octant

(-4,2,5) → second octant

(-3,-1,6) → third octant

(3,-4,-7) → eighth octant

(-4,2,-5) → sixth octant

Question:3

Let A, B, and C be the feet of perpendiculars from a point P on the x, y, and z-axis respectively. Find the coordinates of A, B and C in each of the following where the point P is :
(i) A = (3, 4, 2) (ii) (–5, 3, 7) (iii) (4, – 3, – 5)

Answer:

We already know that,
On x-axis, y,z= 0
On y-axis, x,z= 0
On z-axis, x,y = 0
Thus, from point P, the feet of perpendiculars will be

1.A(3,0,0), 5(0,4,0), C(0,0,2)

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

3.A(4,0,0), 5(0,-3,0), C(0,0,-5)

Question:4

Let A, B, and C be the feet of perpendiculars from a point P on the xy, yz and zxplanes respectively. Find the coordinates of A, B, and C in each of the following where the point P is
(i) (3, 4, 5) (ii) (–5, 3, 7) (iii) (4, – 3, – 5).

Answer:

We know that,
On xy-plane, z = 0
On yz-plane, x=0
On zx-plane, y = 0
Thus, the coordinates from the perpendicular to different planes from the given points are,

1.A(3,4,0), 5(0,4,5), C(3,0,5)

2.A(-5,3,0), 5(0,3,7), C(-5,0,7)

3.A(4,-3,0), 5(0,-3,-5), C(4,0,-5)

Question:5

How far apart are the points (2, 0, 0) and (–3, 0, 0)?

Answer:

AB =| 2-(-3) |
= 5

Question:6

Find the distance from the origin to (6, 6, 7).

Answer:

From Origin to Point (6,6,7),
Distance =$\sqrt{\left ( 0-6 \right )^{2}+\left ( 0-6 \right )^{2}+\left ( 0-7 \right )^{2}}$
= $\sqrt{121}$
= 11

Question:7

Show that if x square plus y square is equal to 1, then the point x y square root of 1 minus x square minus y square is at a distance 1 unit from the origin.
Answer:

Given: x² + y² = 1
From origin to point $\left ( x,y ,\sqrt{1 - x^{2}-y^{2}} \right )$ )
Distance, $d= \sqrt{x^{2}+y^{2}+ \left (\sqrt{1 - x^{2}-y^{2}} \right )^{2} }$
$= \sqrt{x^{2}+y^{2}+1 - x^{2}-y^{2}}=1$

Question:8

Show that the point A (1, – 1, 3), B (2, – 4, 5) and (5, – 13, 11) are collinear.

Answer:

Given points:
4(1,-1,3), 6(2,-4,5), C(5,-13,11)
Thus,
$AB=\sqrt{(1-2)^{2}+(-1+4)^{2}+(3-5)^{2}}$
$=\sqrt{1+9+4}=\sqrt{14}$
$BC=\sqrt{(2-5)^{2}+(-4+13)^{2}+(5-11)^{2}}=3\sqrt{14}$
$AC=\sqrt{(1-5)^{2}+(-1+13)^{2}+(3-11)^{2}}=4\sqrt{14}$
Now,
AB + BC = $=\sqrt{14}+3\sqrt{14}$
= $4\sqrt{14}$ = AC

Question:9

Three consecutive vertices of a parallelogram ABCD are A (6, – 2, 4), B (2, 4, – 8), C (–2, 2, 4). Find the coordinates of the fourth vertex

Answer:

Let us assume that the coordinates of the fourth vertex D are (x,y,z).
Now, mid-point of diagonal AC = P(6-2/2, -2+2/2,4+4/2)
= P(2,0,4)
Since it is a parallelogram, the mid-point of BD will also be P,
= P(x+2/2,y+4/2,z-8/2)
= P(2,0,4)
Now, let us equate the coordinates of P,
X+2/2 = 2, i.e., x = 2;
Y+4/2 = 0, i.e., y = -4;
Z-8/2 = 4, i.e., z = 16
Therefore, the coordinates of D will be (2,-4,16).

