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Suppose that the area of a parallelogram with vertices (0, 0, 0), (1, 0, 0), (1, 9, a) and (2, 9, a) is sqrt*97 . Suppose also that the third vertex, (1, 9, a), makes an obtuse angle with the positive z-axis. What is the value of a ?
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familiar wid cross product, right ?
you need to solve : \[\langle 1,0,0 \rangle \times \langle 1,9,a \rangle = \sqrt{97}\]
no, havent had much practice in cross products
\(||\langle 1,0,0 \rangle \times \langle 1,9,a \rangle|| = \sqrt{97}\) *
cross product of two vectors is defined as : \(\vec{A} \times \vec{B} = |\vec{A}||\vec{B}|\sin \theta \)
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