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Vectors 3veca-5vecb and 2veca + vecb are...

Vectors `3veca-5vecb and 2veca + vecb` are mutually perpendicular. If `veca + 4 vecb and vecb - veca` are also mutually perpendicular, then the cosine of the angle between `veca nad vecb` is

A

`19/(5sqrt43)`

B

`19/(3sqrt43)`

C

`19/(sqrt45)`

D

`19/(6sqrt43)`

Text Solution

Verified by Experts

The correct Answer is:
a

`(3veca-5vecb).(2veca+vecb)=0`
`or 6|veca|^(2)-5|vecb|^(2)= 7 veca.vecb`
`Also , (veca+4vecb).(vecb-veca)=0`
`or -|veca|^(2)+ 4|vecb|^(2)= 3 veca.vecb`
`or 6/7 |veca|^(2)-5/7|vecb|^(2)= -1/3 |veca|^(2)+ 4/3|vecb|^(2)`
` or 25|veca|^(2) = 43 |vecb|^(2)`
` Rightarrow 3 veca. vecb= - |veca|^92) + 4|vecb|^(2)= 57/25 |vecb|^(2)`
`or 3 |veca||vecb| cos theta = 57/25 |vecb|^(2)`
`or 3sqrt(43/25) |vecb|^(2) cos theta = 57/25 |vecb|^(2)`
` or cos theta = 19/(5sqrt43)`
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