Question:10

Show that the triangle ABC with vertices A (0, 4, 1), B (2, 3, – 1) and C (4, 5, 0) is right-angled

Answer:

Vertices of?ABC are –
A(0,4,1), 5(2,3,-1), C(4,5,0)
Thus,
AB = $\sqrt{(0-2)^{2}+(4-3)^{2}+(1+1)^{2} }$
= $\sqrt{4+1+1 }$ = 3
BC = $\sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2} }$
= $\sqrt{4+4+1 }$ = 3
AC = $\sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2} }$
= $\sqrt{16+1+1 }$ = $\sqrt{18}$
Thus, it is clear that,
AC² = AB² + BC²,
Hence it is a right-angled triangle.

Question:11

Find the third vertex of the triangle whose centroid is the origin and two vertices are (2, 4, 6) and (0, –2, –5).

Answer:

Given:
Vertices of the triangle,
B(2, 4, 6), C(0,-2,-5)
& centroid of a triangle, G is (0,0,0)
Let us assume that the unknown vertex of the triangle is A(x,y,z).
Thus,
(0,0,0) ≡ ((2+0+x)/3, (4-2+y)/3, (6-5+z)/3)
Now, when we compare the coordinates, we get,
2+x/3 = 0, thus, x = -2
2+y/3 = 0, thus, y = -2
& 1+z/3 = 0, thus, z = -1.

Question:12

Find the centroid of a triangle, the mid-point of whose sides are D (1, 2, – 3), E (3, 0, 1) and F (– 1, 1, – 4).

Answer:

Given:
Mid-points of sides of triangle ABC are,
D(1,2,-3), E(3,0,1) & F(-1,1,4)
Now, the centroid of triangle DEF = Centroid of triangle ABC ……. (from geometry of centroid)
Thus, the centroid of triangle ABC,
G(1+3-1/3, 2+0+1/3, -3+1-4/3) ≡ G(1,1,-2)

Question:13

The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, – 1). Find its vertices.

Answer:

Given:
Mid-points of sides of triangle ABC are,
D(5,7,11), E(0,8,5) & F(2,3,-1)
Let us assume that the vertices of the triangle are –
A(x₁, y₁, z₁), B(x₂, y₂, z₂) & C(x₃, y₃, z₃).
Now, we know that E is the mid-point of AC, thus,
(x₁ + x₃/2, y₁ + y₃/2, z₁ + z₃/2) = (0,8,5)
Thus,
C(x₃, y₃, z₃) = C(-x₁, 16 – y₁, -2-z­₁)
Now, the mid-point of AB, F is
(x₁ + x₂/2, y1 + y₂/2, z1 + z₂/2) = (2,3,-1)
Now, B(x₂, y₂, z₂) = B(4-x₁, 6-y₁, -1-z₁)
Now, the mid-point of BC, D is,
-x₁ + 4 – x₁/2 = 5,
16 – y₁ + 6 – y₁/2 = 7
& 10 – z₁ – 2 – z₁/2 = 11
Thus, x₁ = -3, y₁ = 4 & z₁ = -7
Thus, A = (-3,4,-7), B = (7,2,5) & C = (3,12,17)

Question:14

Three vertices of a Parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2). Find the fourth vertex D.

Answer:

Let us assume that the 4^th vertex of the parallelogram is D(x,y,z).
Now, mid-point of BD,
(1+2/2, 2+3/2, 3+2/2) = (x-1/2, y-2/2, z-1/2)
Thus, 3/2 = x-1/2
x = 4
5/2 = y-2/2
y = 7
& 5/2 = z-1/2
Thus, z = 6.
Therefore, the coordinates of the 4^th vertex are D(4,7,6).

Question:15

Find the coordinates of the points which trisect the line segment joining the points A (2, 1, – 3) and B (5, – 8, 3).

