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The orthocenter of triangle whose vertic...

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

A

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

B

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

C

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

D

`(1/a+1/b+1/c)`

Text Solution

Verified by Experts

The correct Answer is:
A

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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