Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11

by

Students can go through AP Inter 1st Year Maths Notes 11th Lesson Introduction to 3D Geometry will help students in revising the entire concepts quickly.

Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths 11th Lesson

→ In three dimensions, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the x, y, and z-axes.

→ The three planes determined by the pair of axes are the coordinate planes, called XY, YZ, and ZX-planes.

→ The three coordinate planes divide the space into eight parts known as octants.

→ The coordinates of a point P in three-dimensional geometry are always written in the form of a triplet like (x, y, z). Here, x, y, and z are the distances of P from the YZ, ZX, and XY-planes.

Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11

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

→ Any point on the y-axis is of the form (0, y, 0)

→ Any point on the z-axis is of the form (0, 0, z).

→ Distance between two points P(x1, y1, z1) and Q(x2, y2, z2) is given by PQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

→ The point (x, y, z) divides the line segmentjoining the points (x1, y1, z1) and (x2, y2, z2) in the ratio \(\frac{-\left(x-x_1\right)}{\left(x-x_2\right)}\) (or) \(\frac{-\left(y-y_1\right)}{\left(y-y_2\right)}\) (or) \(\frac{-\left(z-z_1\right)}{\left(z-z_2\right)}\)

→ The centroid of the triangle whose vertices A(x1, y1, z1), B(x2, y2, z2) & C(x3, y3, z3) is \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)\)

Example Problems

Question 1.
In the Figure, if P is (2, 4, 5), find the coordinates of F.
Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11 Q1
Solution:
For point F, the distance measured along OY is zero.
Therefore, the coordinates of Fare are (2, 0, 5).

Question 2.
Find the octant in which the points (-3, 1, 2) and (-3, 1, -2) lie.
Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11 Q2
Solution:
From the Table, the point (-3, 1, 2) lies in the second octant, and the point (-3, 1, -2) lies in octant VI.

Question 3.
Find the distance between the points P(1, -3, 4) and Q(-4, 1, 2).
Solution:
The distance PQ between the points P(1, -3, 4) and Q(-4, 1, 2) is
PQ = \(\sqrt{(-4-1)^2+(1+3)^2+(2-4)^2}\)
= \(\sqrt{25+16+4}\)
= √45
= 3√5 units

Question 4.
Show that the points P(-2, 3, 5), Q(1, 2, 3) and R(7, 0, -1) are collinear.
Solution:
We know that points are said to be collinear if they lie on a line.
Now, PQ = \(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}\)
= \(\sqrt{9+1+4}\)
= √14
QR = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}\)
= \(\sqrt{36+4+16}\)
= √56
= 2√14
and PR = \(\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}\)
= \(\sqrt{81+9+36}\)
= √126
= 3√14
Thus, PQ + QR = PR.
Hence, P, Q, and R are collinear.

Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11

Question 5.
Are the points A(3, 6, 9), B(10, 20, 30), and C(25, -41, 5), the vertices of a right angled triangle?
Solution:
By the distance formula, we have
AB2 = (10 – 3)2 + (20 – 6)2 + (30 – 9)2
= 49 + 196 + 441
= 686
BC2 = (25 – 10)2 + (-41 – 20)2 + (5 – 30)2
= 225 + 3721 + 625
= 4571
CA2 = (3 – 25)2 + (6 + 41)2 + (9 – 5)2
= 484 + 2209 + 16
= 2709
We find that CA2 + AB2 ≠ BC2
Hence, the triangle ABC is not right-angled.

Question 6.
Find the equation of the set of points P such that PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (-1, 3, -7), respectively.
Solution:
Let the coordinates of point P be (x, y, z).
Here PA2 = (x – 3)2 + (y – 4)2 + (z – 5)2
PB2 = (x + 1)2 + (y – 3)2 + (z + 7)2
By the given condition PA2 + PB2 = 2k2
we have (x – 3)2 + (y – 4)2 + (z – 5)2 + (x + 1)2 + (y – 3)2 + (z + 7)2 = 2k2
⇒ 2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 109

Question 7.
Show that the points A(1, 2, 3), B(-1, 2, -1), C(2, 3, 2), and D(4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
Solution:
To show ABCD is a parallelogram, we need to show that opposite sides are equal.
Note that AB = \(\sqrt{(-1-1)^2+(-2-2)^2+(-1-3)^2}\)
= \(\sqrt{4+16+16}\)
= 6
BC = \(\sqrt{(2+1)^2+(3+2)^2+(2+1)^2}\)
= \(\sqrt{9+25+9}\)
= √43
CD = \(\sqrt{(4-2)^2+(7-3)^2+(6-2)^2}\)
= \(\sqrt{4+16+16}\)
= 6
DA = \(\sqrt{(1-4)^2+(2-7)^2+(3-6)^2}\)
= \(\sqrt{9+25+9}\)
= √43
Since AB = CD and BC = AD, ABCD is a parallelogram.
Now, it is required to prove that ABCD is not a rectangle.
For this, we show that diagonals AC and BD are unequal.
We have AC = \(\sqrt{(2-1)^2+(3-2)^2+(2-3)^2}\)
= \(\sqrt{1+1+1}\)
= √3
BD = \(\sqrt{(4+1)^2+(7+2)^2+(6+1)^2}\)
= \(\sqrt{25+81+49}\)
= √155
Since AC ≠ BD, ABCD is not a rectangle.

