Components, Vector Joining Two Points, and Section Formula

Component Form: Vectors as Triples

Along the three axes sit the standard unit vectors $\hat{i}, \hat{j}, \hat{k}$ — each of magnitude $1$, along $OX, OY, OZ$ respectively. Dropping perpendiculars from a point $P(x, y, z)$ shows that its position vector decomposes as

$$\vec{r} = \overrightarrow{OP} = x\hat{i} + y\hat{j} + z\hat{k}$$

This is the component form. The numbers $x, y, z$ are the scalar components and $x\hat{i}, y\hat{j}, z\hat{k}$ the vector components. Applying Pythagoras twice gives the magnitude:

$$|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$$

Algebra becomes arithmetic

For $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$:

1. Sum: $\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}$
2. Difference: $\vec{a} - \vec{b} = (a_1 - b_1)\hat{i} + (a_2 - b_2)\hat{j} + (a_3 - b_3)\hat{k}$
3. Equality: $\vec{a} = \vec{b} \iff a_1 = b_1,\ a_2 = b_2,\ a_3 = b_3$ — all three components, no exceptions.
4. Scalar multiple: $\lambda\vec{a} = (\lambda a_1)\hat{i} + (\lambda a_2)\hat{j} + (\lambda a_3)\hat{k}$

with the distributive laws $k\vec{a} + m\vec{a} = (k + m)\vec{a}$, $k(m\vec{a}) = (km)\vec{a}$, and $k\left(\vec{a} + \vec{b}\right) = k\vec{a} + k\vec{b}$.

Collinearity, in components

$\vec{a}$ and $\vec{b}$ are collinear if and only if $\vec{b} = \lambda\vec{a}$ for some nonzero scalar $\lambda$ — equivalently

$$\frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda$$

(the components are proportional). The sign of $\lambda$ tells same direction ($\lambda > 0$) or opposite ($\lambda < 0$).

Key Point: the components $a_1, a_2, a_3$ of $\vec{a}$ are its direction ratios. And if $l, m, n$ are direction cosines, then $l\hat{i} + m\hat{j} + n\hat{k}$ is the unit vector in that direction — cosines are the components of the unit vector.

Vector Joining Two Points, and the Section Formula

The joining vector

For points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$, the triangle law in triangle $OP_1P_2$ gives $\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1}$:

$$\overrightarrow{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$$

Terminal minus initial — always. Its magnitude is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$, the distance formula.

Section formula

Let $P$ and $Q$ have position vectors $\vec{a}$ and $\vec{b}$, and let $R$ divide $PQ$ in the ratio $m : n$.

1. Internal division ($R$ between $P$ and $Q$ with $\dfrac{PR}{RQ} = \dfrac{m}{n}$):

$$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$$

2. External division ($R$ outside the segment with $\dfrac{PR}{QR} = \dfrac{m}{n}$):

$$\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$$

3. Midpoint ($m = n$):

$$\vec{r} = \frac{\vec{a} + \vec{b}}{2}$$

Key Point (memory hook): in the internal formula the coefficient $m$ (the first number of the ratio) multiplies $\vec{b}$ (the far point $Q$) — the ratio counts from $P$, so $m$ parts lie on the $P$-side and the weight $m$ lands on the opposite end. Cross-pairing $m\vec{a} + n\vec{b}$ is the single most common section-formula error.

The external formula is the internal one with $n \to -n$ — one formula to remember, one sign to flip.

Solved Examples

Example 1: Forcing equality

Find $x, y, z$ so that the vectors $x\hat{i} + 2\hat{j} + z\hat{k}$ and $2\hat{i} + y\hat{j} + \hat{k}$ are equal.

Solution:

1. Componentwise matching: equality of vectors means equality of all corresponding components.
2. Read off: $x = 2$, $y = 2$, $z = 1$.

Answer: $x = 2, y = 2, z = 1$ — one vector equation is three scalar equations.

Example 2: Same magnitude, unequal vectors

Let $\vec{a} = \hat{i} + 2\hat{j}$ and $\vec{b} = 2\hat{i} + \hat{j}$. Is $|\vec{a}| = |\vec{b}|$? Are the vectors equal?

Solution:

1. Magnitudes: $|\vec{a}| = \sqrt{1 + 4} = \sqrt{5}$ and $|\vec{b}| = \sqrt{4 + 1} = \sqrt{5}$ — equal. ✓
2. Components: $a_1 = 1 \neq 2 = b_1$ — the vectors are not equal.

Answer: $|\vec{a}| = |\vec{b}|$ but $\vec{a} \neq \vec{b}$: equal length says nothing about direction.

Example 3: Unit vector from components

Find the unit vector in the direction of $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$.

Solution:

1. Magnitude: $|\vec{a}| = \sqrt{4 + 9 + 1} = \sqrt{14}$.
2. Divide: $\hat{a} = \dfrac{1}{\sqrt{14}}\left(2\hat{i} + 3\hat{j} + \hat{k}\right) = \dfrac{2}{\sqrt{14}}\hat{i} + \dfrac{3}{\sqrt{14}}\hat{j} + \dfrac{1}{\sqrt{14}}\hat{k}$.

