Determine α and β from the vector equality c = αa+βb , where…

Determine $\alpha$ and $\beta$ from the vector equality $\vec{c} = \alpha\vec{a}+\beta\vec{b}$, where $\vec{a} = 4\vec{i}+\vec{j}$, $\vec{b} = -2\vec{i}+3\vec{j}$, $\vec{c} = \vec{i}+4\vec{j}$, and $\vec{i}$, $\vec{j}$ are unit vectors of a rectangular coordinate system. Then decompose $\vec{a}$ analytically and geometrically in terms of $\vec{b}$ and $\vec{c}$.

Step 1: Find $\alpha$ and $\beta$

Set up the system

\[ \vec{i}+4\vec{j} = \alpha(4\vec{i}+\vec{j}) + \beta(-2\vec{i}+3\vec{j}) = (4\alpha - 2\beta)\vec{i} + (\alpha + 3\beta)\vec{j} \]

Equating the coefficients of $\vec{i}$ and $\vec{j}$: \[\begin{cases} 4\alpha - 2\beta = 1 \\ \alpha + 3\beta = 4 \end{cases}\]

Solve the system

From the second equation $\alpha = 4 - 3\beta$. Substituting into the first: \[4(4-3\beta) - 2\beta = 1 \;\Longrightarrow\; 16 - 14\beta = 1 \;\Longrightarrow\; \beta = \frac{15}{14}, \qquad \alpha = 4 - \frac{45}{14} = \frac{11}{14}.\]

\[\alpha = \frac{11}{14}, \qquad \beta = \frac{15}{14}\]

Verification

\[ \frac{11}{14}(4\vec{i}+\vec{j}) + \frac{15}{14}(-2\vec{i}+3\vec{j}) = \frac{1}{14}\big[(44-30)\vec{i}+(11+45)\vec{j}\big] = \vec{i}+4\vec{j} = \vec{c} \;\checkmark \]

Step 2: Decompose $\vec{a}$ in terms of $\vec{b}$ and $\vec{c}$

Analytical decomposition

We seek $p$ and $q$ such that $\vec{a} = p\vec{b} + q\vec{c}$: $4\vec{i}+\vec{j} = p(-2\vec{i}+3\vec{j}) + q(\vec{i}+4\vec{j}) = (-2p+q)\vec{i} + (3p+4q)\vec{j}$.

\[ \begin{cases} -2p + q = 4 \\ 3p + 4q = 1 \end{cases} \]

From the first equation $q = 4+2p$. Substituting: \[3p + 16 + 8p = 1 \;\Longrightarrow\; p = -\frac{15}{11}, \qquad q = 4 - \frac{30}{11} = \frac{14}{11}.\]

\[\vec{a} = -\frac{15}{11}\,\vec{b} + \frac{14}{11}\,\vec{c}\]

Verification

\[ -\frac{15}{11}(-2\vec{i}+3\vec{j}) + \frac{14}{11}(\vec{i}+4\vec{j}) = \frac{1}{11}\big[(30+14)\vec{i}+(-45+56)\vec{j}\big] = 4\vec{i}+\vec{j} = \vec{a} \;\checkmark \]

Geometric decomposition

$\vec a$ is the diagonal of the parallelogram built on $p\vec b$ and $q\vec c$. Because $p\lt 0$, the component along $\vec b$ points opposite to $\vec b$; because $q\gt 1$, the component along $\vec c$ is a little longer than $\vec c$.