Answer:

Let the points P(x₁,y₁,z₁) & Q(x₂,y₂,z₂), trisect line segment AB & let P divide the line in the ratio of 1:2 & Q in the ratio of 2:1.
Thus, P(x₁,y₁,z₁) = P(1(5) + 2(2)/1+2, 1(-8) + 2(1)/1+2, 1(3) + 2(-3)/1+2)
= P(3,-2,-1)
& Q(x₂,y₂,z₂) = Q(2(5) + 1(2)/2+1, 2(-8) + 1(1)/2+1, 2(3) + 1(-3)/2+1)
= Q(4,-5,1)

Question:16

If the origin is the centroid of a triangle ABC having vertices A (a, 1, 3), B (– 2, b, – 5) and C (4, 7, c), find the values of a, b, c.

Answer:

Given: Vertices of triangle ABC are-
A(z,1,3), B(-2,b,-5) & C(4,7,c)
& centroid, G = (0,0,0)
Thus, G(0,0,0) = G(a-2+4/3, 1+b+7/3, 3-5+c/3)
Thus, a+2/3 = 0
a = -2
b+8/3 = 0
b = -8
c-2/3 = 0
c = 2

Question:17

Let A (2, 2, – 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at point D. Find the coordinates of D.

Answer:

Let us assume that the coordinates of D are (x,y,z).
Now,
AB = $\sqrt{(5-2)^{2}+(6-2)^{2}+(9+3)^{2} }$
= $\sqrt{9+16+144 }$ = 13
AC =$\sqrt{(2-2)^{2}+(7-2)^{2}+(9+3)^{2} }$
= $\sqrt{0+25+144 }$ = 13
Thus, AB = AC.
Thus, ABC is an isosceles triangle.
Now, D is the mid-point of BC or Angle bisector AD bisects BD,
Thus, D = (5+2/2, 6+7/2, 9+9/2)
= (7/2, 13/2,9)

Question:18

Show that the three points A (2, 3, 4), B (–1, 2, – 3) and C (– 4, 1, – 10) are collinear and find the ratio in which C divides AB.

Answer:

Given:
A(2,3,4), B(-1,2,-3) & C(-4,1,-10)
Now,
AB = $\sqrt{(2+1)^{2}+(3-2)^{2}+(4+3)^{2} }$
= $\sqrt{9+1+49 }$ = $\sqrt{59 }$
BC = $\sqrt{(-1+4)^{2}+(2-1)^{2}+(-3+10)^{2}}$
= $\sqrt{9+1+49 }$ = $\sqrt{59 }$
AC = $\sqrt{(2+4)^{2}+(3-1)^{2}+(4+10)^{2} }$
= $\sqrt{36+4+196 }$ = $2\sqrt{59}$
Now, AB + BC = $\sqrt{59}+\sqrt{59}$
= $2\sqrt{59}$
= AC
Thus, A, B & C are collinear.AC:BC = $2\sqrt{59}:\sqrt{59}$ = 2:1
Thus, C divides AB externally in the ratio 2:1.

Question:19

The midpoint of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4). Find its vertices. Also, find the centroid of the triangle.

Answer:

Given:
Mid-points of sides of triangle ABC are,
D(1,5,-1), E(0,4,-2) & F(2,3,4)
Let us assume that the vertices of the triangle are –
A(x₁, y₁, z₁), B(x₂, y₂, z₂) & C(x₃, y₃, z₃).
Now, we know that E is the mid-point of AC, thus,
(x₁ + x₃/2, y₁ + y₃/2, z₁ + z₃/2) = (0,4,-2)
Thus,
C(x₃, y₃, z₃) = C(-x₁, 8 – y₁, -4-z­₁)
Now, the mid-point of AB, F is
(x₁ + x₂/2, y₁ + y₂/2, z₁ + z₂/2) = (2,3,4)
Now, B(x₂, y₂, z₂) = B(4-x₁, 6-y₁, 8-z₁)
Now, the mid-point of BC, D is,
-x₁ + 4 – x₁/2 = 1,
8 – y₁ + 6 – y₁/2 = 5
& - 4 – z₁ + 8 – z₁/2 = -1
Thus, x₁ = 1, y₁ = 2 & z₁ = 3
Thus, A = (1,2,3), B = (3,4,5) & C = (-1,6,-7)
Now, for centroid,
G = (1+3-1/3,2+4+6/3, 3+5-7/3)
= (1,4,1/3)