Answer: $\hat{a} = \dfrac{2}{\sqrt{14}}\hat{i} + \dfrac{3}{\sqrt{14}}\hat{j} + \dfrac{1}{\sqrt{14}}\hat{k}$.

Example 4: Prescribed magnitude in a given direction

Find a vector of magnitude $7$ units in the direction of $\hat{i} - 2\hat{j}$.

Solution:

1. Normalise: $|\hat{i} - 2\hat{j}| = \sqrt{5}$, so the unit vector is $\dfrac{1}{\sqrt{5}}\hat{i} - \dfrac{2}{\sqrt{5}}\hat{j}$.
2. Scale by $7$: the required vector is $\dfrac{7}{\sqrt{5}}\hat{i} - \dfrac{14}{\sqrt{5}}\hat{j}$.

Answer: $\dfrac{7}{\sqrt{5}}\hat{i} - \dfrac{14}{\sqrt{5}}\hat{j}$ — normalise first, then stretch.

Example 5: Unit vector of a sum

Find the unit vector in the direction of the sum of $2\hat{i} + 2\hat{j} - 5\hat{k}$ and $2\hat{i} + \hat{j} + 3\hat{k}$.

Solution:

1. Add first: the sum is $4\hat{i} + 3\hat{j} - 2\hat{k}$.
2. Magnitude: $\sqrt{16 + 9 + 4} = \sqrt{29}$.
3. Divide: $\dfrac{4}{\sqrt{29}}\hat{i} + \dfrac{3}{\sqrt{29}}\hat{j} - \dfrac{2}{\sqrt{29}}\hat{k}$.

Answer: $\dfrac{1}{\sqrt{29}}\left(4\hat{i} + 3\hat{j} - 2\hat{k}\right)$. Never normalise the two vectors separately and then add — unit vectors of parts do not sum to the unit vector of the whole.

Example 6: Direction ratios to direction cosines

Write the direction ratios of $\vec{a} = \hat{i} + \hat{j} - 2\hat{k}$ and hence its direction cosines.

Solution:

1. Ratios = components: $a = 1$, $b = 1$, $c = -2$.
2. Normalise: $|\vec{a}| = \sqrt{1 + 1 + 4} = \sqrt{6}$, so $l = \dfrac{1}{\sqrt{6}}$, $m = \dfrac{1}{\sqrt{6}}$, $n = -\dfrac{2}{\sqrt{6}}$.

Answer: direction ratios $(1, 1, -2)$; direction cosines $\left(\dfrac{1}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}, -\dfrac{2}{\sqrt{6}}\right)$.

Example 7: Joining vector and the section formula

(i) Find the vector from $P(2, 3, 0)$ to $Q(-1, -2, -4)$. (ii) Find the point $R$ dividing the join of $A\left(\hat{i} + 2\hat{j} - \hat{k}\right)$ and $B\left(-\hat{i} + \hat{j} + \hat{k}\right)$ in the ratio $2 : 1$ internally and externally.

Solution:

1. (i) Terminal minus initial: $\overrightarrow{PQ} = (-1 - 2)\hat{i} + (-2 - 3)\hat{j} + (-4 - 0)\hat{k} = -3\hat{i} - 5\hat{j} - 4\hat{k}$.
2. (ii) Internal, $m : n = 2 : 1$: $\vec{r} = \dfrac{2\vec{b} + 1\vec{a}}{3} = \dfrac{2\left(-\hat{i} + \hat{j} + \hat{k}\right) + \left(\hat{i} + 2\hat{j} - \hat{k}\right)}{3} = \dfrac{-\hat{i} + 4\hat{j} + \hat{k}}{3}$.
3. External: $\vec{r} = \dfrac{2\vec{b} - \vec{a}}{2 - 1} = -2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k} = -3\hat{i} + 3\hat{k}$.

Answer: (i) $-3\hat{i} - 5\hat{j} - 4\hat{k}$; (ii) internal $\dfrac{1}{3}\left(-\hat{i} + 4\hat{j} + \hat{k}\right)$, external $3\left(\hat{k} - \hat{i}\right)$.

Example 8: A right triangle, detected by vectors

Show that $A(2\hat{i} - \hat{j} + \hat{k})$, $B(\hat{i} - 3\hat{j} - 5\hat{k})$, $C(3\hat{i} - 4\hat{j} - 4\hat{k})$ are the vertices of a right-angled triangle.

Solution:

1. Side vectors: $\overrightarrow{AB} = -\hat{i} - 2\hat{j} - 6\hat{k}$, $\overrightarrow{BC} = 2\hat{i} - \hat{j} + \hat{k}$, $\overrightarrow{CA} = -\hat{i} + 3\hat{j} + 5\hat{k}$.
2. Squared lengths: $\left|\overrightarrow{AB}\right|^2 = 1 + 4 + 36 = 41$, $\left|\overrightarrow{BC}\right|^2 = 6$, $\left|\overrightarrow{CA}\right|^2 = 35$.
3. Pythagoras: $41 = 6 + 35$. ✓

Answer: since $\left|\overrightarrow{AB}\right|^2 = \left|\overrightarrow{BC}\right|^2 + \left|\overrightarrow{CA}\right|^2$, the triangle is right-angled (at $C$, where the two shorter sides meet). Also check the sides close up: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0}$. ✓