The vector vec n is coplanar with vec p and vec q , where…

The vector $\vec n$ is coplanar with $\vec p$ and $\vec q$, where $|\vec p|=2$, $|\vec q|=4$, $\angle(\vec p,\vec q)=\dfrac\pi3$, $\vec n\cdot\vec p=8$, $\vec n\cdot\vec q=16$. Determine: $1^\circ$ the unit vector of $\vec n$ through $\vec p,\vec q$; $2^\circ$ $|\vec n+\vec q|$; $3^\circ$ $\angle(\vec n,\vec p)$.

Step 1: Finding $\vec n$

\[ \vec p^{\,2}=4,\quad \vec q^{\,2}=16,\quad \vec p\cdot\vec q=4;\qquad \vec n=\alpha\vec p+\beta\vec q. \]

\[ 4\alpha+4\beta=8,\quad 4\alpha+16\beta=16\ \Longrightarrow\ \beta=\tfrac23,\ \alpha=\tfrac43,\qquad \vec n=\tfrac43\vec p+\tfrac23\vec q. \]

Step 2: $1^\circ$ Unit vector

\[ |\vec n|^2=\vec n\cdot\left(\tfrac43\vec p+\tfrac23\vec q\right)=\tfrac43\cdot8+\tfrac23\cdot16=\frac{64}{3},\qquad |\vec n|=\frac{8}{\sqrt3}=\frac{8\sqrt3}{3}. \]

\[\vec n_0=\frac{\vec n}{|\vec n|}=\frac{2\vec p+\vec q}{4\sqrt3}=\frac{\sqrt3}{12}(2\vec p+\vec q)\]

Step 3: $2^\circ$ $|\vec n+\vec q|$

\[ |\vec n+\vec q|^2=|\vec n|^2+2\,\vec n\cdot\vec q+|\vec q|^2=\frac{64}{3}+32+16=\frac{208}{3}. \]

\[|\vec n+\vec q|=\sqrt{\frac{208}{3}}=\frac{4\sqrt{39}}{3}\approx8.33\]

Step 4: $3^\circ$ Angle between $\vec n$ and $\vec p$

\[ \cos\angle(\vec n,\vec p)=\frac{8}{\frac{8}{\sqrt3}\cdot2}=\frac{\sqrt3}{2}. \]

\[\angle(\vec n,\vec p)=\frac\pi6=30^\circ\]

Remark. Also $\cos\angle(\vec n,\vec q)=\dfrac{16}{\frac{8}{\sqrt3}\cdot4}=\dfrac{\sqrt3}{2}$. So $\vec n$ bisects the $60^\circ$ angle between $\vec p$ and $\vec q$.