NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions
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Question : 7
Total: 20
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.
(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) Let A(0, 7, –10), B(1, 6, –6) and C(4, 9, –6) be three vertices of triangle ABC. Then
AB =√ ( 1 − 0 ) 2 + ( 6 − 7 ) 2 + ( − 6 + 10 ) 2 √ 1 + 1 + 16 √ 18 3 √ 2
BC =√ ( 4 − 1 ) 2 + ( 9 − 6 ) 2 + ( − 6 + 6 ) 2 √ 9 + 9 + 0 √ 18 3 √ 2
AC =√ ( 4 − 0 ) 2 + ( 9 − 7 ) 2 + ( − 6 + 10 ) 2 √ 16 + 4 + 16 √ 36
Now, AB = BC
Thus, ABC is an isosceles triangle.
(ii) Let A(0, 7, 10), B(–1, 6, 6) and C(–4, 9, 6) be three vertices of triangle ABC. Then
AB =√ ( − 1 − 0 ) 2 + ( 6 − 7 ) 2 + ( 6 − 10 ) 2 √ 1 + 1 + 16 √ 18 3 √ 2
BC =√ ( − 4 + 1 ) 2 + ( 9 − 6 ) 2 + ( 6 − 6 ) 2 √ 9 + 9 + 0 √ 18 3 √ 2
AC =√ ( − 4 − 0 ) 2 + ( 9 − 7 ) 2 + ( 6 − 10 ) 2 √ 16 + 4 + 16 √ 36
Now,A C 2 A B 2 + B C 2
Thus, ABC is a right angled triangle.
(iii) Let A(–1, 2, 1), B(1, –2, 5) and C(4, –7, 8) and D(2, –3, 4) be four vertices of quadrilateral ABCD. Then
AB =√ ( 1 + 1 ) 2 + ( − 2 − 2 ) 2 + ( 5 − 1 ) 2 √ 4 + 16 + 16 √ 36
BC =√ ( 4 − 1 ) 2 + ( − 7 + 2 ) 2 + ( 8 − 5 ) 2 √ 9 + 25 + 9 √ 43
CD =√ ( 2 − 4 ) 2 + ( − 3 + 7 ) 2 + ( 4 − 8 ) 2 √ 4 + 46 + 16 √ 36
AD =√ ( 2 + 1 ) 2 + ( − 3 − 2 ) 2 + ( 4 − 1 ) 2 √ 9 + 25 + 9 √ 43
AC =√ ( 4 + 1 ) 2 + ( − 7 − 2 ) 2 + ( 8 − 1 ) 2 √ 25 + 81 + 49 √ 155
BD =√ ( 2 − 1 ) 2 + ( − 3 + 2 ) 2 + ( 4 − 5 ) 2 √ 1 + 1 + 1 √ 3
Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.
AB =
BC =
AC =
Now, AB = BC
Thus, ABC is an isosceles triangle.
(ii) Let A(0, 7, 10), B(–1, 6, 6) and C(–4, 9, 6) be three vertices of triangle ABC. Then
AB =
BC =
AC =
Now,
Thus, ABC is a right angled triangle.
(iii) Let A(–1, 2, 1), B(1, –2, 5) and C(4, –7, 8) and D(2, –3, 4) be four vertices of quadrilateral ABCD. Then
AB =
BC =
CD =
AD =
AC =
BD =
Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.
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