If p and q be any two unit vectors ... And (p+2q) and (q+2p) are the diagonals of a parallelogram ...the angle between p and q is 30 degrees, then AREA of Parallelogram is?
The desired area is = |(p+2q)x(q+2p)| /2
= |pxq+2pxp+2qxq+ 4qxp|/2
= |pxq+0+0-4pxq|/ 2
= (3/2)|pxq|
= (3/ 2)|p||q|*sin(30)
= (3/2)*1*1*(1/2)
= 3/4.
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