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Question

The orthocenter of a triangle formed by the points (2,1,5),(3,2,3),(4,0,4)

A
(2,1,5)
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B
(3,2,3)
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C
(4,0,4)
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D
(3,1,4)
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Solution

The correct option is A (2,1,5)
Given,
Points of the triangle are A(2,1,5),B(3,2,3) and C(4,0,4)
Let the orthocenter be H(a,b,c)

Direction ratios of line AB=(3-2,2-1,3-5)=(1,1,-2)

Similarly Direction ratios of BC=(1,-2,1)

Direction ratios of AC=(2,-1,-1)

Now Direction ratios of OA=(a-2,b-1,c-5)

Direction ratios of OB=(a-3,b-2,c-3)

Direction ratios of OC=(a-4,b,c-4)

AS we know OABC
1(a4)2(b1)+1(c5)=0
a42b+2+c5=0
a2b+c=7 (1)

Similarly OBAC
2(a3)1(b2)1(c3)=02abc=1 (2)

Also OCAB
1(a4)+1(b)2(c4)=0a+b2c=4 (3)
On solvinf eqs.(1),(2) and (3) we get a=2,b=1 and c=5 respectively.

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