Question

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

• A. ($2$,$1$,$5$)
• B. ($3$,$2$,$3$)
• C. ($4$,$0$,$4$)
• D. ($3$,$1$,$4$)

Answer 1

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 OA$\perp$BC

$\therefore 1(a-4)-2(b-1)+1(c-5)=0$

$\to a-4-2b+2+c-5=0$

$\to a-2b+c=7-\left(1\right)$

Similarly OB$\perp$AC

$\to 2(a-3)-1(b-2)-1(c-3)=0\to 2a-b-c=1-\left(2\right)$

Also OC$\perp$AB

$\therefore 1(a-4)+1\left(b\right)-2(c-4)=0\to a+b-2c=-4-\left(3\right)$

On solvinf eqs.(1),(2) and (3) we get a=2,b=1 and c=5 respectively.