Answer
$a=\frac{(i-j-k)}{\sqrt 3}$
Work Step by Step
$a=xi+yj+zk$
Since unit vector $a$ is orthogonal to both $ i+j$ and $i+k$, then the product equals zero in both cases.
$x \cdot 1+y \cdot 1=0$
$x \cdot 1+z \cdot 1=0$
From these inequalities, we get $x=-y$ and $x=-z$
Thus, $a=xi-xj-xk$
Since $a$ is a unit vector and $|a|=\sqrt {3x^2}=1$
Thus, $x=\frac{1}{\sqrt 3}$
$a=\frac{1}{\sqrt 3}(i-j-k)$
Hence, $a=\frac{(i-j-k)}{\sqrt 3}$
