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NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions

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Question : 7 of 20
Marks: +1, -0
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) 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\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2} = 1+1+16\sqrt{1+1+16} = 18\sqrt{18} = 323\sqrt{2} units
BC = (4−1)2+(9−6)2+(−6+6)2\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2} = 9+9+0\sqrt{9+9+0} = 18\sqrt{18} = 323\sqrt{2} units
AC = (4−0)2+(9−7)2+(−6+10)2\sqrt{(4-0)^2+(9-7)^2+(-6+10)^2} = 16+4+16\sqrt{16+4+16} = 36\sqrt{36} = 6 units
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\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2} = 1+1+16\sqrt{1+1+16} = 18\sqrt{18} = 323\sqrt{2} units
BC = (−4+1)2+(9−6)2+(6−6)2\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2} = 9+9+0\sqrt{9+9+0} = 18\sqrt{18} = 323\sqrt{2} units
AC = (−4−0)2+(9−7)2+(6−10)2\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2} = 16+4+16\sqrt{16+4+16} = 36\sqrt{36} = 6 units
Now, AC2AC^2 = AB2+BC2AB^2 + BC^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\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2} = 4+16+16\sqrt{4+16+16} = 36\sqrt{36} = 6 units
BC = (4−1)2+(−7+2)2+(8−5)2\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2} = 9+25+9\sqrt{9+25+9} = 43\sqrt{43} units
CD = (2−4)2+(−3+7)2+(4−8)2\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2} = 4+46+16\sqrt{4+46+16} = 36\sqrt{36} = 6 units
AD = (2+1)2+(−3−2)2+(4−1)2\sqrt{(2+1)^2+(-3-2)^2+(4-1)^2} = 9+25+9\sqrt{9+25+9} = 43\sqrt{43} units
AC = (4+1)2+(−7−2)2+(8−1)2\sqrt{(4+1)^2+(-7-2)^2+(8-1)^2} = 25+81+49\sqrt{25+81+49} = 155\sqrt{155} units
BD = (2−1)2+(−3+2)2+(4−5)2\sqrt{(2-1)^2+(-3+2)^2+(4-5)^2} = 1+1+1\sqrt{1+1+1} = 3\sqrt{3} units
Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.
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