Find the unit vector vec n_0 of the vector vec n coplanar…

Find the unit vector $\vec n_0$ of the vector $\vec n$ coplanar with $\vec p$ and $\vec q$, given $|\vec p|=2$, $|\vec q|=3$, $\angle(\vec p,\vec q)=\dfrac\pi3$, $\vec n\cdot\vec p=7$, $\vec n\cdot\vec q=3$.

Explanation of Method

The unit vector coplanar with p and q from given scalar products.

Step 1: Ansatz

\[ \vec n=\alpha\vec p+\beta\vec q,\qquad \vec p^{\,2}=4,\quad \vec q^{\,2}=9,\quad \vec p\cdot\vec q=3. \]

Step 2: System

\[ \vec n\cdot\vec p=4\alpha+3\beta=7,\qquad \vec n\cdot\vec q=3\alpha+9\beta=3. \]

From the second equation $\alpha=1-3\beta$, so $4-9\beta=7$: \[\beta=-\tfrac13,\qquad \alpha=2,\qquad \vec n=2\vec p-\tfrac13\vec q.\]

Step 3: Length and unit vector

\[ |\vec n|^2=4\cdot4-2\cdot2\cdot\tfrac13\cdot3+\tfrac19\cdot9=16-4+1=13. \]

\[\vec n_0=\frac{1}{\sqrt{13}}\left(2\vec p-\tfrac13\vec q\right)=\frac{6\vec p-\vec q}{3\sqrt{13}}\]

Check: $\vec n\cdot\vec p=8-1=7$ and $\vec n\cdot\vec q=6-3=3$ $\checkmark$

A shortcut for the length is $|\vec n|^2=\vec n\cdot(2\vec p-\tfrac13\vec q)=2\cdot7-\tfrac13\cdot3=13$.

$\tfrac{\pi}{3}$ $O$ $\vec p$ $\vec q$ $\vec n$ $\vec n_0$ $3.5$ $1$

$\vec n=2\vec p-\tfrac13\vec q,\ |\vec n|=\sqrt{13}$
$\vec n_0=\vec n/\sqrt{13}$

Dashed: the projections of $\vec n$, $\mathrm{Pr}_{\vec p}\vec n=\tfrac{7}{2}$ and $\mathrm{Pr}_{\vec q}\vec n=\tfrac{3}{3}=1$. They encode $\vec n\cdot\vec p=7$ and $\vec n\cdot\vec q=3$.

Result

\[\boxed{\,\vec n_0=\frac{1}{\sqrt{13}}\left(2\vec p-\tfrac13\vec q\right)=\frac{6\vec p-\vec q}{3\sqrt{13}}\,}\]