Let the position vectors of the vertices A , B and C of a triangle be $2 \hat{i}+2 \hat{j}+\hat{k}, \quad \hat{i}+2 \hat{j}+2 \hat{k}$ and $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ 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 $\mathrm{AB}, \mathrm{BC}$ and CA respectively, then $l_{1}^{2}+l_{2}^{2}+l_{3}^{2}$ equals :

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Vector Algebra

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 :

Answer: (B)

A = 2 \hat{i} + 2 \hat{j} + \hat{k}

B = \hat{i} + 2 \hat{j} + 2 \hat{k}

C = 2 \hat{i} + \hat{j} + 2 \hat{k}

A B = - \hat{i} + \hat{k}

B C = \hat{i} - \hat{j}

C A = \hat{j} - \hat{k}

, each of magnitude

2

, so triangle

A B C

is equilateral.

In an equilateral triangle the orthocentre coincides with the centroid:

G = \frac{1}{3} [(2, 2, 1) + (1, 2, 2) + (2, 1, 2)] = (\frac{5}{3}, \frac{5}{3}, \frac{5}{3}) .

The distance from the centroid of an equilateral triangle to any side, with side length

s

, is

\frac{s}{2 3}

. With

s = 2

l_{1} = l_{2} = l_{3} = \frac{2}{2 3} = \frac{1}{6} .

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

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option