Let position vectors of points A, B and C of triangle ABC respectively be i^+j^+2k^, i^+2j^+k^ and 2i^+j^+k^. Let l1 , l2 , l3 be the lengths of perpendiculars drawn from the orthocentre ‘O’ on the sides AB, BC and CA. then the value of 18l1+l2+l32 is

Clearly triangle ABC is equilateral⇒l1=l2=l3= in-radius =16∴l1+l2+l3=36=62Since centroid=43(i^+j^+k^)⇒ In-radius =43−322+43−322+43−12=16

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Let position vectors of points A, B and C of triangle ABC respectively be $\hat{i} + \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} + \hat{k} and 2 \hat{i} + \hat{j} + \hat{k} .$ Let l₁ , l₂ , l₃ be the lengths of perpendiculars drawn from the orthocentre ‘O’ on the sides AB, BC and CA. then the value of $18 {(l_{1} + l_{2} + l_{3})}^{2}$ is

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answer is 27.

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Detailed Solution

Clearly triangle ABC is equilateral
$\begin{array}{l} \Rightarrow l_{1} = l_{2} = l_{3} = in-radius = \frac{1}{\sqrt{6}} \\ ∴ l_{1} + l_{2} + l_{3} = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2} \end{array}$
Since centroid $= \frac{4}{3} (\hat{i} + \hat{j} + \hat{k}) \Rightarrow In-radius = \sqrt{{(\frac{4}{3} - \frac{3}{2})}^{2} + {(\frac{4}{3} - \frac{3}{2})}^{2} + {(\frac{4}{3} - 1)}^{2}} = \frac{1}{\sqrt{6}}$