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Question
Let `vec"a", vec"b", vec"c"` be three non-zero vectors such that `vec"c"` is a unit vector perpendicular to both `vec"a"` and `vec"b"`. If the angle between `vec"a"` and `vec"b"` is `pi/6`, show that `[(vec"a", vec"b", vec"c")]^2 = 1/4|vec"a"|^2|vec"b"|^2`
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Solution
Given `vec"c"` is perpendicular to both `vec"a"` and `vec"b"`
So `vec"c"` is parallel to `vec"a" xx vec"b"`
`[(vec"a", vec"b", vec"c")] = vec"a"*(vec"b" xx vec"c")`
`|[(vec"a", vec"b", vec"c")]| = |vec"a"||vec"b" xx vec"c"|`
= `|vec"a"|vec"b"|vec"c"| sin(pi/6)`
`|[(vec"a", vec"b", vec"c")]| = |vec"a"||vec"b"||vec"c"|(1/2)`
Squaring on both sides `[(vec"a", vec"b", vec"c")]^2 = |vec"a"||vec"b"||vec"c"|^2 1/4` ..........`("since" |vec"c"| = 1)`
`[(vec"a", vec"b", vec"c")]^2 = 1/4|vec"a"|^2|vec"b"|^2`
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