Let `vecA=5veci-4vecj and vecB=-7.5veci+6vecj`. Do we have `vecB=kvecA`? Can we say `vecB/vecA=k`?

For three vectors veca, vecb, vecc satisfies veca+ vecb + vecc = vec0 and |veca| = 3 , |vecb| = 4, |vecc| =2 then veca. vecb + vecb. vecc + vecc.veca = _____________.

If veca and vecb are two unit vectors such that veca+2vecb and 5veca-4vecb are perpendicular to each other, then the angle between veca and vecb is

Assertion : The direction of a zero (null) vector is indeteminate. Reason : We can have vecAxx vecB = vecA *vecB with vecA ne vec0 and vecB ne vec0 .

Three vectors veca,vecbandvecc satisfy the condition veca+vecb+vecc=vec0 . Evaluate the quantity mu=veca*vecb+vecb*vecc+vecc*veca , if |veca|=3,|vecb|=4and|vecc|=2 .

Let veca=(1,1,-1), vecb=(5,-3,-3) and vecc=(3,-1,2) . If vecr is collinear with vecc and has length (|veca+vecb|)/(2) , then vecr equals

Let veca=hati+2hatjandvecb=2hati+hatj . Is |veca|=|vecb| ? Are the vectors vecaandvecb equal?

If veca+vecb+vecc=vec0, |veca| = 3, |vecb| = 5, |vecc| = 7 , then angle between veca and vecb is

vecA, vecB and vecC are vectors each having a unit magneitude. If vecA + vecB+vecC=0 , then vecA. vecB+vecB.vecC+ vecC.vecA will be:

The magnitudes of vectors vecA,vecB and vecC are respectively 12,5 and 13 unit and vecA+vecB=vecC , then the angle between vecA and vecB is :