MHT CET202611 April 2026Morning ShiftMathematicsVector AlgebraActual
Let A(2,3,0) , B(0,3,2) and C(4,0,3) be vertices of a triangle, then the area of the triangle is
Options
- A171 2 sq. units
- B172 2 sq. units
- C173 2 sq. units
- D174 2 sq. units
Correct answer
B. 172 2 sq. units
Step-by-step solution
The position vectors of the vertices are A(2,3,0) , B(0,3,2) , and C(4,0,3) . The vectors representing the sides AB and AC are: AB = (0 - 2) i + (3 - 3) j + (2 - 0) k = -2 i + 0 j + 2 k AC = (4 - 2) i + (0 - 3) j + (3 - 0) k = 2 i - 3 j + 3 k The area of the triangle is given by 1 2 | AB AC | . Calculating the cross product: AB AC = vmatrix i & j & k -2 & 0 & 2 2 & -3 & 3 vmatrix AB AC = i (0 - (-6)) - j (-6 - 4) + k (6 - 0) AB AC = 6 i + 10 j + 6 k Now, finding the magnitude of the cross product: | AB AC | = 6^2 +