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

Q: Wait, why do we reject the standard orthocenter formula? Isn't it safer to just use the general method?

In JEE Advanced, time is your most precious currency. While the general method works, it is prone to calculation errors. Recognizing the equilateral nature of the triangle is a 'shortcut' that is actually a fundamental geometric insight. Always look for symmetry first!

Q: How do I know for sure that the orthocenter coincides with the centroid?

This is a classic property of equilateral triangles. In such a triangle, the median, altitude, angle bisector, and perpendicular bisector from any vertex are all the same line segment. Therefore, all centers—centroid, orthocenter, circumcenter, and incenter—must coincide at the same point.

Q: Is it possible that the triangle is not equilateral? What if I miscalculated the side lengths?

That is a great question. Always double-check your distance calculations. If the sides were not equal, you would indeed have to find the orthocenter using the intersection of altitudes. But in competitive exams, if you see a problem that seems 'too hard' to solve directly, there is almost always a geometric property waiting to be discovered.

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 ortho center of the triangle on the sides $AB, BC$ and $CA$ respectively, then $l_1^2 + l_2^2 + l_3^2$ equals:

Options:

(A) $\frac{1}{5}$

(B) $\frac{1}{2}$

(C) $\frac{1}{4}$

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system, looking at three points suspended in space: $A(2, 2, 1)$, $B(1, 2, 2)$, and $C(2, 1, 2)$.

At first glance, this might seem like just another coordinate geometry problem, a test of your ability to grind through tedious calculations. But wait—before you start calculating the equations of lines and planes to find the orthocenter, let's pause.

In the world of JEE Advanced, the most powerful tool in your arsenal is not just your ability to calculate, but your ability to observe. Let's start by calculating the lengths of the sides of this triangle using the 3D distance formula:

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

For side $AB$, we have:

$$AB = \sqrt{(1-2)^2 + (2-2)^2 + (2-1)^2} = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$$

Now, let's check $BC$:

$$BC = \sqrt{(2-1)^2 + (1-2)^2 + (2-2)^2} = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$$

And finally, $CA$:

$$CA = \sqrt{(2-2)^2 + (2-1)^2 + (1-2)^2} = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$$

### The Magic of Symmetry

What do we see? All three sides are equal to $\sqrt{2}$. We have stumbled upon an **equilateral triangle**!

In an equilateral triangle, the orthocenter ($H$), which is the intersection of altitudes, is identical to the centroid ($G$), the intersection of medians. This is the 'magic' of symmetry.

Instead of solving for the intersection of complex lines in 3D, we can simply find the centroid, which is the average of the coordinates:

$$G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$$

Plugging in our values, we get:

$$G = \left( \frac{2+1+2}{3}, \frac{2+2+1}{3}, \frac{1+2+2}{3} \right) = \left( \frac{5}{3}, \frac{5}{3}, \frac{5}{3} \right)$$

### Final Calculation

The problem asks for the sum of the squares of the perpendicular distances from the orthocenter to the sides: $l_1^2 + l_2^2 + l_3^2$. Because the triangle is equilateral, the distance from the centroid to each side is identical, so $l_1 = l_2 = l_3$.

We only need to calculate one of them. The perpendicular from the centroid to side $AB$ meets at the midpoint $D$ of $AB$:

$$D = \left( \frac{2+1}{2}, \frac{2+2}{2}, \frac{1+2}{2} \right) = \left( \frac{3}{2}, 2, \frac{3}{2} \right)$$

Now, we calculate the distance $l_1$ between $G(\frac{5}{3}, \frac{5}{3}, \frac{5}{3})$ and $D(\frac{3}{2}, 2, \frac{3}{2})$:

$$l_1 = \sqrt{\left( \frac{5}{3} - \frac{3}{2} \right)^2 + \left( \frac{5}{3} - 2 \right)^2 + \left( \frac{5}{3} - \frac{3}{2} \right)^2}$$

Simplifying the terms:

$$l_1 = \sqrt{\left( \frac{1}{6} \right)^2 + \left( -\frac{1}{3} \right)^2 + \left( \frac{1}{6} \right)^2} = \sqrt{\frac{1}{36} + \frac{1}{9} + \frac{1}{36}} = \sqrt{\frac{1+4+1}{36}} = \sqrt{\frac{6}{36}} = \frac{1}{\sqrt{6}}$$

Therefore, $l_1^2 = \frac{1}{6}$. Since $l_1 = l_2 = l_3$, the sum is:

$$l_1^2 + l_2^2 + l_3^2 = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \mathbf{\frac{1}{2}}$$

We have arrived at our answer, **$\frac{1}{2}$**, through the elegance of symmetry rather than the brute force of coordinate geometry. Always remember: in JEE, look for the hidden structure before you start the calculation.

Select Answer:

Visualized Solution

Identify Vertices $A, B, C$

The Distance Formula Tool

Calculate Length of Side $A B$

Calculate Lengths of $B C$ and $C A$

Property of Equilateral Triangles

Calculate Centroid $G$

Distance to Sides $l_{1}, l_{2}, l_{3}$

Find Midpoint of $A B$

Calculate Distance $l_{1}$

Simplify Distance $l_{1}$

Sum of Squares $l_{1}^{2} + l_{2}^{2} + l_{3}^{2}$

Final Conclusion

The Sigma Insight: Scalar and Vector Quantities

Analyzing the Setup

The Magic of Symmetry

Final Calculation

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Visualized Solution

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The Distance Formula Tool

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