Question

Find the area of the parallelogram determined by the vectors i +j+k and –2i + 3j + K.​

Answer 1

The area of the parallelogram spanned by the given vectors is equal to the magnitude of their cross product.

(i + j + k) × (-2i + 3j + k)

= -2 (i × i) - 2 (j × i) - 2 (k × i)

… + 3 (i × j) + 3 (j × j) + 3 (k × j)

… + (i × k) + (j × k) + (k × k)

= - 2 (j × i) - 2 (k × i) + 3 (i × j) + 3 (k × j) + (i × k) + (j × k)

= 2 (i × j) - 2 (k × i) + 3 (i × j) - 3 (j × k) - (k × i) + (j × k)

= 5 (i × j) - 3 (k × i) - 2 (j × k)

= 5 k - 3 j - 2 i

Then the area of the parallelogram is

||-2 i - 3 j + 5 k|| = √((-2)² + (-3)² + 5²) = √38