Let position vector of point A, B and C of triangle ABC respectively be i^+j^+2k^, i^+2j^+k^ and 2i^+j^+k^, let l1, l2, l3 are the length of perpendiculars drawn from the orthocenter ' O ' on the side AB, BC and CA, then the value of 6l1+l2+l3 is?

Clearly the triangle is an equilateral triangle such that AB = BC = CA = 2 . So distance of centre ‘O’ from the sides is equal to in radius of the triangle. So, l1=l2=l3=Δs=34×22322=16So, 6 l1+l2+l3=6×36=3

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Let position vector of point A, B and C of triangle ABC respectively be $\hat{i} + \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} + \hat{k} a n d 2 \hat{i} + \hat{j} + \hat{k}, l e t l_{1}, l_{2}, l_{3}$ are the length of perpendiculars drawn from the orthocenter ' $O$ ' on the side AB, BC and CA, then the value of $\sqrt{6} (l_{1} + l_{2} + l_{3})$ is?

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answer is 3.

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Detailed Solution

Clearly the triangle is an equilateral triangle such that AB = BC = CA = $\sqrt{2}$ . So distance of centre ‘O’ from the sides is equal to in radius of the triangle.
$S o, l_{1} = l_{2} = l_{3} = \frac{Δ}{s} = \frac{\frac{\sqrt{3}}{4} \times {(\sqrt{2})}^{2}}{\frac{3}{2} \sqrt{2}} = \frac{1}{\sqrt{6}}$

$S o, \sqrt{6} (l_{1} + l_{2} + l_{3}) = \sqrt{6} \times \frac{3}{\sqrt{6}} = 3$

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