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

The Correct Option is B

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: \( \frac{1}{2} \)