Mathematics · Vector Algebra

JEE Main 2025 — 23 January, Morning Shift — Question 17

Let the position vectors of the vertices $A, B$ and $C$ of a tetrahedron $A B C D$ be $\hat{i}+2 \hat{j}+\hat{k}, \hat{i}+3 \hat{j}-2 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through $A$ of the triangle $A B C$ at the point $E$ . If the length of $A D$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6 \sqrt{2}}$ , then the position vector of E is

Option A:
$\frac{1}{2}(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$
Option B:
$\frac{1}{12}(7 \hat{i}+4 \hat{j}+3 \hat{k})$
Option C:
$\frac{1}{6}(12 \hat{i}+12 \hat{j}+\hat{k})$
Option D:
$\frac{1}{6}(7 \hat{i}+12 \hat{j}+\hat{k})$
Correct

Answer: D

Step-by-step solution

Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$

$=\frac{1}{2}|5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}|=\frac{1}{2} \sqrt{35}$

volume of tetrahedron

$=\frac{1}{3} \times$ Base area $\times \mathrm{h}=\frac{\sqrt{805}}{6 \sqrt{2}}$

$\frac{1}{3} \times \frac{1}{2} \sqrt{35} \times \mathrm{h}=\frac{\sqrt{805}}{6 \sqrt{2}}$

$\mathrm{h}=\sqrt{\frac{23}{2}}$

$\mathrm{AE}^{2}=\mathrm{AD}^{2}-\mathrm{DE}^{2}=\frac{13}{18} \therefore \mathrm{AE}=\sqrt{\frac{13}{18}}$

$\overrightarrow{\mathrm{AE}}=|\mathrm{AE}| \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right)$

$=\sqrt{\frac{13}{18}} \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right)$

$=\sqrt{\frac{13}{18}} \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right)=\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{6}$

P.V. of $E=\frac{\hat{i}-5 \hat{k}}{6}+\hat{i}+2 \hat{j}+\hat{k}=\frac{1}{6}(7 \hat{i}+12 \hat{j}+\hat{k})$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Volume of parallelopiped, tetrahedron.