What must be the mutual position and properties of vectors a…

What must be the mutual position and properties of vectors $\vec{a}$ and $\vec{b}$ so that: $1^\circ\quad \frac{\vec{a}}{|\vec{a}|}=\frac{\vec{b}}{|\vec{b}|} \qquad\quad 2^\circ\quad \vec{a}=-2\lambda\vec{b} \qquad\quad 3^\circ\quad \vec{a}+\vec{b}=\lambda(\vec{a}-\vec{b})$.

Step 1: Case $1^\circ$: $\dfrac{\vec{a}}{|\vec{a}|}=\dfrac{\vec{b}}{|\vec{b}|}$

The expression $\dfrac{\vec{a}}{|\vec{a}|}$ is the unit vector in the direction of $\vec{a}$, often written $\hat{a}$; similarly $\dfrac{\vec{b}}{|\vec{b}|}=\hat b$. Two unit vectors are equal exactly when they have

• the same direction (codirectional), and
• the same magnitude (both equal to $1$, which is automatic).

\[ \hat{a} = \hat{b} \quad\Longrightarrow\quad \vec{a} \uparrow\uparrow \vec{b} \]

So $\vec{a}$ and $\vec{b}$ are codirectional, while their magnitudes $|\vec{a}|$ and $|\vec{b}|$ can be arbitrary positive numbers.

\[\vec{a} \uparrow\uparrow \vec{b}, \qquad |\vec{a}|,\ |\vec{b}| > 0 \ \text{arbitrary}\]

Note: the zero vector is excluded, since $|\vec{a}|$ and $|\vec{b}|$ appear in denominators.

Step 2: Case $2^\circ$: $\vec{a} = -2\lambda\vec{b}$

$\vec{a}$ is a scalar multiple of $\vec{b}$, so the vectors are collinear (parallel). The scalar factor is $-2\lambda$:

• $\lambda > 0$: $-2\lambda \lt 0$, so $\vec{a}\uparrow\downarrow\vec b$ (antiparallel) and $|\vec{a}| = 2\lambda|\vec{b}|$.
• $\lambda \lt 0$: $-2\lambda > 0$, so $\vec{a}\uparrow\uparrow\vec b$ (codirectional) and $|\vec{a}| = 2|\lambda||\vec{b}|$.
• $\lambda = 0$: $\vec{a} = \vec{0}$.

In every case the magnitude of $\vec a$ is $2|\lambda|$ times that of $\vec b$: \[\vec{a} \parallel \vec{b}, \qquad |\vec{a}| = 2|\lambda|\,|\vec{b}|, \qquad \text{direction given by the sign of } \lambda\]

Step 3: Case $3^\circ$: $\vec{a}+\vec{b}=\lambda(\vec{a}-\vec{b})$

Expand and collect terms: \[\vec{a}+\vec{b} = \lambda\vec{a} - \lambda\vec{b} \quad\Longrightarrow\quad (1-\lambda)\vec{a} = -(1+\lambda)\vec{b} \quad\Longrightarrow\quad (1-\lambda)\vec{a} + (1+\lambda)\vec{b} = \vec{0}.\]

If $\vec a$ and $\vec b$ were non-zero and non-parallel (linearly independent), both coefficients would have to vanish: $1-\lambda=0$ and $1+\lambda=0$, i.e. $\lambda=1$ and $\lambda=-1$ at the same time — a contradiction. Therefore $\vec{a}$ and $\vec{b}$ must be collinear.

Sub-case $\lambda \neq \pm 1$

\[ \vec{a} = -\frac{1+\lambda}{1-\lambda}\,\vec{b}, \qquad \frac{|\vec{a}|}{|\vec{b}|} = \left|\frac{1+\lambda}{1-\lambda}\right|. \]

The direction depends on the sign of $-\dfrac{1+\lambda}{1-\lambda}$:

Special case $\lambda = 1$

\[ \vec{a}+\vec{b} = \vec{a}-\vec{b} \quad\Longrightarrow\quad 2\vec{b} = \vec{0} \quad\Longrightarrow\quad \vec{b} = \vec{0}, \quad \vec a \text{ arbitrary.} \]

Special case $\lambda = -1$

\[ \vec{a}+\vec{b} = -\vec{a}+\vec{b} \quad\Longrightarrow\quad 2\vec{a} = \vec{0} \quad\Longrightarrow\quad \vec{a} = \vec{0}, \quad \vec b \text{ arbitrary.} \]

\[\vec{a} \parallel \vec{b}:\quad \vec{a} = -\frac{1+\lambda}{1-\lambda}\,\vec{b}\ \ (\lambda \neq \pm1); \qquad \vec{b}=\vec{0}\ (\lambda=1); \qquad \vec{a}=\vec{0}\ (\lambda=-1)\]