1.

if `veca,vecb and vecc` are mutally perpendicular vectors of equal magnitudes, then find the angle between vectors and` veca+ vecb=vecc`.

Answer» since `veca,vecb and vecc` are mutually perpendicular , `veca.vecb=vecb.vecc=vecc.veca=0`
Angle between `veca and veca+vecb+vecc` is
`cos theta=(veca.(veca+vecb+vecc))/(|veca||veca+vecb=vecc|)`
`now |veca|=|vecb|=|vecc|=a`
`|veca+vecb=vecc|^(2)= |veca|^(2)|vecb|^(2)+|vecc|^(2)+2veca.vecb+2vecb.vecc+vecc.veca`
`=a^(2)=a^(2)+a^(2)+0+0+0`
`3a^(2)`
`|veca+vecb+vecc|=sqrt3a`
Putting this value in (i) , we get `theta = cos^(-1) 1/sqrt3`


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