If the area of a parallelogram whose diagonals are represented by vectors 3 i + j +2 k and i -2 j +3 k is 117 2 sq. units, then =

Step-by-step solution

Given diagonal vectors are d₁ = 3 i + j +2 k and d₂ = i -2 j +3 k . The area of the parallelogram formed by diagonals d₁ and d₂ is 1 2 | d₁ d₂ | . Compute the cross product: d₁ d₂ = (3 + 4) i - 7 j + (-6 - ) k . Its magnitude is | d₁ d₂ | = (3 + 4)^2 + 49 + (-6 - )^2 = 10 ^2 + 36 + 101 . Given the area is 117 2 , equate: 117 2 = 1 2 10 ^2 + 36 + 101 Multiplying both sides by 2 and squaring: 117 = 10 ^2 + 36 + 101 . Rearranged as the quadratic 5 ^2 + 18 - 8 = 0 . Solving, the discriminant is 324 + 160 = 484 = 22 , s