Question

Let position vector of points A,B and C of triangle ABC respectively will be $\hat{i}+\hat{j}2+\hat{k}$ , $\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ . let $l_1.l_{2}and{l}_3$ be the lengths of perpendiculars drawn from the orthocentre 'O' on the side AB . BC and CA . then $(l_1+l_2+l_3)$ equals-

• A. $\frac{2}{\sqrt{6}}$
• B. $\frac{3}{\sqrt{6}}$
• C. $\frac{\sqrt{6}}{2}$
• D. $\frac{\sqrt{6}}{3}$

Answer 1

The correct option is B $\frac{3}{\sqrt{6}}$

$\begin{array}{c}Givenreactionof\Delta ABCon\\ A(1,1,2)B(1,2,1)C(2,1,1)\\ AB=BC=CA=\sqrt{2}\\ \Delta ABCisequational\Delta \\ orthocentre,centroid,circumaxies,incentrecoinds\\ Dis\tan cefromorthocentreofsideAB\\ \frac{1}{3}\left(htofequalatend\Delta \right)\\ \frac{1}{3}\left(\frac{\sqrt{3}}{2}a\right)\left[a=sideofequalated\Delta =\sqrt{2}\right]\\ \frac{1}{3}\frac{\sqrt{3}}{2}\sqrt{2}=\frac{1}{\sqrt{3}\sqrt{2}}=\frac{1}{\sqrt{6}}\\ l_1=l_2=l_3=\frac{1}{\sqrt{6}}\\ l_1+l_2+l_3=\frac{3}{\sqrt{6}}\end{array}$

so that the correct option is B.