Can vectors a=i-3j , b=αi+j and c=i+βj all be collinear (α…

Can vectors $\vec{a}=\vec{i}-3\vec{j}$, $\vec{b}=\alpha\vec{i}+\vec{j}$ and $\vec{c}=\vec{i}+\beta\vec{j}$ all be collinear ($\alpha$ and $\beta$ are parameters)? When will $\vec{b}$ and $\vec{c}$ be collinear independently of $\vec{a}$?

Explanation of Method

Collinearity of three vectors with parameters $\alpha$ and $\beta$.

Step 1: Collinearity of all three vectors

Three vectors are collinear when each pair is collinear, i.e. all of them lie on one line through the origin. For $\vec{u}=(u_x, u_y)$ and $\vec{v}=(v_x,v_y)$ the condition is $u_x v_y - u_y v_x = 0$.

Condition $\vec{a} \parallel \vec{b}$

\[ \vec{a} = (1,-3),\ \vec{b} = (\alpha, 1): \qquad (1)(1) - (-3)(\alpha) = 0 \;\Longrightarrow\; 1 + 3\alpha = 0 \;\Longrightarrow\; \alpha = -\frac{1}{3} \]

Condition $\vec{a} \parallel \vec{c}$

\[ \vec{a} = (1,-3),\ \vec{c} = (1,\beta): \qquad (1)(\beta) - (-3)(1) = 0 \;\Longrightarrow\; \beta = -3 \]

Check $\vec{b} \parallel \vec{c}$ with these values

\[ \vec{b} = \left(-\tfrac{1}{3},\,1\right),\ \vec{c} = (1,\,-3): \qquad \left(-\tfrac{1}{3}\right)(-3) - (1)(1) = 1 - 1 = 0 \;\checkmark \]

So all three vectors can be collinear, for

\[\alpha = -\frac{1}{3}, \qquad \beta = -3\]

In scalar-multiple form: \[\vec{b} = \left(-\tfrac{1}{3},\,1\right) = -\frac{1}{3}(1,-3) = -\frac{1}{3}\,\vec{a}, \qquad \vec{c} = (1,-3) = \vec{a}.\]

So $\vec c$ coincides with $\vec a$, and $\vec b$ is antiparallel to both, one third as long.

$\vec a$ $\vec b$ $\vec c$

$\vec a=\vec i-3\vec j$
$\vec b=-\tfrac13\vec i+\vec j=-\tfrac13\vec a$
$\vec c=\vec i-3\vec j=\vec a$

$\alpha=-\tfrac13,\ \beta=-3$: all three are collinear. $\vec c$ equals $\vec a$, so it is drawn as a parallel copy starting at $(1.6,0)$ to keep it visible.

Step 2: Collinearity of $\vec{b}$ and $\vec{c}$ independently of $\vec{a}$

Apply the condition directly to $\vec{b} = (\alpha,\,1)$ and $\vec{c} = (1,\,\beta)$: $\alpha\beta - (1)(1) = 0$.

\[\alpha\beta = 1 \qquad \left(\beta = \frac{1}{\alpha},\ \alpha \neq 0\right)\]

This is the necessary and sufficient condition for $\vec{b} \parallel \vec{c}$. Then

\[ \vec{c} = \left(1, \tfrac{1}{\alpha}\right) = \frac{1}{\alpha}(\alpha, 1) = \frac{1}{\alpha}\,\vec{b}. \]

For $\alpha \gt 0$ the vectors are codirectional; for $\alpha \lt 0$ they are antiparallel.

$\vec b_1$ $\vec c_1$ $\vec b_2$ $\vec c_2$

$\alpha=-1:\ \vec b_1=-\vec i+\vec j$
$\beta=-1:\ \vec c_1=\vec i-\vec j$
$\alpha=-2:\ \vec b_2=-2\vec i+\vec j$
$\beta=-\tfrac12:\ \vec c_2=\vec i-\tfrac12\vec j$

Two examples with $\alpha\beta=1$: each pair $\vec b_k,\vec c_k$ lies on one line (dashed). For $\alpha\gt 0$ the pair is codirectional instead.

Result

Question	Answer
Condition for $\vec{a}\parallel\vec{b}$	$\alpha = -\dfrac{1}{3}$
Condition for $\vec{a}\parallel\vec{c}$	$\beta = -3$
Can all three be collinear?	Yes, for $\alpha = -\dfrac{1}{3}$, $\beta = -3$
$\vec{b}\parallel\vec{c}$ independently of $\vec{a}$	$\alpha\beta = 1$, then $\vec c=\dfrac1\alpha\vec b$