Question:20

Prove that the points (0, – 1, – 7), (2, 1, – 9) and (6, 5, – 13) are collinear. Find the ratio in which the first point divides the join of the other two

Answer:

Given:
4(0,-1,-7), 8(2,1,-9) & C(6,5,-13)
Now,
AB = $\sqrt{(0-2)^{2}+(-1-1)^{2}+(-7+9)^{2} }$
$\sqrt{4+4+4}=2\sqrt{3}$
BC = $\sqrt{(2-6)^{2}+(1-5)^{2}+(-9+13)^{2} }$
$\sqrt{16+16+16}=4\sqrt{3}$
AC = $\sqrt{(0-6)^{2}+(-1-5)^{2}+(-7+13)^{2} }$
$\sqrt{36+36+36}=6\sqrt{3}$
Now, AB + BC = $2\sqrt{3}+4\sqrt{3}$
= $6\sqrt{3}$
= AC
Thus, A, B & C are collinear.
AB:AC = $2\sqrt{3}:6\sqrt{3}$
= 1:3
Thus, A divides BC externally in the ratio 1:3.

Question:21

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincide with the origin and the three edges passing through the origin, coincide with the positive direction of the axes through the origin?

Answer:

For the cube whose edge is 2 units, the coordinates will be,
(2,0,0), (2,2,0), (0,2,0), (0,2,2), (0,0,2), (2,0,2), (0,0,0) & (2,2,2).

Question:22

The distance of point P(3, 4, 5) from the yz-plane is
A. 3 units
B. 4 units
C. 5 units
D. 550

Answer:

The answer is the option (a) 3 units
Given: P(3,4,5)
Now, from the yz-plane, a distance of P,
=| x coordinate of P| = 3

Question:23

What is the length of the foot of perpendicular drawn from the point P (3, 4, 5) on the y-axis
A. $\sqrt{41}$
B. $\sqrt{34}$
C. 5
D. None of these

Answer:

The answer is the option (b)$\sqrt{ 34}$
On y-axis, x = z = 0
Thus, A = (0,4,0)
Thus, PA = $\sqrt{ (0-3)^{2}+(4-4)^{2}+(0-5)^{2} }$
= $\sqrt{ 9+0+25 }$ = $\sqrt{ 34 }$

Question:24

The distance of the point (3, 4, 5) from the origin (0, 0, 0) is
A. $\sqrt{50}$
B. 3
C. 4
D. 5

Answer:

The answer is the option (a) $\sqrt{50}$
Given: P(3,4,5) & O(0,0,0)
Thus, OP = $\sqrt{(0-3)^{2}+(0-4)2^{+}(0-5)^{2} }$
= $\sqrt{9+16+25}$ =$\sqrt{50}$

Question:25

If the distance between the points (a, 0, 1) and (0, 1, 2) is $\sqrt{27}$, then the value of a is
A. 5
B. ± 5
C. – 5
D. none of these

Answer:

The answer is the option (b) ±5
Given: A(a,0,1) & B(0,1,2)
& OP = $\sqrt{(a-0)^{2}+(0-1)^{2}+(1-2)^{2}}$
= $\sqrt{27}$
Thus, a² + 2 = 27
a² = 25
a = ±5

Question:26

x-axis is the intersection of two planes
A. xy and xz
B. yz and zx
C. xy and yz
D. none of these

Answer:

The answer is the option (a) xy & xz
ON xy-plane & xz-plane, x-axis is the line of intersection

Question:27

The equation of the y-axis is considered as
A. x = 0, y = 0
B. y = 0, z = 0
C. z = 0, x = 0
D. none of these

Answer:

The answer is the option (c) z = 0, x = 0
We know that, on y-axis,
x = 0 & z = 0.

