Let the position vector of the vertices A, B and C of a triangle be , and respectively. Let , and be the lengths of perpendiculars drawn from the orthocenter of the triangle on the sides AB, BC and CA respectively, then equals : [2024]

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Solution

Q.
Let the position vector 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 orthocenter of the triangle on the sides AB, BC and CA respectively, then $l_{1}^{2} + l_{2}^{2} + l_{3}^{2}$ equals : [2024]

1 $\frac{1}{5}$

2 $\frac{1}{3}$

3 $\frac{1}{2}$

4 $\frac{1}{4}$

Ans.
(3)

Position vector of $A = 2 \hat{i} + 2 \hat{j} + \hat{k}$

Position vector of $B = \hat{i} + 2 \hat{j} + 2 \hat{k}$

Position vector of $C = 2 \hat{i} + \hat{j} + 2 \hat{k}$

$\Rightarrow$ Coordinates of A, B and C of $△ A B C$ are A(2, 2, 1), B(1, 2, 2) and C(2, 1, 2)

$A B = \sqrt{{(1 - 2)}^{2} + {(2 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{2} units$

$B C = \sqrt{{(2 - 1)}^{2} + {(1 - 2)}^{2} + {(2 - 2)}^{2}} = \sqrt{2} units$

$A C = \sqrt{{(2 - 2)}^{2} + {(1 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{2} units$

$\Rightarrow$ $△ A B C$ is an equilateral triangle with side $\sqrt{2}$ units. In an equilateral triangle, orthocenter and centroid will be same.

Centroid, $G (x, y, z) \equiv G (\frac{2 + 1 + 2}{3}, \frac{2 + 2 + 1}{3}, \frac{2 + 2 + 1}{3}) \equiv G (\frac{5}{3}, \frac{5}{3}, \frac{5}{3})$

Mid-point of AB is $D (\frac{3}{2}, 2, \frac{3}{2})$

$l_{1} = \sqrt{{(\frac{5}{3} - \frac{3}{2})}^{2} + {(\frac{5}{3} - 2)}^{2} + {(\frac{5}{3} - \frac{3}{2})}^{2}} = \sqrt{\frac{1}{36} + \frac{1}{9} + \frac{1}{36}} = \sqrt{\frac{1}{6}}$

$l_{1} = l_{2} = l_{3} = \sqrt{\frac{1}{6}}$

$∴ l_{1}^{2} + l_{2}^{2} + l_{3}^{2} = 3 \times \frac{1}{6} = \frac{1}{2}$

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