If |a|=2,|b|=3 and a,b are mutually perpendicular, then the area of the triangle whose vertices are0,a+b,a–bis?
5
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6
8
Find the area of the triangle whose vertices are0,a+b,a–b:
Suppose,A,B and Care the position vectors of vertices 0,a+band a-b.
Area of triangle ,
=12|(a+b)x(a-b)|=12|2bxa|=|b||a|sinθ=3×2sin900=6(∵sin900=1)formula=12|ABxAC|
Hence, the correct option is (C).
The area of triangle whose vertices are (0, 0), (a2, 0) and (0, b2) is: