question_answer

A plane makes intercepts OA, OB, OC whose measures are a, b, c on the axes x, y, z respectively. The area of the triangle ABC is

A) \[\frac{1}{2}\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]

B) \[\frac{1}{2}\sqrt{{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}+{{a}^{2}}{{b}^{2}}}\]

C) \[\frac{1}{2}\sqrt{ab+bc+ac}\]       

D) \[\frac{1}{2}\sqrt{{{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}a}\]

Correct Answer: B

Solution :

Let OP be the normal from O to the plane of \[\Delta ABC,\] Let l, m, n be the direction      cosines of OP Let s be the area of \[\Delta ABC\] Therefore,              OAC = Projection of \[\Delta ABC\] on zx plane = s.m. But OAC is a right angled triangle, so that area of \[\Delta OAC=\frac{1}{2}\] \[\therefore \] \[\frac{1}{2}ac=s.m.\Rightarrow m=\frac{ac}{2s}.\] Similarly, in \[\Delta OBC,\] \[l=\frac{bc}{2s}\]and in \[\Delta OAB,n=\frac{ab}{2s}\] Since, \[{{\ell }^{2}}+{{m}^{2}}+{{n}^{2}}=1\] \[\therefore \] \[{{\left( \frac{ac}{2s} \right)}^{2}}+{{\left( \frac{bc}{2s} \right)}^{2}}+{{\left( \frac{ab}{2s} \right)}^{2}}=1\] \[\Rightarrow \] \[s=\frac{1}{2}\sqrt{{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}+{{a}^{2}}{{b}^{2}}}\]