Question 8.
Find the equation of the set of points P such that its distances from the points A(3, 4, -5) and B(-2, 1, 4) are equal.
Solution:
If P(x, y, z) is a point such that PA = PB.
then \(\sqrt{(x-3)^2+(y-4)^2+(z+5)^2}\) = \(\sqrt{(x+2)^2+(y-1)^2+(z-4)^2}\)
⇒ (x – 3)2 + (y – 4)2 + (z + 5)2 = (x + 2)2 + (y – 1)2 + (z – 4)2
⇒ 10x + 6y – 18z – 29 = 0

Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11

Question 9.
The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, -5, 7) and (-1, 7, -6), respectively, find the coordinates of point C.
Solution:
Let the coordinates of C be (x, y, z) and the coordinates of the centroid G be (1, 1, 1).
Then \(\frac{x+3-1}{3}\) = 1; \(\frac{y-5+7}{3}\) = 1; \(\frac{z+7-6}{3}\) = 1
⇒ x = 1; y = 1; z = 2
Hence, the coordinates of C are (1, 1, 2).

Multiple Choice Questions

Question 1.
The co-ordinates of the foot of the perpendicular from a point P(6, 7, 8) on the y-axis are
(1) (6, 0, 0)
(2) (0, 7, 0)
(3) (0, 0, 8)
(4) (0, 7, 8)
Answer:
(2) (0, 7, 0)

Question 2.
The length of the perpendicular drawn from the point P(13, 5, 12) on the x-axis is
(1) 13
(2) 2√18
(3) √194
(4) √313
Answer:
(1) 13

Question 3.
If a parallelepiped is formed by planes drawn through the points (1, 2, 5) and (4, 5, 8) parallel to the coordinate planes, then the length of the diagonal of the parallelepiped is
(1) 2√2
(2) 3
(3) √3
(4) 3√3
Answer:
(4) 3√3

Question 4.
The Locus of the point for which y = 0 is
(1) xy-plane
(2) yz-plane
(3) zx-plane
(4) None of these
Answer:
(3) zx-plane

Question 5.
If the distance between the points P(0, a, 3) and Q(3, 0, 7) is √41, then a = ____
(1) -4
(2) 4
(3) ±4
(4) 4
Answer:
(3) ±4

Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11

Question 6.
The ratio in which the line joining (2, 4, 5) and (3, 5, -9) is divided by the yz-plane
(1) 2 : 3
(2) 3 : 2
(3) -2 : 3
(4) 4 : -3
Answer:
(3) -2 : 3

Question 7.
The points (1, 1, 1), (4, 1, 1), (4, 5, 1), and (1, 5, 1) are the vertices of
(1) a rectangle
(2) a square
(3) parallelogram
(4) None of these
Answer:
(3) parallelogram

Question 8.
What is the locus of a point for which y = 0, z = 0?
(1) x-axis
(2) y-axis
(3) z-axis
(4) yz-plane
Answer:
(1) x-axis

Question 9.
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
(1) 3 : 7
(2) 2 : 7
(3) -3 : 7
(4) -2 : 4
Answer:
(3) -3 : 7

Question 10.
The co-ordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz-plane are
(1) (3, 4, 0)
(2) (0, 4, 5)
(3) (3, 0, 5)
(4) (3, 0, 0)
Answer:
(2) (0, 4, 5)

Introduction to 3D Geometry Class 11 Notes AP Inter 1st Year Maths Chapter 11

Question 11.
Statement 1 (Assertion): Point (3, a2 – 1, a2 – 3a + 2) lies on the x-axis if a = 1.
Statement 2 (Reason): On the x-axis, the y and z co-ordinates of every point are each equal to zero.
(1) Statement-1 and statement-2 are true, statement-2 is a correct explanation for statement-1
(2) Statement-1 and statement-2 are true, statement-2 is not a correct explanation for statement-1
(3) Statement-1 is true, statement-2 is false
(4) Statement-1 is false, statement-2 is true
Answer:
(1) Statement-1 and statement-2 are true, statement-2 is a correct explanation for statement-1

Leave a Comment