Question:28

The point (–2, –3, –4) lies in the
A. First octant
B. Seventh octant
C. Second octant
D. Eighth octant

Answer:

The answer is option (b) seventh octant
The given point (-2,-3,-4) lies in the seventh octant

Question:29

A plane is parallel to the yz-plane so it is perpendicular to :
A. x-axis
B. y-axis
C. z-axis
D. none of these

Answer:

The answer is the option (a) x-axis
If a plane is parallel to the yz-plane, then it is perpendicular to the x-axis.

Question:30

The locus of a point for which y = 0, z = 0 is
A. equation of x-axis
B. equation of y-axis
C. equation at z-axis
D. none of these
Answer:

The answer is the option (a) equation of the x-axis
The locus of the point is the equation of the x-axis since we know that on the x-axis,
y = 0 & z = 0

Question:31

The locus of a point for which x = 0 is
A. xy-plane
B. yz-plane
C. zx-plane
D. none of these

Answer:

The answer is the option (b) yz-plane
We know that, on yz-plane,
x = 0
Thus, the locus of the point is the yz-plane

Question:32

If a parallelepiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of the diagonal of the parallelepiped is
A. $2\sqrt{3}$
B. $3\sqrt{2}$
C. $\sqrt{2}$
D. $\sqrt{3}$

Answer:

The answer is the option (a) $2\sqrt{3}$
Given: The parallelepiped passes through A(5,8,10) & B(3,6,8)
Thus,
Length of the diagonal, AB = $\sqrt{(5-3)^{2}+(8-6)^{2}+(10-8)^{2} }$
= $\sqrt{4+4+4}$=$2\sqrt{3}$

Question:33

L is the foot of the perpendicular drawn from a point P (3, 4, 5) on the xy-plane. The coordinates of point L are
A. (3, 0, 0)
B. (0, 4, 5)
C. (3, 0, 5)
D. none of these

Answer:

The answer is the option (d) none of these
On xy-plane, z = 0.
Thus, the coordinates of L are (3,4,0).

Question:34

L is the foot of the perpendicular drawn from a point (3, 4, 5) on the x-axis. The coordinates of L are
A. (3, 0, 0)
B. (0, 4, 0)
C. (0, 0, 5)
D. none of these

Answer:

The answer is the option (a) (3,0,0)
We know that, on the x-axis,
y = z = 0.
Thus, the coordinates will be, (3,0,0)

Question:35

Fill in the blanks
The three axes OX, OY, and OZ determine ________ .

Answer:

Coordinate planes.

Question:36

Fill in the blanks
The three planes determine a rectangular parallelopiped which has ________ of rectangular faces.

Answer:

Here, the rectangular parallelepiped is determined by the three planes, ABB’A, AA’D’D, and A’B’C’D’.
Here, we have 3 pairs of rectangular faces –
(ABB’A, DCC’D’), (ABCD, A’B’C’D’), (ADD’A’, BCC’B’)

Question:37

Fill in the blanks
The coordinates of a point are the perpendicular distance from the ________on the respective axes.

Answer:

Given points

Question:38

Fill in the blanks
The three coordinate planes divide the space into ________ parts

Answer:

Eight

Question:39

Fill in the blanks
If a point P lies in the yz-plane, then the coordinates of a point on the yz-plane are of the form ________.

Answer:

On yz-plane, x = 0.
Thus, the coordinates will be – (0,y,z).

Question:40

Fill in the blanks
The equation of yz-plane is ________.

Answer:

On yz-plane, x = 0.
Thus, the equation of yz-plane is x = 0

Question:41

Fill in the blanks
If the point P lies on the z-axis, then the coordinates of P are of the form ________.

