1.

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

Answer»

`(1/a^(2)+1/b^(2)+1/c^(2))^(-1)`
`(1/a+1/b+1/c)^(-1)`
`(1/a^(2)+1/b^(2)+1/c^(2))`
`(1/a+1/b+1/c)`

Solution :Let `O(alpha,BETA,GAMMA)` be orthocenter.
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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