Linear Combinations, Collinearity, and Coplanarity

A vector $\vec{r}$ is a linear combination of vectors $\vec{a}_1, \vec{a}_2, \dots, \vec{a}_k$ if there exist scalars $c_1, c_2, \dots, c_k$ such that $\vec{r} = c_1\vec{a}_1 + c_2\vec{a}_2 + \dots + c_k\vec{a}_k$. In the standard basis you build a combination componentwise: $m\vec{a} + n\vec{b}$ has $\hat{i}$-component $ma_1 + nb_1$, $\hat{j}$-component $ma_2 + nb_2$, and so on. The scalars are called the coefficients of the combination.

• Collinear vectors: $\vec{a}$ and $\vec{b}$ ($\vec{b} \neq \vec{0}$) are collinear iff $\vec{a} = k\vec{b}$ for some scalar $k$. In components, this means the components are proportional: $\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}$.
• Collinear points: $A, B, C$ (position vectors $\vec{a}, \vec{b}, \vec{c}$) are collinear iff $\vec{AB} \parallel \vec{AC}$, i.e. $\vec{b} - \vec{a} = k(\vec{c} - \vec{a})$ for some scalar $k$.
• \"is collinear with\": the phrase $\vec{u}$ is collinear with $\vec{v}$ means $\vec{u} = t\vec{v}$ — introduce ONE unknown scalar and solve.

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar iff one is a linear combination of the other two: $\vec{a} = m\vec{b} + n\vec{c}$ for some scalars $m, n$ (assuming $\vec{b}, \vec{c}$ non-collinear). Equivalently, their scalar triple product vanishes: $[\vec{a}\ \vec{b}\ \vec{c}] = 0$, i.e. the determinant of their components is zero. \"Lies in the plane of $\vec{b}$ and $\vec{c}$\" is the phrase that signals this: write the unknown vector as $m\vec{b} + n\vec{c}$ and solve for $m, n$.

Once a linear combination $\vec{r} = m\vec{a} + n\vec{b} + p\vec{c}$ is built componentwise, the follow-up is routine:

• Unit vector along $\vec{r}$: $\hat{r} = \vec{r}/|\vec{r}|$.
• Vector of magnitude $M$ parallel to $\vec{r}$: $M\,\hat{r} = \dfrac{M}{|\vec{r}|}\vec{r}$.
• Angle between two combinations $\vec{u}, \vec{v}$: $\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}|\,|\vec{v}|}$.

Points $P, Q, R$ with position vectors $\vec{p}, \vec{q}, \vec{r}$ are collinear iff $\vec{QR} = k\,\vec{PQ}$ (or any two displacements are proportional), where $\vec{PQ} = \vec{q} - \vec{p}$, $\vec{QR} = \vec{r} - \vec{q}$. Reading the multiplier: if $\vec{QR} = k\,\vec{PQ}$ with $k > 0$, the line goes $P \to Q \to R$ so $Q$ lies between $P$ and $R$; a negative $k$ means $R$ doubles back, putting a different point in the middle.

To write $\vec{a} = m\vec{b} + n\vec{c}$: equate each component to get the system $a_1 = mb_1 + nc_1$, $a_2 = mb_2 + nc_2$, $a_3 = mb_3 + nc_3$. Pick the two simplest equations, solve for $m$ and $n$, then verify with the remaining one. If the third equation also holds, the representation is valid (the three vectors are coplanar); if it fails, no such $m, n$ exist.

Vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent if there exist scalars $x, y, z$, not all zero, with $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$. They are independent if $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$ forces $x = y = z = 0$. Test in 3-D: dependent ⟺ coplanar ⟺ the determinant of their components is zero:

• $[\vec{a}\ \vec{b}\ \vec{c}] = 0$ ⇒ dependent / coplanar.
• $[\vec{a}\ \vec{b}\ \vec{c}] \neq 0$ ⇒ independent / non-coplanar (they form a basis).

Turn each \"collinear with\" into a scalar multiple, then use independence to match coefficients:

• $\vec{a} + 2\vec{b} = t\vec{c}$ (one scalar $t$).
• $\vec{b} + 3\vec{c} = \lambda\vec{a}$ (one scalar $\lambda$).

Substitute one into the other and compare coefficients of the linearly-independent $\vec{a}, \vec{b}, \vec{c}$ (each side's coefficient of a given basis vector must match). This pins down $t, \lambda$ — typically $t = -6,\ \lambda = -\tfrac{1}{2}$ — and yields the target, e.g. $\vec{a} + 2\vec{b} = -6\vec{c}$, so $\vec{a} + 2\vec{b} + 6\vec{c} = \vec{0}$.

A vector $\vec{v}$ coplanar with non-collinear $\vec{b}, \vec{c}$ has the form $\vec{v} = m\vec{b} + n\vec{c}$. Common second conditions and how to use them:

• Bisects the angle between $\vec{b}, \vec{c}$: $\vec{v}$ is parallel to $\hat{b} + \hat{c}$ (sum of the UNIT vectors).
• Perpendicular to $\vec{q}$: impose $\vec{v}\cdot\vec{q} = 0$.
• Projection on $\vec{c}$ equals $k$: impose $\dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|} = k$.
• Magnitude $M$: scale the resulting direction to length $M$.

If $\vec{p}, \vec{q}, \vec{r}$ are non-coplanar and $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$ (the given components), then writing $\vec{s} = x(-\vec{p}+\vec{q}+\vec{r}) + y(\vec{p}-\vec{q}+\vec{r}) + z(-\vec{p}-\vec{q}+\vec{r})$ and collecting like terms gives a system: the coefficient of $\vec{p}$ must equal $4$, of $\vec{q}$ equal $3$, of $\vec{r}$ equal $5$. Solve the $3\times 3$ system for $x, y, z$; independence of $\vec{p}, \vec{q}, \vec{r}$ is what makes the coefficient-matching valid.