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) and R (4, 9, - 6)
If any 2 sides are equal, hence it will be an isosceles triangle. So, first, let us calculate the length of the sides.
Calculating PQ:
P = (0, 7, - 10) and Q = (1, 6, - 6)
By using the 3d distance formula,
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
PQ = √(1 - 0)² + (6 - 7)² + (- 6 - (- 10))²
= √1² + (- 1)² +(4)²
= √1 + 1 + 16
= √18
Calculating QR:
Q = (1, 6, - 6) and R = (4, 9, - 6)
By using the formula,
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
QR = √(4 - 1)² + (9 - 6)² + (- 6 - (- 6))²
= √3² + 3² + 0²
= √9 + 9
= √18
Here, PQ = QR = √18
Since two sides are equal, ΔPQR is an isosceles triangle.
Thus, (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.
Let the points be P (0, 7, 10), Q (- 1, 6, 6) and R (- 4, 9, 6)
First, let us calculate the length of sides PQ, QR and PR.
Calculating PQ:
P = (0, 7, 10) and Q = (- 1, 6, 6)
By using the formula,
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
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 = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
QR = √(- 4 - (- 1))² + (9 - 6)² + (6 - 6)²
= √(- 3)² + 3² + 0²
= √9 + 9
= √18
Here, PQ = QR = √18
Since, two sides are equal, ΔPQR is an isosceles triangle.
Thus, (0, 7, - 10), (1, 6, - 6) and (4, 9, - 6) are the vertices of an isosceles triangle.
Calculating PR:
P = (0, 7, 10) and R = (- 4, 9, 6)
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
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 Pythagoras theorem, the given vertices P, Q and R are the vertices of a right-angled triangle at Q.
Thus, (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.
Let the points be A(- 1, 2, 1), B (1, - 2, 5), C (4, - 7,8) and D (2, - 3, 4).
If pairs of opposite sides are equal then only ABCD can be a parallelogram.
First, let us calculate the lengths of the sides.
Calculating AB:
A(- 1, 2, 1) and B (1, - 2, 5)
By using Distance formula,
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
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 = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
BC = √(4 - 1)² + (7 - (- 2))² + (8 - 5)²
= √(3)² + (- 5)² + 3²
= √9 + 25 + 9
= √43
Calculating CD:
C (4, - 7,8) and D (2, - 3, 4)
By using the distance formula,
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
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 distance formula,
Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²
Using this,
DA = √(- 1 - 2)² + (2 - (- 3))² + (1 - 4)²
= √(- 3)² + (5)² + (- 3)²
= √9 + 25 + 9
= √43
Since in quadrilateral ABCD both the pairs of opposite sides are equal i.e., AB = CD and BC = AD, ABCD is a parallelogram.
Thus, (- 1, 2, 1), (1, - 2, 5), (4, - 7,8) and (2, - 3, 4) are the vertices of a parallelogram
NCERT Solutions Class 11 Maths Chapter 12 Exercise 12.2 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.
Summary:
(i) Yes, (0, 7, - 10), (1, 6, - 6) and (4, 9, - 6) are the vertices of an isosceles triangle.
(ii) Yes, (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of a right angled triangle.
(iii) Yes, (- 1, 2, 1), (1, - 2, 5), (4, - 7,8) and (2, - 3, 4) are the vertices of a parallelogram.
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