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If vec aa n d vec b are two vectors and...

If ` vec aa n d vec b` are two vectors and angle between them is `theta,` then `| vec axx vec b|^2+( vec adot vec b)^2=| vec a|^2| vec b|^2` `| vec axx vec b|=( vec adot vec b),iftheta=pi//4` ` vec axx vec b=( vec adot vec b) hat n ,(w h e r e hat n` is unit vector,`)` if `theta=pi//4` `( vec axx vec b)dot( vec a+ vec b)=0`

A

`|vecaxxvecb|^(2)+ (veca.vecb)^(2)= |veca|^(2)|vecb|^(2)`

B

`|vecaxxvecb|^(2)+ (veca.vecb)^(2), if theta= pi//4`

C

`veca xx vecb = (veca. Vecb) hatn` ( where `hatn` is a normal unit vector ) `if theta f= pi//4`

D

`(veca xx vecb ) . (veca + vecb) =0`

Text Solution

Verified by Experts

The correct Answer is:
a,b,c,d

`vecaxx vecb= |veca||vecb| sin theta hatn `
`or |vecaxx vecb|=|veca||vecb|sintheta`
`or sin theta (|vecaxxvecb|)/(|veca||vecb|)`
`veca. vecb= |veca||vecb|cos theta`
`Rightarrow cos theta= (|veca.vecb|)/(|veca||vecb|)`
From (i) and (ii) .
`sin^(2)theta + cos ^(2) theta=1`
`if theta= pi//4 "then" sintheta=costheta= 1//sqrt2.` Therefore,
`|vecaxxvecb|= (|veca||vecb|)/sqrt2 and veca.vecb= (|veca||vecb|)/sqrt2`
`|vecaxxvecb|= veca.vecb`
`vecaxxvecb= |veca||vecb|sinthetahatn = (|veca||vecb|)/sqrt2hatn`
`(veca.vecb)hatn`
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