Find the area of the parallelogram whose diagonals are vec…

Find the area of the parallelogram whose diagonals are $\vec d_1=3\vec m+3\vec n$ and $\vec d_2=\vec m-\vec n$, where $\vec m,\vec n$ are unit vectors with $\angle(\vec m,\vec n)=\dfrac\pi6$.

Explanation of Method

The area of a parallelogram from its diagonal vectors.

Step 1: Area from the diagonals

If the sides are $\vec a,\vec b$, then $\vec d_1=\vec a+\vec b$ and $\vec d_2=\vec a-\vec b$, so

\[ \vec d_1\times\vec d_2=(\vec a+\vec b)\times(\vec a-\vec b)=-\vec a\times\vec b+\vec b\times\vec a=-2\,\vec a\times\vec b\ \Longrightarrow\ S=|\vec a\times\vec b|=\tfrac12|\vec d_1\times\vec d_2|. \]

(The source uses this formula without proof.)

Step 2: Computation

\[ \vec d_1\times\vec d_2=(3\vec m+3\vec n)\times(\vec m-\vec n)=-3\,\vec m\times\vec n+3\,\vec n\times\vec m=-6\,\vec m\times\vec n, \]

\[ |\vec m\times\vec n|=\sin\tfrac\pi6=\tfrac12,\qquad |\vec d_1\times\vec d_2|=3. \]

\[S=\frac32\]

Step 3: Check through the sides

\[ \vec a=\tfrac12(\vec d_1+\vec d_2)=2\vec m+\vec n,\qquad \vec b=\tfrac12(\vec d_1-\vec d_2)=\vec m+2\vec n, \]

\[ \vec a\times\vec b=(2\cdot2-1\cdot1)\,\vec m\times\vec n=3\,\vec m\times\vec n,\qquad S=3\cdot\tfrac12=\tfrac32.\ \checkmark \]

$\tfrac\pi6$ $O$ $\vec m$ $\vec n$ $\vec a=2\vec m+\vec n$ $\vec b=\vec m+2\vec n$ $\vec d_1$ $\vec d_2$

$\vec d_1=3\vec m+3\vec n=\vec a+\vec b$
$\vec d_2=\vec m-\vec n=\vec a-\vec b$

The sides are recovered from the diagonals: $\vec a=\tfrac12(\vec d_1+\vec d_2)$, $\vec b=\tfrac12(\vec d_1-\vec d_2)$. Area $=|\vec a\times\vec b|=\tfrac32$.

Result

\[\boxed{\,S=\frac32\,}\]