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How do you find the Area of the Parallelogram with Vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)?
We will be using the concept of cross-product of two adjacent vectors, a and b to find the area of the parallelogram when vertices given.
Answer: The Area of the Parallelogram with Vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3) is √265.
Let's solve this step by step.
Explanation:
Given that, vertices k(1, 2, 3), l(1, 3, 6), m(3, 8, 6), and n(3, 7, 3)
Area of Parallelogram = |KL × KN|
KL = (1, 2, 3) - (1, 3, 6) = (0, 1, 3)
KN = (1, 2, 3) - (3, 7, 3) = (2, 5, 0)
Area of the parallelogram = cross product of two vectors, represented by the adjacent sides.
Area of Parallelogram = |(0, 1, 3) × (2, 5, 0)|
= |i(1 × 0 - 5 × 3) - j(0 × 0 - 2 × 3) + k(0 × 5 - 2 × 1)|
= |i(-15) - j(-6) + k(-2)|
= |-15i + 6j -2 k|
= √(-15)2 + (6)2 + (-2)2
= √(225 + 36 + 4)
= √265
Thus, the area of the parallelogram with vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3) is √265.
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