If `veca,vecb,vecc` are mutually perpendicular vectors of equal magnitudes , show that the vector `vecc*vecd=15` is equally inclined to `veca,vecbandvecc`.

Explanation

Since `veca,vecb and vecc` are mutually perpendicular vectrors, we have `veca.vecb=vecb.vecc=vecc.veca=0`
it is given that
`|veca|=|vecb|=|vecc|`
Let vector `veca+vecb+vecc` be inclined to `veca,vecb and vecc` at angle `theta_(1),theta_(2)and theta_(3)` respectively. then , we have
`costheta_(1)=((veca+vecb+vecc).veca)/(|veca+vecb+vecc||veca|)=(veca.veca+vecb.veca+vecc.veca)/(|veca+vecb+vecc||veca|)`
`(|veca|^(2))/(|veca+vecb+vecc||veca|)[vecb.veca.vecc.veca=0]`
`=(|veca|)/(|veca+vecb+vecc|)`
`costheta_(2)=((veca+vecb+vecc).vecb)/(|veca+vecb+vecc||vecb|)=(veca.veca+vecb.veca+vecc.vecb)/(|veca+vecb+vecc|.|vecb|)`
`=(|vecb|^(2))/(|veca+vecb+vecc|.|vecb|)[veca.vecb=vecc.vecc.vecb=0]`
`(|vecb|)/(|veca+vecb+vecc|)`
`costheta_(3)=((veca+vecb+vecc).vecc)/(|veca+vecb+vecc||vecc|)=(veca.vecc+vecb.vecc+vecc.vecc)/(|veca+vecb+vecc|.|vecc|)`
`(|vecc|^(2))/(|veca+vecb+vecc|.|vecc|)[veca.vecc=vecb.vecc=0]`
`= (|vecc|)/(|veca+vecb+vecc|)`
now as `|veca|=|vecb|=|vecc|, costheta_(1)=costheta_(2)=costheta_(3)`
`theta_(1)=theta_(2)=theta_(3)`
Hence, the vector `(veca+vecb+vecc)` is equally inclined to `veca,vecb and vecc`.