JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 2)
Let the position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ 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 $$\mathrm{AB}, \mathrm{BC}$$ and $$\mathrm{CA}$$ respectively, then $$l_1^2+l_2^2+l_3^2$$ equals:
$$\frac{1}{4}$$
$$\frac{1}{5}$$
$$\frac{1}{3}$$
$$\frac{1}{2}$$
Explanation
$$\triangle \mathrm{ABC}$$ is equilateral
Orthocentre and centroid will be same
$$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$$
_27th_January_Evening_Shift_en_2_1.png)
Mid-point of $$\mathrm{AB}$$ is $$\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)$$
$$\begin{aligned} & \therefore \ell_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}} \\ & \ell_1=\sqrt{\frac{1}{6}}=\ell_2=\ell_3 \\ & \therefore \ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2} \end{aligned}$$
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