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The orthocenter of triangle whose vertices are A(a,0,0),B(0,b,0) and C(0,0,c) is (k/a,k/b,k/c) then k is equal to |
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Answer» `(1/a^(2)+1/b^(2)+1/c^(2))^(-1)` Now, `AO BOT BC` Similarly, `a(alpha)=b(beta)=cgamma=k` `THEREFORE alpha=k/a, beta=k/b,gamma=k/c` The plane `x/a+y/b+z/c=1` contains `(alpha,beta,gamma)` `therefore k(1/a^(2)+1/b^(2)+1/c^(2))=1` |
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