Question

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.

Class X1 - Maths -Introduction to Three Dimensional Geometry Page 273

Answer 1

(i) Let
A ≡ (0, 7, -10)
B ≡ ( 1, 6, -6)
C ≡ ( 4, 9, -6 )
now, use distance formula,
AB = √{(1-0)² + (6-7)² + (-6+10)²}
= √{ 1² + (-1)² + 4²}
= √18 = 3√2

BC = √{(4 -1)² + (9-6)²+(-6+6)² }
= √{3² + 3² + 0² }
= √18 = 3√2

here it is clear that
AB = BC ,hence, triangle is an isosceles triangle .


(ii)
Let
A ≡ ( 0, 7, 10)
B ≡ ( -1, 6, 6)
C ≡ ( -4, 9, -6)

use distance formula,
AB = √{(-1-0)² + ( 6-7)² + (6-10)²}
= √{ 1 + 1 + 16 } = 3√2

BC = √{(-4+1)² + (9-6)² + (6-6)² }
= √{ 9 + 9 } = 3√2

CA = √{(10-6)² + (7-9)² + (0+4)²}
= √{ 16 + 4 + 16 } = 6

here, it is clear that ,
AB² = 18
BC² = 18
CA² = 36
e.g CA² = AB² + BC²
from Pythagoras theorem,
∆ABC is a right angled∆.


(iii) Let
A ≡ ( -1, 2, 1) = (x1, y1, z1)
B ≡ ( 1 , -2, 5) = (x2, y2, z2)
C ≡ ( 4, -7 , 8) = (x3, y3, z3)
D ≡ ( 2, -3 , 4) =(x4, y4, z4)

The midpoint of AC = {(x1 +x2 )/2 , (y1 + y2 )/2 + (z1 + z2 )/2 }
= { (-1+4)/2, (2-7)/2, (1+8)/2 }
= { 3/2, -5/2, 9/2 }

The midpoint of BD = {(x2 + x4)/2 , (y2 + y4)/2 , (z2+z4)/2 }
= {(1+2)/2, (-2-3)/2, (5+4)/2}
= { 3/2, -5/2, 9/2 }

here it is clear that
midpoint of AC = midpoint of BD .
hence, midpoint of both diagonals are same so, ABCD is a parallelogram .