Find the magnitude of vector a=p-2q given that |p|=2…

Find the magnitude of vector $\vec{a}=\vec{p}-2\vec{q}$ given that $|\vec{p}|=2$, $|\vec{q}|=\sqrt{3}$ and $\angle(\vec{p},\vec{q})=\dfrac{\pi}{6}$.

Explanation of Method

The length of $\vec a=\vec p-2\vec q$.

Step 1: Basic products

\[ \vec{p}\cdot\vec{p} = 4, \qquad \vec{q}\cdot\vec{q} = 3, \qquad \vec{p}\cdot\vec{q} = 2\cdot\sqrt{3}\cdot\cos\frac{\pi}{6} = 2\cdot\sqrt3\cdot\frac{\sqrt3}{2} = 3 \]

Step 2: Scalar square of $\vec a$

\[ |\vec{a}|^2 = (\vec{p}-2\vec{q})\cdot(\vec{p}-2\vec{q}) = \vec{p}\cdot\vec{p} - 4\,\vec{p}\cdot\vec{q} + 4\,\vec{q}\cdot\vec{q} = 4 - 12 + 12 = 4 \]

\[|\vec{a}| = |\vec{p}-2\vec{q}| = 2\]

Step 3: Coordinate check

Take $\vec p=(2,0)$ and $\vec q=\sqrt3(\cos30^\circ,\sin30^\circ)=\left(\tfrac32,\tfrac{\sqrt3}2\right)$. Then

\[ \vec a = (2,0) - (3,\sqrt3) = (-1,\,-\sqrt3), \qquad |\vec a| = \sqrt{1+3} = 2 \;\checkmark \]

$\tfrac\pi6$ $O$ $\vec p$ $\vec q$ $2\vec q$ $\vec a\,(-1,-\sqrt3)$ $-2\vec q$

$\vec a=\vec p-2\vec q$
$-2\vec q\ \text{from the tip of }\vec p$
$\text{circle of radius }2$

Drawn with $\vec p=(2,0)$, $\vec q=\left(\tfrac32,\tfrac{\sqrt3}2\right)$. The tips of $\vec p$ and $\vec a$ both lie on the circle of radius $2$: $|\vec a|=|\vec p|=2$.

Result

\[\boxed{\,|\vec{a}| = |\vec{p}-2\vec{q}| = 2\,}\]