Vector Algebra
Area using cross product
Grade 12
Question:
<p>The area (in sq. units) of the parallelogram, whose diagonals are along the vectors \(8\hat{i}-6\hat{j}\) and \(3\hat{i}+4\hat{j}-12\hat{k}\), is</p>
<p>65</p>
<p>52</p>
<p>26</p>
<p>20</p>
Step-by-Step Solution
Key Concept: The area of a parallelogram with diagonals d₁ and d₂ is (1/2)|d₁ × d₂|. The cross product magnitude gives twice the area because diagonals of a parallelogram bisect each other.
Step 1: Identify the diagonal vectors. d_1 = 8î - 6ĵ + 0k̂ d_2 = 3î + 4ĵ - 12k̂ Step 2: Compute the cross product d_1 × d_2. d_1 × d_2 = |î ĵ k̂| |8 -6 0| |3 4 -12| = î[(-6)(-12) - (0)(4)] - ĵ[(8)(-12) - (0)(3)] + k̂[(8)(4) - (-6)(3)] = î(72) - ĵ(-96) + k̂(32 + 18) = 72î + 96ĵ + 50k̂ Step 3: Find the magnitude of the cross product. |d_1 × d_2| = √(72^2 + 96^2 + 50^2) = √(5184 + 9216 + 2500) = √16900 = 130 Step 4: Apply the parallelogram area formula. Area = (1/2)|d_1 × d_2| = (1/2)(130) = 65 ∴ Answer: 65 sq. units
Correct Answer: A