If ax+by+cz=d intersects the coordinate axis at A,B,C and the area of the triangle ABC is (a2+b2)√2(a2+c2+b2)a2b2(a2+b2). If a2,c2,b2 are in A.P. then find the perpendicular distance of the plane from the origin.
A
23
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B
1
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C
2√33
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D
2
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Solution
The correct option is C2√33 A≡(da,0,0),B≡(0,db,0),C≡(0,0,dc)−−→AB=da^i−db^j+0^k−−→BC=0^i+db^j−dc^k
Area of the triangle ABC=12|−−→AB×−−→BC|=12d2√(a2+b2+c2)abc∴12d2√(a2+b2+c2)abc=(a2+b2)√2(a2+c2+b2)a2b2(a2+b2)⇒d22abc=(2c2)√(2)abc√(2)d2c2=4 dc=2 The distance of plane from origin =∣∣∣0+0+0−d√(a2+b2+c2)∣∣∣=d√3c=2√33