## Calculus: Early Transcendentals 8th Edition

$a=\frac{(i-j-k)}{\sqrt 3}$
$a=xi+yj+zk$ 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, $a=\frac{(i-j-k)}{\sqrt 3}$