Vectors — The Language

A vector has magnitude and direction; a scalar has magnitude only. In components, a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} with magnitude ∣a⃗∣=a12+a22+a32\vert\vec{a}\vert = \sqrt{a_1^2 + a_2^2 + a_3^2}

The type-checklist (one-markers live here): the zero vector 0⃗\vec{0} has magnitude 00 and no fixed direction; a unit vector has magnitude 11, and the unit vector along a⃗\vec{a} is a^=a⃗∣a⃗∣\hat{a} = \dfrac{\vec{a}}{\vert\vec{a}\vert}; collinear vectors satisfy b⃗=λa⃗\vec{b} = \lambda\vec{a} for some scalar λ\lambda; equal vectors have the same magnitude and direction, wherever they start.

Direction cosines: the unit vector's components are the direction cosines: l=a1∣a⃗∣l = \frac{a_1}{\vert\vec{a}\vert}, m=a2∣a⃗∣m = \frac{a_2}{\vert\vec{a}\vert}, n=a3∣a⃗∣n = \frac{a_3}{\vert\vec{a}\vert}, and always l2+m2+n2=1l^2 + m^2 + n^2 = 1. The components a1,a2,a3a_1, a_2, a_3 themselves are direction ratios.

Position vectors and the joining vector: if AA and BB have position vectors a⃗\vec{a} and b⃗\vec{b}, then AB→=b⃗−a⃗(head minus tail)\overrightarrow{AB} = \vec{b} - \vec{a} \quad \text{(head minus tail)}

Section formula: the point dividing ABAB in ratio m:nm : n internally has position vector r⃗=mb⃗+na⃗m+n\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} (externally: mb⃗−na⃗m−n\frac{m\vec{b} - n\vec{a}}{m - n}). The midpoint is a⃗+b⃗2\frac{\vec{a} + \vec{b}}{2}. Memory hook: the ratio parts attach to the far ends — mm multiplies b⃗\vec{b}.

The Dot (Scalar) Product

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=a1b1+a2b2+a3b3\vec{a}\cdot\vec{b} = \vert\vec{a}\vert\vert\vec{b}\vert\cos\theta = a_1b_1 + a_2b_2 + a_3b_3 The output is a scalar. Key consequences:

  1. cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{a}\vert\vert\vec{b}\vert} — the angle formula.
  2. a⃗⊥b⃗  ⟺  a⃗⋅b⃗=0\vec{a} \perp \vec{b} \iff \vec{a}\cdot\vec{b} = 0 (for nonzero vectors).
  3. a⃗⋅a⃗=∣a⃗∣2\vec{a}\cdot\vec{a} = \vert\vec{a}\vert^2 — the workhorse for expanding ∣a⃗+b⃗∣2=∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2\vert\vec{a} + \vec{b}\vert^2 = \vert\vec{a}\vert^2 + 2\vec{a}\cdot\vec{b} + \vert\vec{b}\vert^2.
  4. Projection of a⃗\vec{a} on b⃗\vec{b} =a⃗⋅b⃗∣b⃗∣= \dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert} — divide by the magnitude of the vector being projected onto.
  5. Sign of a⃗⋅b⃗\vec{a}\cdot\vec{b} = sign of cos⁡θ\cos\theta: positive for acute, zero for right, negative for obtuse angles.

The Cross (Vector) Product

a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣,∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}, \qquad \vert\vec{a}\times\vec{b}\vert = \vert\vec{a}\vert\vert\vec{b}\vert\sin\theta The output is a vector, perpendicular to both a⃗\vec{a} and b⃗\vec{b} (right-hand rule). Key facts:

  1. a⃗×b⃗=−b⃗×a⃗\vec{a}\times\vec{b} = -\vec{b}\times\vec{a} — anticommutative, unlike the dot product.
  2. a⃗×a⃗=0⃗\vec{a}\times\vec{a} = \vec{0}, and parallel vectors have zero cross product.
  3. Cycle of units: i^×j^=k^\hat{i}\times\hat{j} = \hat{k}, j^×k^=i^\hat{j}\times\hat{k} = \hat{i}, k^×i^=j^\hat{k}\times\hat{i} = \hat{j} (reverse order flips the sign).
  4. Areas: parallelogram with adjacent sides a⃗,b⃗\vec{a}, \vec{b} has area ∣a⃗×b⃗∣\vert\vec{a}\times\vec{b}\vert; the triangle has area 12∣a⃗×b⃗∣\frac{1}{2}\vert\vec{a}\times\vec{b}\vert.
  5. Lagrange's identity: ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2\vert\vec{a}\times\vec{b}\vert^2 + (\vec{a}\cdot\vec{b})^2 = \vert\vec{a}\vert^2\vert\vec{b}\vert^2 — links the two products when the angle is unknown.

