Introduction to Three Dimensional Geometry NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download
Exercise 11.1
Q1: A point is on the x-axis. What are its y-coordinate and z-coordinates?
Ans: If a point is on the x-axis, then the coordinates of y and z are 0.
So the point is (x, 0, 0).
Q2: A point is in the XZ-plane. What can you say about its y-coordinate?
Ans: If a point is in the XZ plane, then its y-co-ordinate is 0.
Q3: Name 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).
Ans: Here is the table which represents the octants:
(i) (1, 2, 3)
Here, x is positive, y is positive, and z is positive.
So, it lies in the I octant.
(ii) (4, -2, 3)
Here, x is positive, y is negative, and z is positive.
So, it lies in the IV octant.
(iii) (4, -2, -5)
Here, x is positive, y is negative, and z is negative.
So, it lies in the VIII octant.
(iv) (4, 2, -5)
Here, x is positive, y is positive, and z is negative.
So, it lies in the V octant.
(v) (-4, 2, -5)
Here, x is negative, y is positive, and z is negative.
So, it lies in VI octant.
(vi) (-4, 2, 5)
Here, x is negative, y is positive, and z is positive.
So, it lies in the II octant.
(vii) (-3, -1, 6)
Here, x is negative, y is negative, and z is positive.
So, it lies in the III octant.
(viii) (2, -4, -7)
Here, x is positive, y is negative, and z is negative.
So, it lies in the VIII octant.
Q4: Fill 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.
Ans: (i) The x-axis and y-axis, taken together, determine a plane known as XY Plane.
(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).
(iii) Coordinate planes divide the space into eight octants.
Exercise 11.2
Q1: Find 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)
Ans: (i) (2, 3, 5) and (4, 3, 1)
Let P be (2, 3, 5) and Q be (4, 3, 1)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 2, y1 = 3, z1 = 5
x2 = 4, y2 = 3, z2 = 1
Distance PQ = √[(4 – 2)2 + (3 – 3)2 + (1 – 5)2]
= √[(2)2 + 02 + (-4)2]
= √[4 + 0 + 16]
= √20
= 2√5
∴ The required distance is 2√5 units.
(ii) (–3, 7, 2) and (2, 4, –1)
Let P be (– 3, 7, 2) and Q be (2, 4, –1)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = – 3, y1 = 7, z1 = 2
x2 = 2, y2 = 4, z2 = – 1
Distance PQ = √[(2 – (-3))2 + (4 – 7)2 + (-1 – 2)2]
= √[(5)2 + (-3)2 + (-3)2]
= √[25 + 9 + 9]
= √43
∴ The required distance is √43 units.
(iii) (–1, 3, –4) and (1, –3, 4)
Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 1, y1 = 3, z1 = – 4
x2 = 1, y2 = – 3, z2 = 4
Distance PQ = √[(1 – (-1))2 + (-3 – 3)2 + (4 – (-4))2]
= √[(2)2 + (-6)2 + (8)2]
= √[4 + 36 + 64]
= √104
= 2√26
∴ The required distance is 2√26 units.
(iv) (2, –1, 3) and (–2, 1, 3)
Let P be (2, – 1, 3) and Q be (– 2, 1, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 1, z1 = 3
x2 = – 2, y2 = 1, z2 = 3
Distance PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
∴ The required distance is 2√5 units.
Q2: Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
Ans: If three points are collinear, then they lie on the same line.
First, let us calculate the distance between the 3 points
i.e., PQ, QR and PR
Calculating PQ
P ≡ (–2, 3, 5) and Q ≡ (1, 2, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = – 2, y1 = 3, z1 = 5
x2 = 1, y2 = 2, z2 = 3
Distance PQ = √[(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]
= √[(3)2 + (-1)2 + (-2)2]
= √[9 + 1 + 4]
= √14
Calculating QR
Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 2, z1 = 3
x2 = 7, y2 = 0, z2 = – 1
Distance QR = √[(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]
= √[(6)2 + (-2)2 + (-4)2]
= √[36 + 4 + 16]
= √56
= 2√14
Calculating PR
P ≡ (–2, 3, 5) and R ≡ (7, 0, – 1)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = –2, y1 = 3, z1 = 5
x2 = 7, y2 = 0, z2 = –1
Distance PR = √[(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]
= √[(9)2 + (-3)2 + (-6)2]
= √[81 + 9 + 36]
= √126
= 3√14
Thus, PQ = √14, QR = 2√14 and PR = 3√14
So, PQ + QR = √14 + 2√14
= 3√14
= PR
∴ The points P, Q and R are collinear.
Q3: 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.
Ans: (i) (0, 7, –10), (1, 6, –6), and (4, 9, – 6) are the vertices of an isosceles triangle.
Let us consider the points,
P(0, 7, –10), Q(1, 6, –6) and R(4, 9, –6)
If any 2 sides are equal, it will be an isosceles triangle
So, first, let us calculate the distance of PQ, QR
Calculating PQ
P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here,
x1 = 0, y1 = 7, z1 = – 10
x2 = 1, y2 = 6, z2 = – 6
Distance PQ = √[(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]
= √[(1)2 + (-1)2 + (4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 6, z1 = – 6
x2 = 4, y2 = 9, z2 = – 6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Hence, PQ = QR
18 = 18
2 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.
