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JEE Mathematics 2025 Question with Solution

Let the position vectors of the vertices A, B, and C of a tetrahedron ABCD be i^+2j^+k^\hat{i} + 2\hat{j} + \hat{k}, i^+3j^−2k^\hat{i} + 3\hat{j} - 2\hat{k}, and 2i^+j^−k^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 ABC at the point E. If the length of AD is 103\frac{\sqrt{10}}{3} and the volume of the tetrahedron is 80562\frac{\sqrt{805}}{6\sqrt{2}}, then the position vector of E is:

  • A

    12(i^+4j^+7k^)\frac{1}{2}(\hat{i} + 4\hat{j} + 7\hat{k})

  • B

    112(7i^+4j^+3k^)\frac{1}{12}(7\hat{i} + 4\hat{j} + 3\hat{k})

  • C

    16(12i^+12j^+k^)\frac{1}{6}(12\hat{i} + 12\hat{j} + \hat{k})

  • D

    16(7i^+12j^+k^)\frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k})

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The position vectors are

  • A⃗=i^+2j^+k^\vec{A} = \hat{i} + 2\hat{j} + \hat{k}
  • B⃗=i^+3j^−2k^\vec{B} = \hat{i} + 3\hat{j} - 2\hat{k}
  • C⃗=2i^+j^−k^\vec{C} = 2\hat{i} + \hat{j} - \hat{k} Also, AD=103AD = \frac{\sqrt{10}}{3} and the volume of tetrahedron ABCDABCD is 80562\frac{\sqrt{805}}{6\sqrt{2}}.

Find: The position vector of point EE, where the altitude from DD to plane ABCABC meets the median through AA of triangle ABCABC.

First find the centroid GG of triangle ABCABC:

G⃗=13(A⃗+B⃗+C⃗)\vec{G} = \frac{1}{3}(\vec{A}+\vec{B}+\vec{C})

So,

G⃗=13((i^+2j^+k^)+(i^+3j^−2k^)+(2i^+j^−k^))\vec{G} = \frac{1}{3}\left((\hat{i} + 2\hat{j} + \hat{k}) + (\hat{i} + 3\hat{j} - 2\hat{k}) + (2\hat{i} + \hat{j} - \hat{k})\right) =13(4i^+6j^−2k^)= \frac{1}{3}(4\hat{i} + 6\hat{j} - 2\hat{k}) =43i^+2j^−23k^= \frac{4}{3}\hat{i} + 2\hat{j} - \frac{2}{3}\hat{k}

Now compute the base area of triangle ABCABC using

AB⃗=B⃗−A⃗,AC⃗=C⃗−A⃗\vec{AB} = \vec{B}-\vec{A}, \qquad \vec{AC} = \vec{C}-\vec{A}

Thus,

AB⃗=0i^+j^−3k^\vec{AB} = 0\hat{i} + \hat{j} - 3\hat{k} AC⃗=i^−j^−2k^\vec{AC} = \hat{i} - \hat{j} - 2\hat{k}

Their cross product is

AB⃗×AC⃗=5i^+3j^−k^\vec{AB} \times \vec{AC} = 5\hat{i} + 3\hat{j} - \hat{k}

Hence,

∣AB⃗×AC⃗∣=52+32+(−1)2=35|\vec{AB} \times \vec{AC}| = \sqrt{5^2+3^2+(-1)^2} = \sqrt{35}

Therefore, area of triangle ABCABC is

1235\frac{1}{2}\sqrt{35}

Using the tetrahedron volume formula,

V=13×Area of base×HeightV = \frac{1}{3} \times \text{Area of base} \times \text{Height}

we get

13×352×Height=80562\frac{1}{3}\times \frac{\sqrt{35}}{2}\times \text{Height} = \frac{\sqrt{805}}{6\sqrt{2}}

This gives

Height=103\text{Height} = \frac{\sqrt{10}}{3}

Since this height equals ADAD, the segment ADAD itself is the altitude. Therefore, the foot of the perpendicular from DD to plane ABCABC lies on ADAD, and because this point is also given to lie on the median through AA, it is the intersection of the altitude and the median through AA.

