A parallelogram is built on the vectors vec m=2vec a-vec…
A parallelogram is built on the vectors $\vec m=2\vec a-\vec b+\vec c$ and $\vec n=-3\vec a-\vec b+\vec c$, where $\vec a,\vec b,\vec c$ are unit vectors with $\angle(\vec a,\vec b)=\angle(\vec a,\vec c)=30^\circ$ and $\angle(\vec b,\vec c)=150^\circ$. Find the lengths of its diagonals.
Explanation of Method
Diagonals of the parallelogram built on m and n.
Step 1: Basic products
\[ \vec a\cdot\vec b=\vec a\cdot\vec c=\frac{\sqrt3}{2},\qquad \vec b\cdot\vec c=-\frac{\sqrt3}{2}. \]
Step 2: The diagonals as vectors
\[ \vec d_1=\vec m+\vec n=-\vec a-2\vec b+2\vec c,\qquad \vec d_2=\vec m-\vec n=5\vec a. \]
Step 3: Lengths
\[ |\vec d_1|^2=1+4+4+2\left[2\,\vec a\cdot\vec b-2\,\vec a\cdot\vec c-4\,\vec b\cdot\vec c\right]=9+2\left[\sqrt3-\sqrt3+2\sqrt3\right]=9+4\sqrt3, \]
\[ |\vec d_2|=|5\vec a|=5. \]
\[|\vec d_1|=\sqrt{9+4\sqrt3}\approx3.97,\qquad |\vec d_2|=5\]
Step 4: Consistency check
Similarly, \[|\vec m|^2=6+\sqrt3,\qquad |\vec n|^2=11+\sqrt3,\qquad \vec m\cdot\vec n=\sqrt3-4.\]
These values satisfy the parallelogram law: $|\vec d_1|^2+|\vec d_2|^2=34+4\sqrt3=2\left(|\vec m|^2+|\vec n|^2\right)$.
- $\vec d_1=\vec m+\vec n,\ |\vec d_1|=\sqrt{9+4\sqrt3}$
- $\vec d_2=\vec m-\vec n=5\vec a,\ |\vec d_2|=5$
Result
\[\boxed{\,|\vec d_1|=\sqrt{9+4\sqrt3}\approx3.97,\qquad |\vec d_2|=5\,}\]
Remark
Warning — the data are inconsistent. Three unit vectors with the angles $30^\circ,30^\circ,150^\circ$ do not exist: the angles between three directions must satisfy $\angle(\vec b,\vec c)\le\angle(\vec a,\vec b)+\angle(\vec a,\vec c)$, and $150^\circ\gt60^\circ$. Equivalently, the Gram determinant is negative:
\[ \det\begin{pmatrix}1&\tfrac{\sqrt3}2&\tfrac{\sqrt3}2\\ \tfrac{\sqrt3}2&1&-\tfrac{\sqrt3}2\\ \tfrac{\sqrt3}2&-\tfrac{\sqrt3}2&1\end{pmatrix}=-\frac{5+3\sqrt3}{4}\approx-2.55\lt0. \]
The answers above are therefore only formal results of the algebra. $|\vec d_2|=5$ holds for any unit vector $\vec a$, whatever the other angles are.