Let the position vectors of the vertices A, B and C of a tetrahedron A B C D be i +2 j + k , i +3 j -2 k and 2 i + j - k respectively. The altitude from the vertex D to the opposite face A B C meets the median line

Correct answer: D. (1)/(6)(7 i +12 j + k ) Area of ABC=(1)/(2)| AB × AC | =(1)/(2)|^ 5 i +3 j + k |=(1)/(2) 35 volume of tetrahedron aligned & =(1)/(3) × Base area × h= 805 6 2 & (1)/(3) × (1)/(2) 35 × h= 805 6 2 & h= (23)/(2) aligned AE^2=AD^2-DE^2=(13)/(18) AE= (13)/(18) aligned & AE =|AE| ·( i -5 k 26 ) & = (13)/(18) ·( i -5 k 26 ) & = (13)/(18) ·( i -5 k 26 )= i -5 k 6 & P.V. of E= i -5 k 6 + i +2 j + k =(1)/(6)(7 i +12 j + k ) aligned

JEE Main 202523 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Vector Algebra

Let the position vectors of the vertices A, B and C of a...

Let the position vectors of the vertices

A, B

and

C

of a tetrahedron

A B C D

\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{k}

and

2 \hat{i}+\hat{j}-\hat{k}

respectively. The altitude from the vertex

D

to the opposite face

A B C

meets the median line segment through

A

of the triangle

A B C

at the point

E

. If the length of

A D

\frac{\sqrt{110}}{3}

and the volume of the tetrahedron is

\frac{\sqrt{805}}{6 \sqrt{2}}

, then the position vector of

E

Options

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

Correct answer

D. $\frac{1}{6}(7 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

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

\begin{aligned} & =\frac{1}{3} \times \text{ 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}} \end{aligned}

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

\begin{aligned} & \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} \\ & \text{ P.V. of } \mathrm{E}=\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{6}+\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}=\frac{1}{6}(7 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+\hat{\mathrm{k}})\end{aligned}

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