NCERT Solutions
Class 11 Maths
Introduction to Three Dimensional Geometry
Ex.12.2 Q.3
Verify the following:
(1) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.
(2) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right-angled triangle.
(3) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.
(1) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively.
Now,
AB = √ {(1 - 0)2 + (6 - 7)2 + (-6 + 10)2}
= √ {12 + (-1)2 + 42}
= √ (1 + 1 + 16)
= √18
= 3√2
BC = √ {(4 - 1)2 + (9 - 6)2 + (-6 + 6)2}
= √ {32 + 32 + 0}
= √ (9 + 9)
= √18
= 3√2
CA = √ {(0 - 4)2 + (7 - 9)2 + (-10 + 6)2}
= √ {(-4)2 + (-2)2 +(-4)2}
= √ (16 + 4 + 16)
= √36
= 6
Here, AB = BC ≠ CA
Thus, the given points are the vertices of an isosceles triangle.
(2) Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.
AB = √ {(-1 - 0)2 + (6 - 7)2 + (6 - 10)2}
= √ {(-1)2 + (-1)2 + (-4)2}
= √ (1 + 1 + 16)
= √18
= 3√2
BC = √ {(-4 + 1)2 + (9 - 6)2 + (6 - 6)2}
= √ {(-3)2 + 32 + 0}
= √ (9 + 9)
= √18
= 3√2
CA = √ {(0 + 4)2 + (7 - 9)2 + (10 - 6)2}
= √ {42 + (-2)2 + 42}
= √ (16 + 4 + 16)
= √36
= 6
Now, AB2 + BC2 = (3√2)2 + (3√2)2 = 18 + 18 = 36 = AC2
Therefore, by Pythagoras theorem, ABC is a right triangle.
Hence, the given points are the vertices of a right-angled triangle.
(3) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively.
AB = √ {(1 + 1)2 + (-2 + 2)2 + (5 - 1)2}
= √ {22 + (-4)2 + (4)2}
= √ (4 + 16 + 16)
= √36
= 6
BC = √ {(4 - 1)2 + (-7 + 2)2 + (8 - 5)2}
= √ {32 + (-5)2 + 32}
= √ (9 + 25 + 9)
= √43
CD = √ {(2 - 4)2 + (-3 + 7)2 + (4 - 8)2}
= √ {(-2)2 + 42 + (-4)2}
= √ (4 + 16 + 16)
= √36
= 6
DA = √ {(-1 - 2)2 + (2 + 3)2 + (1 - 4)2}
= √ {(-3)2 + 52 + (-3)2}
= √ (9 + 25 + 9)
= √43
= 6
Here, AB = CD = 6 and BC = AD = √43
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.