Vector Algebra
Vector Algebra
nta_pyq_2025_jan
Grade 12
Question:
Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be ^i + 2^j + k, ^ ^ i + 3 j - 2k and ^ ^ ^ ^ ^ 2i + j - k respectively. The altitude from the vertex D to the opposite face ABC meets the median line \sqrt110 segment through A of the triangle ABC at the point E. If the length of AD is 3 and the volume of the \sqrt805 tetrahedron is , then the position vector of E is 6\sqrt2
1 12 ^ ^ ^ (7 i + 4 j + 3k)
1 2 ^ ^ ^ ( i + 4 j + 7k)
1 6 ^ ^ ^ (12 i + 12 j + k)
1 6 ^ ^ ^ (7 i + 12 j + k)
Step-by-Step Solution
Key Concept: Apply the core result for dot product, cross product and projections and simplify using the given constraints.
(4) - -\to - -\to Area of △ABC = 1 2 |AB \times AC| 1 ^ ^ ^ 1 = |5 i + 3 j + k| = \sqrt35 2 2 volume of tetrahedron 1 \sqrt805 = \times Base area \times h = 3 6\sqrt2 1 1 \sqrt805 \times \sqrt35 \times h = 3 2 6\sqrt 2 23 h = \sqrt 2 2 2 2 13 13 AE = AD - DE = \therefore AE = \sqrt 18 18 - -\to ^ ^ i - 5k AE = |AE| ⋅ ( ) \sqrt26 ^ ^ 13 i - 5k = \sqrt ⋅ ( ) 18 \sqrt26 ^ ^ ^ ^ 13 i - 5k i - 5k = \sqrt ⋅ ( ) = 18 \sqrt26 6 ^ ^ i - 5k 1 ^ ^ ^ ^ ^ ^ P.V. of E = + i + 2j + k = (7 i + 12 j + k) 6 6
Correct Answer: 4