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Question
If `veca` and `vecb` are unit vectors inclined at an angle 30° to each other, then find the area of the parallelogram with `(veca + 3vecb)` and `(3veca + vecb)` as adjacent sides.
Solution
We know, Area of parallelogram with adjacents sides `vecp` and `vecq` is given by A = `|vecp xx vecq|`
Here, Area = `|(veca + 3vecb) xx (3veca + vecb)|`
= `|3(veca xx veca) + (veca xx vecb) + 9(vecb xx veca) + 3(vecb xx vecb)|`
= `|3 xx 0 + (veca xx vecb) -9(veca xx vecb) + 3 xx 0|` ...`[∵ veca xx veca = 0 = vecb xx vecb and vecb xx veca = -veca xx vecb]`
= `|-8(veca xx vecb)|`
= `8|veca xx vecb|`
= `8|veca|.|vecb| sin θ`
= 8.1.1. sin 30° ...[Given, `|veca|` = 1 = `|vecb|` and θ = 30°]
= `8. 1/2`
= 4 sq.units
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