Given vectors a=αi+j+4k , b=i-2αj and c=3αi-3j+4k , where i…

Given vectors $\vec{a}=\alpha\vec{i}+\vec{j}+4\vec{k}$, $\vec{b}=\vec{i}-2\alpha\vec{j}$ and $\vec{c}=3\alpha\vec{i}-3\vec{j}+4\vec{k}$, where $\vec{i}$, $\vec{j}$ and $\vec{k}$ are non-coplanar (linearly independent) unit vectors. Determine the scalar $\alpha$ so that the vectors are coplanar.

Step 1: Coplanarity condition

$\vec a$ and $\vec b$ are never collinear ($\vec a$ has a $\vec k$ component, $\vec b$ does not), so by Problem 21 the three vectors are coplanar if and only if

\[ \vec{c} = \lambda\vec{a} + \mu\vec{b} \] for some scalars $\lambda,\mu$. Since $\vec{i}$, $\vec{j}$, $\vec{k}$ are linearly independent, we may compare their coefficients.

Step 2: Set up the system

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

\[ \begin{cases} \lambda\alpha + \mu = 3\alpha & (i)\\ \lambda - 2\alpha\mu = -3 & (ii)\\ 4\lambda = 4 & (iii) \end{cases} \]

Step 3: Solve

From $(iii)$: $\lambda = 1$. Then $(i)$ gives $\alpha + \mu = 3\alpha$, so $\mu = 2\alpha$. Substituting into $(ii)$: \[1 - 2\alpha\cdot 2\alpha = -3 \;\Longrightarrow\; 4\alpha^2 = 4 \;\Longrightarrow\; \alpha^2 = 1\]

\[\alpha = 1 \qquad \text{or} \qquad \alpha = -1\]

Step 4: Verification for $\alpha = 1$

$\lambda=1,\ \mu=2$; $\vec{a} = \vec{i}+\vec{j}+4\vec{k}$, $\vec{b} = \vec{i}-2\vec{j}$, $\vec{c} = 3\vec{i}-3\vec{j}+4\vec{k}$: \[\vec{a}+2\vec{b} = (1+2)\vec{i}+(1-4)\vec{j}+4\vec{k} = 3\vec{i}-3\vec{j}+4\vec{k} = \vec{c} \;\checkmark\]

Step 5: Verification for $\alpha = -1$

$\lambda=1,\ \mu=-2$; $\vec{a} = -\vec{i}+\vec{j}+4\vec{k}$, $\vec{b} = \vec{i}+2\vec{j}$, $\vec{c} = -3\vec{i}-3\vec{j}+4\vec{k}$: \[\vec{a}-2\vec{b} = (-1-2)\vec{i}+(1-4)\vec{j}+4\vec{k} = -3\vec{i}-3\vec{j}+4\vec{k} = \vec{c} \;\checkmark\]