Discussion Forum

If the vertices  of a triangle are A(0,4,1) ,B(2,3,-1) and C(4,5,0) then the orthocentre of $\triangle ABC$ is 

Answer:

Explanation:

 Vertices of $\triangle ABC$ are A(0,4,1) ,B(2,3,-1) and C(4,5,0) 

 AB= $\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}}=\sqrt{4+1+4}=3$

$BC=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}}=\sqrt{4+4+1}=3$

and 

$CA=\sqrt{(4-0)^{2}+(5-4)^{2}+(0+1)^{2}}=\sqrt{16+1+4}=3\sqrt{2}$

 $AB^{2}+BC^{2}=AC^{2}$

$\therefore$  $\triangle$ ABC is a right angled triangle 

We know that  , the orthocentre of a right angled triangle is the vertex containing the right angle 

$\therefore$ Orthocentre is point B(2,3,-1)