Mathematics · 3D Geometry

JEE Main 2024 — 27 January, Shift 2 — Question 20

Let the position vectors of the vertices $\mathrm{A}, \mathrm{B}$ 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 :

Option A:
$\frac{1}{5}$
Option B:
$\frac{1}{2}$
Correct
Option C:
$\frac{1}{4}$
Option D:
$\frac{1}{3}$

Answer: B

Step-by-step solution

$\triangle \mathrm{ABC}$ is equilateral Orthocentre and centroid will be same $\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$

Mid-point of AB is $\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)$

$\therefore \ell_{1}=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}}$

$\ell_{1}=\sqrt{\frac{1}{6}}=\ell_{2}=\ell_{3}$

$\therefore \ell_{1}^{2}+\ell_{2}^{2}+\ell_{3}^{2}=\frac{1}{2}$

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Mathematics
Chapter: 3D Geometry
Topic: Introduction to 3D Geometry

Answer: B

Step-by-step solution

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