Let the position vectors of the vertices A,B,C of a triangle be 2 i+2 j+ k, i+2 j+2 k, and…

Question

Let the position vectors of the vertices $A,B,C$ of a triangle be $2\hat i+2\hat j+\hat k$, $\hat i+2\hat j+2\hat k$, and $2\hat i+\hat j+2\hat k$ respectively. Let $l_1,l_2,l_3$ be the lengths of perpendiculars drawn from the orthocenter of the triangle to the sides $AB, BC, CA$ respectively. Then $l_1^2+l_2^2+l_3^2$ equals:

Accepted Answer

$\vec{AB}=(-1,0,1),\ \vec{BC}=(1,-1,0),\ \vec{CA}=(0,1,-1)$, each of length $\sqrt2$ --- the triangle is equilateral with side $\sqrt2$.

For an equilateral triangle, the orthocenter coincides with the centroid, and the perpendicular distance from the centroid to each side equals one-third of the altitude:
$$h=\frac{\sqrt3}{2}\cdot\sqrt2=\frac{\sqrt6}{2},\qquad l_1=l_2=l_3=\frac{h}{3}=\frac{\sqrt6}{6}.$$
$$l_1^2+l_2^2+l_3^2=3\times\frac{6}{36}=\frac12.$$

Solutions · 1