Find the scalar product of vectors a=2m-n and b=m-2n , where…

Find the scalar product of vectors $\vec{a}=2\vec{m}-\vec{n}$ and $\vec{b}=\vec{m}-2\vec{n}$, where $|\vec{m}|=2$, $|\vec{n}|=4$ and $\angle(\vec{m},\vec{n})=\dfrac{\pi}{3}$.

Explanation of Method

Scalar product of vectors expressed through $\vec m$ and $\vec n$.

Step 1: Basic products

\[ \vec{m}\cdot\vec{m} = 4, \qquad \vec{n}\cdot\vec{n} = 16, \qquad \vec{m}\cdot\vec{n} = 2\cdot4\cdot\cos\frac{\pi}{3} = 4 \]

Step 2: Expand

\[ \vec{a}\cdot\vec{b} = (2\vec{m}-\vec{n})\cdot(\vec{m}-2\vec{n}) = 2\,\vec m\cdot\vec m - 4\,\vec m\cdot\vec n - \vec n\cdot\vec m + 2\,\vec n\cdot\vec n = 2\,\vec m\cdot\vec m - 5\,\vec m\cdot\vec n + 2\,\vec n\cdot\vec n \]

\[ \vec{a}\cdot\vec{b} = 2\cdot 4 - 5\cdot 4 + 2\cdot 16 = 8 - 20 + 32 \]

\[\vec{a}\cdot\vec{b} = 20\]

Step 3: Lengths and angle

\[ |\vec{a}|^2 = 4\cdot4 - 4\cdot4 + 16 = 16, \qquad |\vec{a}| = 4 \]

\[ |\vec{b}|^2 = 4 - 4\cdot4 + 4\cdot16 = 52, \qquad |\vec{b}| = 2\sqrt{13} \]

\[ \cos\angle(\vec{a},\vec{b}) = \frac{20}{4\cdot2\sqrt{13}} = \frac{5\sqrt{13}}{26} \approx 0.693, \qquad \angle(\vec a,\vec b) \approx 46.1^\circ \]

$\tfrac\pi3$ $\approx46^\circ$ $O$ $\vec m$ $\vec n$ $2\vec m$ $\vec a$ $\vec b$

$\vec a=2\vec m-\vec n$
$\vec b=\vec m-2\vec n$
$-\vec n\ \text{from }2\vec m,\ \ -2\vec n\ \text{from }\vec m$

Drawn with $\vec m=(2,0)$, $\vec n=(2,2\sqrt3)$ ($|\vec n|=4$, $60^\circ$). Then $\vec a=(2,-2\sqrt3)$, $\vec b=(-2,-4\sqrt3)$ and $\vec a\cdot\vec b=-4+24=20$.

Result

\[\boxed{\,\vec{a}\cdot\vec{b} = 20\,}\]