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QJEEM24Ma299e: JEE Main 2024 Physics - Class 11 JEEM24Ma299e Solution: Let the position vectors of the vertices and of a triangle be…

Let the position vectors of the vertices A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} of a triangle be 2i^+2j^+k^,i^+2j^+2k^2 \hat{i}+2 \hat{j}+\hat{k}, \hat{i}+2 \hat{j}+2 \hat{k} and 2i^+j^+2k^2 \hat{i}+\hat{j}+2 \hat{k} respectively. Let l1,l2l_1, l_2 and l3l_3 be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides AB,BC\mathrm{AB}, \mathrm{BC} and CA\mathrm{CA} respectively, then l12+l22+l32l_1^2+l_2^2+l_3^2 equals:

Multiple Choice Options

  • A 14\frac{1}{4}
  • B 15\frac{1}{5}
  • C 13\frac{1}{3}
  • D 12\frac{1}{2}

Step-by-Step Verified Solution

△ABC\triangle \mathrm{ABC} is equilateral

Orthocentre and centroid will be same

G(53,53,53)\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)

JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Vector Algebra Question 44 English Explanation

Mid-point of AB\mathrm{AB} is D(32,2,32)\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)

amp;∴ℓ1=136+19+136amp;ℓ1=16=ℓ2=ℓ3amp;∴ℓ12+ℓ22+ℓ32=12\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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