Three unit vectors vec i,vec j,vec k make the angles…

Three unit vectors $\vec i,\vec j,\vec k$ make the angles $\angle(\vec i,\vec j)=45^\circ$, $\angle(\vec j,\vec k)=90^\circ$, $\angle(\vec k,\vec i)=120^\circ$. For $\vec A=\vec i-\vec j+\vec k$ and $\vec B=\vec i+\vec j$, find $|\vec A\times\vec B|$.

Step 1: Scalar products of the basis

\[ \vec i\cdot\vec j=\tfrac{\sqrt2}{2},\qquad \vec j\cdot\vec k=0,\qquad \vec k\cdot\vec i=-\tfrac12,\qquad \vec i^{\,2}=\vec j^{\,2}=\vec k^{\,2}=1. \]

The coordinate formula for the vector product does not apply in an oblique basis, so the Lagrange identity is used instead: $|\vec A\times\vec B|^2=|\vec A|^2|\vec B|^2-(\vec A\cdot\vec B)^2$.

Step 2: Ingredients

\[ |\vec A|^2=3-2\,\vec i\cdot\vec j+2\,\vec i\cdot\vec k-2\,\vec j\cdot\vec k=3-\sqrt2-1-0=2-\sqrt2, \]

\[ |\vec B|^2=2+2\,\vec i\cdot\vec j=2+\sqrt2, \]

\[ \vec A\cdot\vec B=\vec i^{\,2}-\vec j^{\,2}+\vec k\cdot\vec i+\vec k\cdot\vec j=1-1-\tfrac12+0=-\tfrac12. \]

Step 3: Result

\[ |\vec A\times\vec B|^2=(2-\sqrt2)(2+\sqrt2)-\tfrac14=2-\tfrac14=\tfrac74. \]

\[|\vec A\times\vec B|=\frac{\sqrt7}{2}\approx1.32\]

Consistency check (added). Such three unit vectors really exist, because their Gram determinant is positive:

\[ \det\begin{pmatrix}1&\tfrac{\sqrt2}2&-\tfrac12\\\tfrac{\sqrt2}2&1&0\\-\tfrac12&0&1\end{pmatrix}=1-\tfrac12-\tfrac14=\tfrac14\gt0. \]