© All Rights reserved @ LearnWithDash
Step-by-Step Detailed Solution
Step 1: Identify the Vertices
The position vectors of the vertices are given as:
$ \mathbf{A} = 2\hat{i} + 2\hat{j} + \hat{k} $, which corresponds to the point
$ A(2,\, 2,\, 1). $
$ \mathbf{B} = \hat{i} + 2\hat{j} + 2\hat{k} $, which corresponds to the point
$ B(1,\, 2,\, 2). $
$ \mathbf{C} = 2\hat{i} + \hat{j} + 2\hat{k} $, which corresponds to the point
$ C(2,\, 1,\, 2). $
Step 2: Compute the Side Lengths
To check for special properties, find the lengths of the sides of the triangle:
$ \overrightarrow{AB} = (1 - 2,\, 2 - 2,\, 2 - 1) = (-1,\, 0,\, 1). $
Thus,
$ |AB| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2}. $
$ \overrightarrow{BC} = (2 - 1,\, 1 - 2,\, 2 - 2) = (1,\, -1,\, 0). $
Thus,
$ |BC| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}. $
$ \overrightarrow{CA} = (2 - 2,\, 1 - 2,\, 2 - 1) = (0,\, -1,\, 1). $
Thus,
$ |CA| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}. $
All sides have length $ \sqrt{2}, $ so the triangle is equilateral.
Step 3: Recognize Key Properties of an Equilateral Triangle
In an equilateral triangle, the centroid, orthocenter, and circumcenter
coincide. Therefore, the orthocenter $H$ is the same point as the centroid $G.$
The coordinates of the centroid $G$ are the averages of the coordinates of $A,\, B,$ and $C:$
$
G = \left(
\frac{2 + 1 + 2}{3},\, \frac{2 + 2 + 1}{3},\, \frac{1 + 2 + 2}{3}
\right)
= \left(\frac{5}{3},\, \frac{5}{3},\, \frac{5}{3}\right).
$
Step 4: Find the Perpendicular Distance from the Orthocenter to Any Side
The altitude of an equilateral triangle with side length $s$ is
$ \frac{\sqrt{3}}{2}s. $
In an equilateral triangle, the centroid (also the orthocenter) is
$ \tfrac{2}{3} $ of the way from a vertex along each altitude. Thus, the
distance from the orthocenter to any side (i.e., the perpendicular from
$H$ to a side) is $ \tfrac{1}{3} $ of the altitude.
Since $s = \sqrt{2},$ the altitude is:
$
\frac{\sqrt{3}}{2} \times \sqrt{2} \;=\; \frac{\sqrt{6}}{2}.
$
Hence the perpendicular distance from the orthocenter to each side is:
$
l
= \frac{1}{3} \times \frac{\sqrt{6}}{2}
= \frac{\sqrt{6}}{6}.
$
Step 5: Calculate the Sum of Squares of the Three Perpendiculars
In an equilateral triangle, these perpendiculars from the orthocenter to each
side are all equal. Therefore,
$
l_1 = l_2 = l_3 = l = \frac{\sqrt{6}}{6}.
$
Squaring $l$ gives:
$
l^2 = \left(\frac{\sqrt{6}}{6}\right)^2
= \frac{6}{36}
= \frac{1}{6}.
$
The sum of the squares of the three perpendiculars is therefore:
$
l_1^2 + l_2^2 + l_3^2
= 3 \times \frac{1}{6}
= \frac{1}{2}.
$
Final Answer
$ l_1^2 + l_2^2 + l_3^2 = \frac{1}{2}. $
