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Vector Algebra question

2025 · 23 Jan · Shift 1 · Q30
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  5. /2025 · 23 Jan · Shift 1 · Q30

Vector Algebra question

2025 · 23 Jan · Shift 1 · Q30

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let the position vectors of the vertices A,B\mathrm{A}, \mathrm{B}A,B and C of a tetrahedron ABCD be i^+2j^+k^,i^+3j^−2k^\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{k}i^+2j^​+k^,i^+3j^​−2k^ and 2i^+j^−k^2 \hat{i}+\hat{j}-\hat{k}2i^+j^​−k^ respectively. The altitude from the vertex DDD to the opposite face ABCA B CABC meets the median line segment through AAA of the triangle ABCA B CABC at the point EEE. If the length of ADA DAD is 1103\frac{\sqrt{110}}{3}3110​​ and the volume of the tetrahedron is 80562\frac{\sqrt{805}}{6 \sqrt{2}}62​805​​, then the position vector of E is
  1. A
    16(7i^+12j^+k^)\frac{1}{6}(7 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+\hat{\mathrm{k}})61​(7i^+12j^​+k^)
  2. B
    112(7i^+4j^+3k^)\frac{1}{12}(7 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})121​(7i^+4j^​+3k^)
  3. C
    16(12i^+12j^+k^)\frac{1}{6}(12 \hat{i}+12 \hat{j}+\hat{k})61​(12i^+12j^​+k^)
  4. D
    12(i^+4j^+7k^)\frac{1}{2}(\hat{i}+4 \hat{j}+7 \hat{k})21​(i^+4j^​+7k^)
View written solutionFree

Correct answer: A

  1. Given position vectors

Let

A⃗=(1,2,1),B⃗=(1,3,−2),C⃗=(2,1,−1).\vec A=(1,2,1),\quad \vec B=(1,3,-2),\quad \vec C=(2,1,-1).A=(1,2,1),B=(1,3,−2),C=(2,1,−1).

We need the point EEE such that:

  • EEE lies on the median through AAA of triangle ABCABCABC,
  • EEE also lies on the altitude from DDD to plane ABCABCABC.

So EEE is the foot of the perpendicular from DDD onto plane ABCABCABC, and it lies on the median from AAA.

  1. Equation of the median through AAA in triangle ABCABCABC

First find the midpoint of BCBCBC:

M=B⃗+C⃗2=(1+22,3+12,−2+(−1)2)=(32,2,−32).M=\frac{\vec B+\vec C}{2} =\left(\frac{1+2}{2},\frac{3+1}{2},\frac{-2+(-1)}{2}\right) =\left(\frac32,2,-\frac32\right).M=2B+C​=(21+2​,23+1​,2−2+(−1)​)=(23​,2,−23​).

Hence the median through AAA is the line joining A(1,2,1)A(1,2,1)A(1,2,1) and M(32,2,−32)M\left(\frac32,2,-\frac32\right)M(23​,2,−23​).

Direction vector of this median:

AM→=(12,0,−52)=12(1,0,−5).\overrightarrow{AM}=\left(\frac12,0,-\frac52\right)=\frac12(1,0,-5).AM=(21​,0,−25​)=21​(1,0,−5).

So a general point on the median is

E=(1,2,1)+t(1,0,−5)=(1+t,2,1−5t).E=(1,2,1)+t(1,0,-5)=(1+t,2,1-5t).E=(1,2,1)+t(1,0,−5)=(1+t,2,1−5t).
  1. Find the plane ABCABCABC

Compute two vectors in the plane:

AB→=B−A=(0,1,−3),\overrightarrow{AB}=B-A=(0,1,-3),AB=B−A=(0,1,−3), AC→=C−A=(1,−1,−2).\overrightarrow{AC}=C-A=(1,-1,-2).AC=C−A=(1,−1,−2).

A normal vector is

n⃗=AB→×AC→.\vec n=\overrightarrow{AB}\times \overrightarrow{AC}.n=AB×AC.

Now,

AB→×AC→=∣i^j^k^01−31−1−2∣=−5i^−3j^−k^.\overrightarrow{AB}\times \overrightarrow{AC} = \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 1 & -3\\ 1 & -1 & -2 \end{vmatrix} = -5\hat i-3\hat j-\hat k.AB×AC=​i^01​j^​1−1​k^−3−2​​=−5i^−3j^​−k^.

Thus a normal vector is

n⃗=(5,3,1)\vec n=(5,3,1)n=(5,3,1)

(up to sign, either works).

