The vectors $\vec a, \vec b$ and $\vec c$ are equal in length and, taken pairwise, they make equal angles. If $\vec a =\hat i +\hat j,\vec b=\hat j+\hat k$, and $\vec c$ makes an obtuse angle with x-axis, then $\vec c =$

$\frac{1}{3}(-\hat i+4\hat j-\hat k)$

Let $c=x\hat i+y\hat j+z\hat k$. Then,

$|\vec a|=|\vec b|=|\vec c| ⇒ x^2 + y^2+z^2=2$   ...(i)

Since c makes an obtuse angle with x-axis.

$∴\vec c.\hat i<0⇒x<0$   ...(ii)

It is given that $\vec a,\vec b,\vec c$ taken pairwise make equal angles.

$∴\frac{\vec a.\vec b}{|\vec a||\vec b|}=\frac{\vec a.\vec c}{|\vec a||\vec c|}=\frac{\vec b.\vec c}{|\vec b||\vec c|}$

$⇒\frac{1}{2}=\frac{x+y}{2}=\frac{y+z}{2}$

$⇒x+y=1=y+z⇒x=z$ and $y =1-z$

Putting $x =z$ and $y = 1 -z$ in (i), we get

$z^2+(1-z)^2+z^2 = 2$

$⇒3z^2-2z-1=0⇒(3z + 1) (z-1)=0⇒ z=1, -1/3$

Since $x =z$ and $x <0$ (from (ii)). Therefore,

$x=z=-1/3 ⇒ y=4/3$

Hence, $\vec c=\frac{1}{3}(-\hat i+4\hat j-\hat k)$