Vector Algebra — JEE Main · 2025 · 23 Jan 2025 · Shift 1 Question with Solution

Let the position vectors of the vertices

A, B

and C of a tetrahedron ABCD 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

A B C

meets the median line segment through A of the triangle ABC at the point E . If the length of AD is

\frac{110}{3}

and the volume of the tetrahedron is

\frac{805}{6 2}

, then the position vector of E is

Answer: (D)

Solution

With

A = (1, 2, 1)

B = (1, 3, - 2)

C = (2, 1, - 1)

A B = (0, 1, - 3), A C = (1, - 1, - 2)

A B \times A C = \hat{i} 01 \hat{j} 1 - 1 \hat{k} - 3 - 2 = (- 5, - 3, - 1), ∣ A B \times A C ∣ = 35

Area of

△ A B C = \frac{35}{2}

.

From

V = \frac{1}{3} \times Area \times h = \frac{805}{6 2}

h = \frac{3 V}{Area} = \frac{3 \cdot \frac{805}{6 2}}{\frac{35}{2}} = \frac{23}{2}

D lies above the base at perpendicular distance

h

, with foot at E. Since

D E ⊥

plane

A B C

, triangle

A D E

is right-angled at

E

A E^{2} = A D^{2} - D E^{2} = \frac{110}{9} - \frac{23}{2} = \frac{13}{18}

E lies on the median from A to the midpoint of BC,

M = (\frac{3}{2}, 2, - \frac{3}{2})

. The direction

A M = (\frac{1}{2}, 0, - \frac{5}{2})

is parallel to

(1, 0, - 5)

, giving the unit vector

\frac{i ^ - 5 k ^}{26}

A E = \frac{13}{18} \cdot \frac{i ^ - 5 k ^}{26} = \frac{i ^ - 5 k ^}{6}

E = A + A E = (1, 2, 1) + (\frac{1}{6}, 0, - \frac{5}{6}) = (\frac{7}{6}, 2, \frac{1}{6}) = \frac{1}{6} (7 \hat{i} + 12 \hat{j} + \hat{k})

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option