Hence plane ABCABCABC is

5(x−1)+3(y−2)+(z−1)=05(x-1)+3(y-2)+(z-1)=05(x−1)+3(y−2)+(z−1)=0

that is,

5x+3y+z−12=0.5x+3y+z-12=0.5x+3y+z−12=0.
  1. Since EEE is the foot of perpendicular from DDD to plane ABCABCABC

The height of tetrahedron from DDD to face ABCABCABC is the perpendicular distance from DDD to plane ABCABCABC, i.e. DEDEDE.

Using volume formula:

V=13(area of △ABC)⋅DE.V=\frac13(\text{area of }\triangle ABC)\cdot DE.V=31​(area of △ABC)⋅DE.

So first find area of △ABC\triangle ABC△ABC.

We already have

n⃗=AB→×AC→=(−5,−3,−1).\vec n=\overrightarrow{AB}\times\overrightarrow{AC}=(-5,-3,-1).n=AB×AC=(−5,−3,−1).

Its magnitude is

∣n⃗∣=25+9+1=35.|\vec n|=\sqrt{25+9+1}=\sqrt{35}.∣n∣=25+9+1​=35​.

Therefore,

Area(△ABC)=1235.\text{Area}(\triangle ABC)=\frac12\sqrt{35}.Area(△ABC)=21​35​.

Given

V=80562.V=\frac{\sqrt{805}}{6\sqrt2}.V=62​805​​.

Hence

13⋅352⋅DE=80562.\frac13\cdot \frac{\sqrt{35}}{2}\cdot DE = \frac{\sqrt{805}}{6\sqrt2}.31​⋅235​​⋅DE=62​805​​.

Multiply both sides by 666:

35 DE=8052.\sqrt{35}\,DE = \frac{\sqrt{805}}{\sqrt2}.35​DE=2​805​​.

Thus

DE=80570=80570=232.DE=\frac{\sqrt{805}}{\sqrt{70}}=\sqrt{\frac{805}{70}}=\sqrt{\frac{23}{2}}.DE=70​805​​=70805​​=223​​.
  1. Use the given length ADADAD

Since DE⊥DE \perpDE⊥ plane ABCABCABC and AEAEAE lies in plane ABCABCABC, triangle ADEADEADE is right-angled at EEE.

Therefore,

AD2=AE2+DE2.AD^2=AE^2+DE^2.AD2=AE2+DE2.

Given

AD=1103  ⟹  AD2=1109.AD=\frac{\sqrt{110}}{3} \implies AD^2=\frac{110}{9}.AD=3110​​⟹AD2=9110​.

Also,

DE2=232.DE^2=\frac{23}{2}.DE2=223​.

So

AE2=1109−232=220−20718=1318.AE^2=\frac{110}{9}-\frac{23}{2} =\frac{220-207}{18}=\frac{13}{18}.AE2=9110​−223​=18220−207​=1813​.

Hence

AE=1318.AE=\sqrt{\frac{13}{18}}.AE=1813​​.
  1. Express AEAEAE in terms of parameter ttt on the median

A general point on the median is

E=(1+t,2,1−5t).E=(1+t,2,1-5t).E=(1+t,2,1−5t).

Thus

AE→=(t,0,−5t),\overrightarrow{AE}=(t,0,-5t),AE=(t,0,−5t),

so

AE2=t2+25t2=26t2.AE^2=t^2+25t^2=26t^2.AE2=t2+25t2=26t2.

Set this equal to 1318\frac{13}{18}1813​:

26t2=1318  ⟹  t2=136.26t^2=\frac{13}{18} \implies t^2=\frac{1}{36}.26t2=1813​⟹t2=361​.

Thus

t=±16.t=\pm \frac16.t=±61​.

So the two possible points on the median are:

  • For t=16t=\frac16t=61​,
E=(1+16,2,1−56)=(76,2,16).E=\left(1+\frac16,2,1-\frac56\right)=\left(\frac76,2,\frac16\right).E=(1+61​,2,1−65​)=(67​,2,61​).

This is

E⃗=16(7i^+12j^+k^).\vec E=\frac16(7\hat i+12\hat j+\hat k).E=61​(7i^+12j^​+k^).
  • For t=−16t=-\frac16t=−61​,
E=(56,2,116),E=\left(\frac56,2,\frac{11}{6}\right),E=(65​,2,611​),

which is not among the options.

Therefore the required point is

16(7i^+12j^+k^).\boxed{\frac16(7\hat i+12\hat j+\hat k)}.61​(7i^+12j^​+k^)​.
  1. Check with options

This matches Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

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