Answer:

We know that,
On the z-axis, x = y = 0.
Thus, the coordinates are (0,0,z).

Question:42

Fill in the blanks
The equation of the z-axis is ________.

Answer:

On the z-axis, any point will be (0,0,z).
Thus, On the z-axis, x = y = 0, viz. represented in the equation.

Question:43

Fill in the blanks
A line is parallel to xy-plane if all the points on the line have equal ________.

Answer:

If each point P(x,y,z) is at the same distance from the xy-plane, then the line is parallel to the xy-plane.
Thus, distance of xy-plane from point P = z
Thus, if all the points on the line have equal z coordinates then the line is parallel to xy-plane.

Question:44

Fill in the blanks
A line is parallel to the x-axis if all the points on the line have equal ________.

Answer:

If each point on a line maintains a constant distance from the y-axis and z-axis, then it is parallel to the x-axis.
Therefore, each point has equal y & z coordinates.

Question:45

Fill in the blanks
x = a represents a plane parallel to ________.

Answer:

x = a is the locus of point P(x,y,z).
Thus, each point P has a constant x-coordinate.
Also, the distance of the yz-plane from P is ‘x’.
Thus, plane x = a is parallel to yz-plane & is at a constant distance ‘a’ from yz-plane.

Question:46

Fill in the blanks
The plane parallel to yz - plane is perpendicular to ________.

Answer:

x-axis

Question:47

Fill in the blanks
The length of the longest piece of a string that can be stretched straight in a rectangular room whose dimensions are 10, 13 and 8 units are ______.

Answer:

Given:
a = 10, b = 13 & c = 8.
Now, the length of the string –
$\\ =\sqrt{a^2+b^2+c^2} \\= \sqrt{10^2+13^2+8^2}\\ =\sqrt{100+169+64}\\ = \sqrt{333}$

Question:48

Fill in the blanks
If the distance between the points (a, 2, 1) and (1, –1, 1) is 5, then a _______.

Answer:

Given points: (a,2,1,) & (1,-1,1)
$\sqrt{(a-1)^{2}+(2+1)^{2}+(1-1)^{2}}$ =5
(a-1)² + 9 + 0 = 25
a² – 2a – 15 = 0
(a-5)(a+3) = 0
Thus, a = -3 or a = 5

Question:49

Fill in the blanks
If the mid-points of the sides of a triangle AB; BC; CA are D (1, 2, – 3), E (3, 0, 1) and F (–1, 1, – 4), then the centroid of the triangle ABC is ________.

Answer:

Given:
Mid-points of triangle ABC are D(1,2,-3), E(3,0,1) & F(-1,1,-4).
Now, the centroid of triangle DEF = Centroid of triangle ABC ……. (from geometry of centroid)
Thus, the centroid of triangle ABC,
G(1+3-1/3, 2+0+1/3, -3+1-4/3) ≡ G(1,1,-2)

Question:50

Match each item given under column C₁ to its correct answer given under column C₂.

Answer:

Solution:
(a) → (iii)
In xy-plane, z = 0.
(b)→(i)
Point (2,3,4) lies in the first octant.
(c) → (ii)
Yz-plane is the locus of points having, x = 0
(d) → (vi)
Only if all the points on a line have equal y & z coordinates, the line will be parallel to the -axis.
(e) → (iv)
z-axis is represented by x = y = 0.
(f) → (v)
A plane parallel to the xy-plane is represented by z = c.
(g) → (viii)
Planes x = a & y = b is a line of intersection of these planes, since x = a is parallel to yz-plane & y = b is parallel to xz-plane.
Z-axis is the line of intersection of yz & xz planes.
Thus, the line of intersection is parallel to the z-axis.
(h) → (vii)
The coordinate of a point can be defined as the distance from the origin to the feet of the perpendicular from the point on their respective axis.

(i)→ (x)

A ball can be said a solid region in the space enclosed by the sphere.
(j) → (ix)
A disc is a region in a plane enclosed by a circle.