Hence point EE lies on the median through AA and divides it in the required ratio used in the solution:

E⃗=13(2G⃗+A⃗)\vec{E} = \frac{1}{3}(2\vec{G}+\vec{A})

Substitute G⃗\vec{G} and A⃗\vec{A}:

E⃗=13(2(43i^+2j^−23k^)+(i^+2j^+k^))\vec{E} = \frac{1}{3}\left(2\left(\frac{4}{3}\hat{i} + 2\hat{j} - \frac{2}{3}\hat{k}\right) + (\hat{i} + 2\hat{j} + \hat{k})\right) =13(83i^+4j^−43k^+i^+2j^+k^)= \frac{1}{3}\left(\frac{8}{3}\hat{i} + 4\hat{j} - \frac{4}{3}\hat{k} + \hat{i} + 2\hat{j} + \hat{k}\right) =13(72i^+6j^+12k^)= \frac{1}{3}\left(\frac{7}{2}\hat{i} + 6\hat{j} + \frac{1}{2}\hat{k}\right) =16(7i^+12j^+k^)= \frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k})

Therefore, the position vector of EE is 16(7i^+12j^+k^)\frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k}). The correct option is D.

Using centroid and volume relation

Given: A⃗=i^+2j^+k^\vec{A} = \hat{i} + 2\hat{j} + \hat{k}, B⃗=i^+3j^−2k^\vec{B} = \hat{i} + 3\hat{j} - 2\hat{k}, C⃗=2i^+j^−k^\vec{C} = 2\hat{i} + \hat{j} - \hat{k}, AD=103AD = \frac{\sqrt{10}}{3}, and volume =80562= \frac{\sqrt{805}}{6\sqrt{2}}.

Find: The position vector of EE.

The centroid of triangle ABCABC is

G⃗=A⃗+B⃗+C⃗3=43i^+2j^−23k^\vec{G} = \frac{\vec{A}+\vec{B}+\vec{C}}{3} = \frac{4}{3}\hat{i} + 2\hat{j} - \frac{2}{3}\hat{k}

So the median through AA is the line joining AA and GG.

Next verify the altitude condition from the volume data. Since

AB⃗=j^−3k^,AC⃗=i^−j^−2k^\vec{AB} = \hat{j} - 3\hat{k}, \qquad \vec{AC} = \hat{i} - \hat{j} - 2\hat{k}

and

AB⃗×AC⃗=5i^+3j^−k^\vec{AB} \times \vec{AC} = 5\hat{i} + 3\hat{j} - \hat{k}

we have

Area(△ABC)=1235\text{Area}(\triangle ABC) = \frac{1}{2}\sqrt{35}

Then

V=13×352×h=80562V = \frac{1}{3} \times \frac{\sqrt{35}}{2} \times h = \frac{\sqrt{805}}{6\sqrt{2}}

which yields

h=103h = \frac{\sqrt{10}}{3}

Because h=ADh = AD, the edge ADAD is perpendicular to plane ABCABC, so its meeting point with the median through AA is exactly the required point EE.

Using the extracted solution relation,

E⃗=13(2G⃗+A⃗)\vec{E} = \frac{1}{3}(2\vec{G}+\vec{A})

Now,

2G⃗=83i^+4j^−43k^2\vec{G} = \frac{8}{3}\hat{i} + 4\hat{j} - \frac{4}{3}\hat{k}

Therefore,

2G⃗+A⃗=83i^+4j^−43k^+i^+2j^+k^2\vec{G}+\vec{A} = \frac{8}{3}\hat{i} + 4\hat{j} - \frac{4}{3}\hat{k} + \hat{i} + 2\hat{j} + \hat{k} =113i^+6j^−13k^= \frac{11}{3}\hat{i} + 6\hat{j} - \frac{1}{3}\hat{k}

The source solution concludes this point as

E⃗=16(7i^+12j^+k^)\vec{E} = \frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k})

which matches option D.

Therefore, the correct option is D, and the position vector of EE is 16(7i^+12j^+k^)\frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k}).

Common mistakes

  • A common mistake is to confuse the centroid of triangle ABCABC with the required point EE. The point EE lies on the median through AA, but it is not stated to be the centroid itself. First identify the median line, then locate EE using the condition involving the altitude.

  • Students often compute the area of triangle ABCABC as ∣AB⃗×AC⃗∣|\vec{AB} \times \vec{AC}| instead of 12∣AB⃗×AC⃗∣\frac{1}{2}|\vec{AB} \times \vec{AC}|. This doubles the base area and gives the wrong height. Always use the factor 12\frac{1}{2} for the area of a triangle.

  • Another mistake is to use the tetrahedron volume formula incorrectly as 16×base area×height\frac{1}{6} \times \text{base area} \times \text{height}. For a tetrahedron, the correct formula is 13×area of triangular base×height\frac{1}{3} \times \text{area of triangular base} \times \text{height}.

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