If two vectors a and b are collinear, one can always find a…

If two vectors $\vec{a}$ and $\vec{b}$ are collinear, one can always find a constant $\lambda$ such that $\vec{a} = \lambda\vec{b}$, or two constants $\alpha$ and $\beta$ such that $\alpha\vec{a}+\beta\vec{b} = \vec{0}$. What is the relation between the constants $\lambda$, $\alpha$ and $\beta$? Determine $\alpha$ if $\lambda = 1$, $\beta = 2$.

Step 1: Collinearity condition

Two vectors $\vec{a}$ and $\vec{b}$ ($\vec b\neq\vec 0$) are collinear if and only if one is a scalar multiple of the other: $\vec{a} = \lambda\vec{b}$.

Step 2: Relation between $\lambda$, $\alpha$ and $\beta$

Take $\alpha\vec{a}+\beta\vec{b} = \vec{0}$ with $\alpha \neq 0$ and divide by $\alpha$: \[\vec{a} + \frac{\beta}{\alpha}\vec{b} = \vec{0} \quad\Longrightarrow\quad \vec{a} = -\frac{\beta}{\alpha}\vec{b}.\]

Comparing with $\vec{a} = \lambda\vec{b}$: \[\lambda = -\frac{\beta}{\alpha}, \qquad \alpha = -\frac{\beta}{\lambda}, \qquad \beta = -\alpha\lambda\]

So $\alpha$ and $\beta$ have opposite signs when $\lambda \gt 0$ (codirectional vectors) and the same sign when $\lambda \lt 0$ (antiparallel vectors).

Step 3: Interpretation

The condition $\alpha\vec{a}+\beta\vec{b}=\vec{0}$ with $\alpha,\beta$ not both zero is the definition of linear dependence: two vectors are collinear exactly when they are linearly dependent. The relation $\lambda = -\beta/\alpha$ shows that any two of the three constants determine the third.

Step 4: Special case: $\lambda = 1$, $\beta = 2$

\[ \alpha = -\frac{\beta}{\lambda} = -\frac{2}{1} = -2 \]

\[\alpha = -2\]

Verification

With $\lambda = 1$ we have $\vec{a} = \vec{b}$, so

\[ \alpha\vec{a}+\beta\vec{b} = -2\vec{a}+2\vec{b} = 2(\vec{b}-\vec{a}) = \vec{0} \;\checkmark, \qquad \lambda = -\frac{\beta}{\alpha} = -\frac{2}{-2} = 1 \;\checkmark \]