Given vectors p=a+b+c , a=3m-n , b=m-2n and c=-m+7n . Find…

Given vectors $\vec{p}=\vec{a}+\vec{b}+\vec{c}$, $\vec{a}=3\vec{m}-\vec{n}$, $\vec{b}=\vec{m}-2\vec{n}$ and $\vec{c}=-\vec{m}+7\vec{n}$. Find the linear dependence relation between vectors $\vec{p}$, $\vec{a}$ and $\vec{b}$.

Step 1: Compute $\vec{p}$

\[ \vec{p} = (3\vec{m}-\vec{n}) + (\vec{m}-2\vec{n}) + (-\vec{m}+7\vec{n}) = 3\vec{m}+4\vec{n} \]

So, in terms of $\vec m$ and $\vec n$ (assumed non-collinear): \[\vec{p} = 3\vec{m}+4\vec{n}, \qquad \vec{a} = 3\vec{m}-\vec{n}, \qquad \vec{b} = \vec{m}-2\vec{n}.\]

Step 2: Find the dependence

We seek $\alpha,\beta,\gamma$, not all zero, with $\alpha\vec{p} + \beta\vec{a} + \gamma\vec{b} = \vec{0}$: $(3\alpha+3\beta+\gamma)\vec{m} + (4\alpha-\beta-2\gamma)\vec{n} = \vec{0}$.

\[ \begin{cases} 3\alpha + 3\beta + \gamma = 0 & (i) \\ 4\alpha - \beta - 2\gamma = 0 & (ii) \end{cases} \]

Step 3: Solve the homogeneous system

Two equations, three unknowns: a one-parameter family of solutions. Set $\alpha = 1$. From $(i)$: $\gamma = -3 - 3\beta$. Substituting into $(ii)$: \[4 - \beta + 6 + 6\beta = 0 \;\Longrightarrow\; \beta = -2, \qquad \gamma = -3 + 6 = 3.\]

Step 4: Linear dependence relation

\[ \vec{p} - 2\vec{a} + 3\vec{b} = \vec{0} \]

\[\vec{p} = 2\vec{a} - 3\vec{b}\]

Verification

\[ 2(3\vec{m}-\vec{n}) - 3(\vec{m}-2\vec{n}) = 6\vec m-2\vec n-3\vec m+6\vec n = 3\vec{m}+4\vec{n} = \vec{p} \;\checkmark \]

So $\vec{p}$, $\vec{a}$, $\vec{b}$ are linearly dependent: $\vec p$ lies in the plane spanned by $\vec a$ and $\vec b$.

Step 5: Graphical construction

With $\vec{m}=(1,0)$, $\vec{n}=(0,1)$: $\vec{p} = (3,4)$, $\vec{a} = (3,-1)$, $\vec{b} = (1,-2)$, and $2\vec{a}-3\vec{b} = (6,-2)+(-3,6) = (3,4)$ $\checkmark$