Question

Let the position vectors of the vertices $A, B$ and $C$ of a triangle be $2\mathbf{i} + 2\mathbf{j} + \mathbf{k}, \quad \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \quad \text{and} \quad 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ respectively. Let $l_1, l_2$ and $l_3$ be the lengths of the perpendiculars drawn from the ortho center of the triangle on the sides $AB, BC$ and $CA$ respectively. Then $l_1^2 + l_2^2 + l_3^2$ equals:

• A. $\frac{1}{5}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

Given that $\Delta ABC$ is equilateral, the orthocenter and centroid coincide.

The coordinates of the centroid $G$ are:
$G = \left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right).$

Considering point $A(2, 2, 1)$, point $B(1, 2, 2)$, and point $C(2, 1, 2)$, the midpoint $D$ of side $AB$ is calculated as:
$D = \left(\frac{3}{2}, 2, \frac{3}{2}\right).$

To find the lengths of perpendiculars from $G$ to the sides, we use the distance formula:
$\ell_1 = \sqrt{\frac{1}{36} + \frac{1}{9} + \frac{1}{36}} = \frac{1}{\sqrt{6}}.$

Since the triangle is equilateral, we have:
$\ell_1 = \ell_2 = \ell_3 = \frac{1}{\sqrt{6}}.$

The sum of the squares of these perpendicular lengths is:
$\ell_1^2 + \ell_2^2 + \ell_3^2 = \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2.$

Simplifying:
$\ell_1^2 + \ell_2^2 + \ell_3^2 = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}.$

The Correct answer is: $\frac{1}{2}$