Step 1

Let \(a=i+j,\ b=i+k\)

The cross product \(a\times b\) is arthogonal to both a and b

\(a\times b=\left|\begin{matrix}i & j & k \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}\right|\)

\(=i(1-0)-j(1-0)+k(0-1)\)

\(a\times b=i-j-k\)

Since the vector we found had magnitude \(\sqrt{3}\)

\(i.e.\ |a\times b|=\sqrt{\check{1}+\check{1}+\check{1}}=\sqrt{3}\)

A unit vector that is arthogonal to both

\(i+j\) and \(i+k\) is

\(u=\frac{(i-j-k)}{\sqrt{3}}\)

or

\(u=\frac{\sqrt{3}(i-j-k)}{3}\)