Determine the parameter α so that vectors a = 2i+3j and b =…

Determine the parameter $\alpha$ so that vectors $\vec{a} = 2\vec{i}+3\vec{j}$ and $\vec{b} = \alpha\vec{i}-\vec{j}$ are collinear.

Step 1: Collinearity condition

Vectors $\vec{a} = a_x\vec{i}+a_y\vec{j}$ and $\vec{b} = b_x\vec{i}+b_y\vec{j}$ are collinear if and only if their components are proportional, i.e. the determinant of their components vanishes:

\[ \frac{a_x}{b_x} = \frac{a_y}{b_y} \qquad\Longleftrightarrow\qquad a_x b_y - a_y b_x = 0. \]

Step 2: Apply the condition

Here $a_x = 2,\ a_y = 3,\ b_x = \alpha,\ b_y = -1$: \[(2)(-1) - (3)(\alpha) = 0 \;\Longrightarrow\; -2 - 3\alpha = 0 \;\Longrightarrow\; 3\alpha = -2\]

\[\alpha = -\frac{2}{3}\]

Step 3: Verification via proportionality

With $\alpha = -\tfrac{2}{3}$, $\vec{b} = -\tfrac{2}{3}\vec{i} - \vec{j}$, and

\[ \frac{a_x}{b_x} = \frac{2}{-\frac{2}{3}} = -3, \qquad \frac{a_y}{b_y} = \frac{3}{-1} = -3 \;\checkmark \]

Step 4: Scalar multiple form

\[ \vec{a} = -3\,\vec{b}, \qquad -3\left(-\frac{2}{3}\vec{i}-\vec{j}\right) = 2\vec{i}+3\vec{j} = \vec{a} \;\checkmark \]

Since $\lambda = -3 \lt 0$, the vectors are antiparallel (opposite directions), and $|\vec a| = 3|\vec b|$.