3D Geometry
Orthocentre Perpendicular Lengths in 3D Triangle
DAILY_CHALLENGE
Grade 12
Question:
Let the position vectors of the vertices $A$, $B$ and $C$ 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 $l_1,l_2$ and $l_3$ be the lengths of perpendiculars drawn from the orthocentre of the triangle on the sides $AB$, $BC$ and $CA$ respectively, then $l_1^2+l_2^2+l_3^2$ equals:
$\dfrac{1}{5}$
$\dfrac{1}{2}$
$\dfrac{1}{4}$
$\dfrac{1}{3}$
Step-by-Step Solution
Key Concept: Triangle is equilateral (all sides $=\sqrt2$). Orthocentre = centroid = $(5/3,5/3,5/3)$. Distance from centroid to each side (altitude $h$ in equilateral): use formula $l=\frac{2\cdot\text{Area}}{\text{side}}$. Area $=\frac{\sqrt3}{4}(\sqrt2)^2=\frac{\sqrt3}{2}$... $l=\frac{2\cdot\sqrt3/2}{\sqrt2}=\frac{\sqrt3}{\sqrt2}$. $l^2=3/2$. $l_1^2+l_2^2+l_3^2=3\times1/6=1/2$.
$l_1^2+l_2^2+l_3^2=1/2$.
Correct Answer: 2