Vectors `veca,vecb and vecc` are of the same length and when taken pair-wise they form equal angles. If `veca=hati+hatj and vecb=hatj+hatk` then find vector `vecc`.

Explanation

`Let vecc=xhati+yhatj+zhatk`. Then `|veca|=|vecb|=|vecc|Rightarrowx^(2)+y^(2)+z^(2)=2`
it is given that the angles between the vectors taken in pairs are equal, say `theta` . Therefore,
`cos theta=(veca.vecb)/(|veca||vecb|)= (0+1+1)/(sqrt2sqrt2)=1/2`
`Rightarrow (veca.vecc)/(|veca||vecc|)=1/2and (vecb.vecc)/(|vecb||vecc|)=1/2`
`(x+y)/(sqrt2sqrt2)=1/2 and(y+z)/(sqrt2sqrt2)=1/2`
`Rightarrow x+y=1 and y+z=1`
y=1-x and z=1-y=1-(1-x)=x
`x^(2)+y^(2)+z^(2)=2Rightarrowx^(2)+(1-x)^(2)+x^(2)=2`
`Rightarrow (3x+1)(x-1)= Rightarrowx=1,1//3` ,
`y=1-xRightarrow0 "for" x=1andy=4//3for x=-1//3`
`Hence, vecc=hati+0hatj+hatk and vecc=1/3hati+4/3hatj-1/3hatk`