Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the BCD is equal to :
- 110
- 12
- 340
- 7 3
Answer
(1) 110
Solution
Take A as origin and AB, AC, AD along the axes: b = p i, c = q j, d = r k.
12pq = 5, 12qr = 6, 12rp = 7
Area of BCD = 12| BC BD|
BC BD = (-p i + q j) (-p i + r k) = qr i + pr j + pq k
Area = 12 (qr)2 + (rp)2 + (pq)2
= ( qr2 )2 + ( rp2 )2 + ( pq2 )2
= 62 + 72 + 52 = 110
Answer: (1)