The position vectors of the vertices A, B, C of a tetrahedron ABCD are î + ĵ + k̂, î and 3î respectively and the altitude from the vertex D to the opposite face ABC meets the face at E. If the volume of tetrahedron is <span class="math-tex">\(\frac {2\sqrt 2} 3\)</span>, then the length of DE is:
A(1, 1, 1), B(1, 0, 0), C(3, 0, 0).
\(\vec{AB} = (0, -1, -1)\), \(\vec{AC} = (2, -1, -1)\)
\(\vec{AB} \times \vec{AC} = (0, -2, 2)\), with magnitude \(2\sqrt{2}\). Area of ABC = \(\sqrt{2}\).
Volume = \(\frac{1}{3} \times \text{area} \times DE\): \(\frac{1}{3}\sqrt{2} \cdot DE = \frac{2\sqrt{2}}{3} \Rightarrow DE = 2\)
Answer: (B) 2
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