Question

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

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

Answer 1

The correct option is D $(3,1,4)$

Orthocentre is point of intersection of altitudes of triangle

Equation of line through $A\&C$

$\frac{x-2}{4-2}=\frac{y-1}{0-1}=\frac{z-5}{4-5}\Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{z-5}{-1}$

Equation of line trough $B\&C\frac{x-3}{4-3}=\frac{y-2}{0-2}=\frac{z-3}{4-3}\Rightarrow \frac{x-3}{1}=\frac{y-2}{-2}=\frac{z-3}{1}$

Equation of altitude from $A$ on $BC$

P is foot of perpendicular genneral equation of $P\equiv (r+3,-2r+2,r++3)$

$\therefore d.r^{\prime }sofAP\equiv (r+3+2,2r+2-1,r+3-5)\equiv (r+1,-2r+1,r-2)AP\perp BC\Rightarrow (r+1)1+(-2r+1)(-2)+(r-2)1=0\Rightarrow r=\frac{1}{2}\therefore P\equiv (\frac{1}{2}+3,\frac{1}{2}\times -2+2,\frac{1}{2}+3)\equiv (\frac{7}{2},1,\frac{7}{2})$

Equation of line through $A\&P\frac{x-2}{2-\frac{7}{2}}=\frac{y-1}{1-1}=\frac{z-5}{5-\frac{7}{2}}\Rightarrow \frac{x-2}{-3}=\frac{y-1}{0}=\frac{z-5}{3}\Rightarrow \frac{x-2}{-1}=\frac{y-1}{0}=\frac{z-5}{1}\to \left(1\right)$

For equation of Altitude from B on AC

Let a be a point on line AC be foot of 1.

$\therefore Q\equiv (2r+2,-r+1,-r+5)d.r^{\prime }sofBQ=(2r+2-3,-r+1-2,-r+5-3)=(2r-1,-r-1,-r+2)BQ\perp AC\Rightarrow (2r-1)2+(-r-1)(-1)+(-r+2)(-1)=0\Rightarrow =\frac{1}{2}\therefore d=(2\times \frac{1}{2}+2,-\frac{1}{2}+1,-\frac{1}{2}+5)=(3,\frac{1}{2},\frac{9}{2})$

Equation of line from B and Q,

$\frac{x-3}{3-3}=\frac{y-2}{2-\frac{1}{2}}=\frac{z-3}{3-\frac{3}{2}}\Rightarrow \frac{x-3}{0}=\frac{y-2}{3}=\frac{z-3}{-3}\to \left(2\right)$

Finding point of intersection of $\left(1\right)\&\left(2\right)$ that will be orthocentre,

Let $H\equiv (3,3r+2-1=\frac{-3r+3-5}{1})\Rightarrow r=-\frac{1}{3}\therefore \frac{3-2}{-1}=\frac{3r+2-1}{0}=\frac{-3r+3-5}{1}\Rightarrow r=-\frac{1}{3}\therefore H=(3,3\times -\frac{1}{3}+2,-3\times -\frac{1}{3}+3)\Rightarrow H=(3,1,4)$

Answer D