If A(3,1,-1), B(5/3, 7/3, 1/3), C(2,2,1) and D(10/3, 2/3, -1/3) are the vertices of a quadrilateral A B C D, then its area is

Solution

Maths sir · 4 steps

For a quadrilateral, skip the four sides: area is half the cross product of the two diagonals.

1. Find the diagonals. The diagonals are AC⃗ = C - A = (-1, 1, 2) and BD⃗ = D - B = (5/3, -5/3, -2/3).

   AC⃗ × BD⃗ = |î, ĵ, k̂; -1, 1, 2; 5/3, -5/3, -2/3| = 8/3î + 8/3ĵ + 0 k̂

2. Take the magnitude. |AC⃗ × BD⃗| = √(64/9 + 64/9) = 8√2/3. Area of a plane quadrilateral is 1/2|d₁⃗ × d₂⃗|, so the area is 1/2 × 8√2/3.
3. Check it is flat. The formula needs all four points in one plane. With AB⃗ = (-4/3, 4/3, 4/3) and AD⃗ = (1/3, -1/3, 2/3), the scalar triple product [AB⃗ AC⃗ AD⃗] = 0, so they are coplanar.
4. So. The area of ABCD is 4√2/3, which is option (a).

The idea to keep. Area of any plane quadrilateral is 1/2|d₁⃗ × d₂⃗|, using the two diagonals. One cross product does the job of splitting it into two triangles.

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