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Question

# →p=w^i+x^j and →q=y^i+z^j are two vectors in the first quadrant such that |→p|=2|→q|=2r,r>0 and →p⋅→q=0. If →a=w^i+2y^j and →b=x2^i+z^j, then

A
|a|=r
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B
|b|=r
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C
ab=0
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D
|a×b|=2r
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Solution

## The correct options are B |→b|=r C →a⋅→b=0Here, |→p|=√w2+x2=2r ...(1), |→q|=√y2+z2=r ...(2), →p⋅→q=wy+xz=0 ...(3) Now, |→a|=√w2+4y2 ...(4) w=−xzy from eqn (3) Putting this value in eqn (1), we get |→p|=xyr=2r ⇒x=2y Similarly, putting x=−wyz in eqn (1), we get w=2z Putting the value of w in eqn (4), |→a|=2r Similarly, we get |→b|=r →a⋅→b=wx2+2yz=wy+xz=0 [From (3)] |→a×→b|=|wz−xy|=|2z2−2y2| =2|z2−y2|≠2r

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