Given vectors a=3p-2q , b=-2p+q and c=7p-4q , where p and q…

Given vectors $\vec{a}=3\vec{p}-2\vec{q}$, $\vec{b}=-2\vec{p}+\vec{q}$ and $\vec{c}=7\vec{p}-4\vec{q}$, where $\vec{p}$ and $\vec{q}$ are unit vectors with $\angle(\vec{p},\vec{q})=\dfrac{\pi}{3}$. Decompose vector $\vec{a}$ along the directions of vectors $\vec{b}$ and $\vec{c}$ (analytically and geometrically).

Step 1: Set up

We seek $\lambda,\mu$ with $\vec{a} = \lambda\vec{b} + \mu\vec{c}$: \[3\vec{p}-2\vec{q} = \lambda(-2\vec{p}+\vec{q}) + \mu(7\vec{p}-4\vec{q}) = (-2\lambda+7\mu)\vec{p} + (\lambda-4\mu)\vec{q}\]

$\vec{p}$ and $\vec{q}$ are non-collinear ($60^\circ$ apart), so we may compare coefficients: \[\begin{cases} -2\lambda + 7\mu = 3 & (i)\\ \lambda - 4\mu = -2 & (ii) \end{cases}\]

Step 2: Solve

From $(ii)$: $\lambda = 4\mu - 2$. Substituting into $(i)$: \[-2(4\mu-2) + 7\mu = 3 \;\Longrightarrow\; -\mu = -1 \;\Longrightarrow\; \mu = 1, \qquad \lambda = 2.\]

\[\vec{a} = 2\vec{b} + \vec{c}\]

Verification

\[ 2(-2\vec{p}+\vec{q}) + (7\vec{p}-4\vec{q}) = 3\vec{p}-2\vec{q} = \vec{a} \;\checkmark \]

Step 3: Magnitudes

Use $|\vec p|=|\vec q|=1$ and $\vec p\cdot\vec q=\cos\frac{\pi}{3}=\frac12$: \[|\vec{b}|^2 = 4 - 4\cdot\tfrac{1}{2} + 1 = 3, \qquad |\vec{b}| = \sqrt{3}\]

\[ |\vec{c}|^2 = 49 - 56\cdot\tfrac{1}{2} + 16 = 37, \qquad |\vec{c}| = \sqrt{37} \]

\[ |\vec{a}|^2 = 9 - 12\cdot\tfrac{1}{2} + 4 = 7, \qquad |\vec{a}| = \sqrt{7} \]

The components have lengths $|2\vec{b}| = 2\sqrt{3}$ and $|\vec{c}| = \sqrt{37}$.

Step 4: Angle between $\vec{b}$ and $\vec{c}$

\[ \vec{b}\cdot\vec{c} = (-2\vec{p}+\vec{q})\cdot(7\vec{p}-4\vec{q}) = -14 + 15\,(\vec p\cdot\vec q) - 4 = -18 + \tfrac{15}{2} = -\tfrac{21}{2} \]

\[ \cos\angle(\vec{b},\vec{c}) = \frac{-\frac{21}{2}}{\sqrt{3}\,\sqrt{37}} = -\frac{21}{2\sqrt{111}} \approx -0.9966, \qquad \angle(\vec{b},\vec{c}) \approx 175.3^\circ \]

$\vec b$ and $\vec c$ are nearly antiparallel — a strongly oblique decomposition.

Step 5: Coordinates for the drawing

Take $\vec{p} = (1,0)$, $\vec{q} = \left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right)$. Then

\[ \vec{a} = (2,\,-\sqrt{3}), \qquad \vec{b} = \left(-\tfrac{3}{2},\,\tfrac{\sqrt{3}}{2}\right), \qquad \vec{c} = (5,\,-2\sqrt{3}), \] and indeed

\[ 2\vec{b}+\vec{c} = (-3,\sqrt{3})+(5,-2\sqrt{3}) = (2,-\sqrt{3}) = \vec{a} \;\checkmark \]