Let the points be
P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)
First, let us calculate the distance of PQ, OR and PR
Calculating PQ
P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 0, y1 = 7, z1 = 10
x2 = –1, y2 = 6, z2 = 6
Distance PQ = √[(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]
= √[(-1)2 + (-1)2 + (-4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, –6) and R ≡ (4, 9, –6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 6, z1 = –6
x2 = 4, y2 = 9, z2 = –6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Calculating PR
P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 0, y1 = 7, z1 = 10
x2 = – 4, y2 = 9, z2 = 6
Distance PR = √[(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]
= √[(-4)2 + (2)2 + (-4)2]
= √[16 + 4 + 16]
= √36
Now,
PQ2 + QR2 = 18 + 18
= 36
= PR2
By using the converse of Pythagoras theorem,
∴ 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.
Let the points: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)
ABCD can be vertices of parallelogram only if opposite sides are equal.
i.e., AB = CD and BC = AD
First, let us calculate the distance
Calculating AB
A ≡ (–1, 2, 1) and B ≡ (1, –2, 5)
By using the formula,
Distance AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = – 1, y1 = 2, z1 = 1
x2 = 1, y2 = – 2, z2 = 5
Distance AB = √[(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]
= √[(2)2 + (-4)2 + (4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating BC
B ≡ (1, –2, 5) and C ≡ (4, –7, 8)
By using the formula,
Distance BC = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = – 2, z1 = 5
x2 = 4, y2 = – 7, z2 = 8
Distance BC = √[(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]
= √[(3)2 + (-5)2 + (3)2]
= √[9 + 25 + 9]
= √43
Calculating CD
C ≡ (4, –7, 8) and D ≡ (2, –3, 4)
By using the formula,
Distance CD = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 4, y1 = –7, z1 = 8
x2 = 2, y2 = –3, z2 = 4
Distance CD = √[(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]
= √[(-2)2 + (4)2 + (-4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating DA
D ≡ (2, –3, 4) and A ≡ (–1, 2, 1)
By using the formula,
Distance DA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 2, y1 = – 3, z1 = 4
x2 = –1, y2 = 2, z2 = 1
Distance DA = √[(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]
= √[(-3)2 + (5)2 + (-3)2]
= √[9 + 25 + 9]
= √43
Since AB = CD and BC = DA (given),
In ABCD, both pairs of opposite sides are equal.
∴ ABCD is a parallelogram.
Q4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Ans: 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
First, let us calculate
Calculating PA
P ≡ (x, y, z) and A ≡ (1, 2, 3)
By using the formula,
Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = x, y1 = y, z1 = z
x2 = 1, y2 = 2, z2 = 3
Distance PA = √[(1 – x)2 + (2 – y)2 + (3 – z)2]
Calculating PB
P ≡ (x, y, z) and B ≡ (3, 2, –1)
By using the formula,
Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = x, y1 = y, z1 = z
x2 = 3, y2 = 2, z2 = –1
Distance PB = √[(3 – x)2 + (2 – y)2 + (-1 – z)2]
Since PA = PB
Square on both sides, we get
PA2 = PB2
(1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2
(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)
(9 + x2 – 6x) + (4 + y2 – 4y) + (1 + z2 + 2z)
– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
4x – 8z = 0
x – 2z = 0
∴ The required equation is x – 2z = 0.
Q5: 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.
Ans: Let A(4, 0, 0) & B(– 4, 0, 0)
Let the coordinates of point P be (x, y, z)
Calculating PA
P ≡ (x, y, z) and A ≡ (4, 0, 0)
By using the formula,
Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = x, y1 = y, z1 = z
x2 = 4, y2 = 0, z2 = 0
Distance PA = √[(4– x)2 + (0 – y)2 + (0 – z)2]
Calculating PB,
P ≡ (x, y, z) and B ≡ (– 4, 0, 0)
By using the formula,
Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = x, y1 = y, z1 = z
x2 = – 4, y2 = 0, z2 = 0
Distance PB = √[(-4– x)2 + (0 – y)2 + (0 – z)2]
It is given that,
PA + PB = 10
PA = 10 – PB
Square on both sides, we get
PA2 = (10 – PB)2
PA2 = 100 + PB2 – 20 PB
(4 – x)2 + (0 – y)2 + (0 – z)2
100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB
(16 + x2 – 8x) + (y2) + (z2)
100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB
20 PB = 16x + 100
5 PB = (4x + 25)
Square on both sides again, we get
25 PB2 = 16x2 + 200x + 625
25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625
25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625
25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625
9x2 + 25y2 + 25z2 – 225 = 0
∴ The required equation is 9x2 + 25y2 + 25z2 – 225 = 0.
|
74 videos|245 docs|91 tests
|
Exam
,study material
,Introduction to Three Dimensional Geometry NCERT Solutions | Mathematics (Maths) Class 11 - Commerce
,Free
,Important questions
,Previous Year Questions with Solutions
,Introduction to Three Dimensional Geometry NCERT Solutions | Mathematics (Maths) Class 11 - Commerce
,Semester Notes
,past year papers
,Introduction to Three Dimensional Geometry NCERT Solutions | Mathematics (Maths) Class 11 - Commerce
,Extra Questions
,Summary
,Sample Paper
,video lectures
,practice quizzes
,Objective type Questions
,MCQs
,Viva Questions
,shortcuts and tricks
,mock tests for examination
,ppt
;
NCERT Solutions: Introduction to Three Dimensional Geometry Free PDF Download
Importance of NCERT Solutions: Introduction to Three Dimensional Geometry
NCERT Solutions: Introduction to Three Dimensional Geometry Notes
NCERT Solutions: Introduction to Three Dimensional Geometry Commerce Questions
Study NCERT Solutions: Introduction to Three Dimensional Geometry on the App
|
© EduRev
|
Education Revolution
|
|
