The area of the quadrilateral ABCD with vertices A(2,1,1), B(1,2,5), C(-2,-3,5) and D(1,-6,-7) is equal to :

Solution

Maths sir · 3 steps

Take the two diagonals, cross them, and half the length of that cross product is the area.

1. Use the diagonals. For a plane quadrilateral, the area is half the magnitude of the cross product of its diagonals. Here the diagonals are AC⃗ = (-4, -4, 4) and BD⃗ = (0, -8, -12). The four points are coplanar, because AB⃗ · (AC⃗ × AD⃗) = -60 - 36 + 96 = 0.

   AC⃗ × BD⃗ = |î, ĵ, k̂; -4, -4, 4; 0, -8, -12| = 80î - 48ĵ + 32k̂

2. Find the magnitude. |AC⃗ × BD⃗| = √(6400 + 2304 + 1024) = √9728 = √(256 × 38) = 16√38.
3. So. Area = 1/2 × 16√38 = 8√38. As a check, triangles ABC and ACD give 2√38 + 6√38 = 8√38, which is option (b).

The idea to keep. Quadrilateral area in 3D is 1/2|d₁⃗ × d₂⃗| for the diagonals. If you are unsure, split it into two triangles along a diagonal and add their areas.

Answer from the NTA final answer key. Solution by PrepWiser.