Question

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

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

Answer 1

The correct option is B $(2,3,-1)$

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

By using the distance formula

$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$

We have

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

$=\sqrt{4+1+4}=3$

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

$=\sqrt{4+4+1}=3$

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

$=\sqrt{16+1+1}=3\sqrt{2}$

$A{B}^2+B{C}^2=A{C}^2$

$\therefore \Delta ABC$ 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 $\equiv (2,3,-1)$.