Dot answers angle and projection questions; cross answers area and perpendicular-direction questions. Choosing the wrong product is the chapter's most common wrong turn.

Worked Examples — Basics, Sections and the Dot Product

Example 1 — Unit vector and rescaling

Find the unit vector along a⃗=2i^+2j^−k^\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}, and a vector of magnitude 66 in the same direction.

Step 1 — magnitude: ∣a⃗∣=4+4+1=3\vert\vec{a}\vert = \sqrt{4 + 4 + 1} = 3.

Step 2 — divide, then rescale: a^=13(2i^+2j^−k^)\hat{a} = \dfrac{1}{3}\left(2\hat{i} + 2\hat{j} - \hat{k}\right).

Answer: a^=23i^+23j^−13k^\hat{a} = \dfrac{2}{3}\hat{i} + \dfrac{2}{3}\hat{j} - \dfrac{1}{3}\hat{k}, and 6a^=4i^+4j^−2k^6\hat{a} = 4\hat{i} + 4\hat{j} - 2\hat{k}.

Example 2 — Direction cosines

Find the direction cosines of a⃗=i^+2j^+2k^\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k} and verify their key property.

Step 1 — magnitude: ∣a⃗∣=1+4+4=3\vert\vec{a}\vert = \sqrt{1 + 4 + 4} = 3.

Step 2 — divide each component: l=13l = \dfrac{1}{3}, m=23m = \dfrac{2}{3}, n=23n = \dfrac{2}{3}.

Answer: (13,23,23)\left(\dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3}\right), and 19+49+49=1\dfrac{1}{9} + \dfrac{4}{9} + \dfrac{4}{9} = 1 as required.

Example 3 — The joining vector

Find AB→\overrightarrow{AB} and its magnitude for A(1,2,3)A(1, 2, 3) and B(4,0,−1)B(4, 0, -1).

Step 1 — head minus tail: AB→=(4−1)i^+(0−2)j^+(−1−3)k^=3i^−2j^−4k^\overrightarrow{AB} = (4-1)\hat{i} + (0-2)\hat{j} + (-1-3)\hat{k} = 3\hat{i} - 2\hat{j} - 4\hat{k}.

Step 2 — magnitude: 9+4+16\sqrt{9 + 4 + 16}.

Answer: AB→=3i^−2j^−4k^\overrightarrow{AB} = 3\hat{i} - 2\hat{j} - 4\hat{k}, ∣AB→∣=29\vert\overrightarrow{AB}\vert = \sqrt{29}.

Example 4 — Collinearity

Show that a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} and b⃗=−4i^−6j^+2k^\vec{b} = -4\hat{i} - 6\hat{j} + 2\hat{k} are collinear.

Step 1 — hunt for a common ratio: −42=−63=2−1=−2\dfrac{-4}{2} = \dfrac{-6}{3} = \dfrac{2}{-1} = -2.

Answer: b⃗=−2a⃗\vec{b} = -2\vec{a}, so the vectors are collinear (parallel, opposite senses) — one scalar λ\lambda must work for all three components.

Example 5 — Section formula

Find the position vector of the point dividing the join of P(i^+2j^−k^)P(\hat{i} + 2\hat{j} - \hat{k}) and Q(4i^−j^+2k^)Q(4\hat{i} - \hat{j} + 2\hat{k}) in the ratio 2:12 : 1 internally.

Step 1 — apply mq⃗+np⃗m+n\dfrac{m\vec{q} + n\vec{p}}{m + n} with m=2m = 2, n=1n = 1: r⃗=2(4i^−j^+2k^)+1(i^+2j^−k^)3=9i^+0j^+3k^3\vec{r} = \frac{2(4\hat{i} - \hat{j} + 2\hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{3} = \frac{9\hat{i} + 0\hat{j} + 3\hat{k}}{3}

Answer: 3i^+k^3\hat{i} + \hat{k} — the ratio's first number multiplies the second point.

Example 6 — Angle between two vectors

Find the angle between a⃗=i^+j^\vec{a} = \hat{i} + \hat{j} and b⃗=j^+k^\vec{b} = \hat{j} + \hat{k}.

Step 1 — the three ingredients: a⃗⋅b⃗=0+1+0=1\vec{a}\cdot\vec{b} = 0 + 1 + 0 = 1, ∣a⃗∣=∣b⃗∣=2\vert\vec{a}\vert = \vert\vec{b}\vert = \sqrt{2}.

Step 2 — angle formula: cos⁡θ=12⋅2=12\cos\theta = \dfrac{1}{\sqrt{2}\cdot\sqrt{2}} = \dfrac{1}{2}.

Answer: θ=π3\theta = \dfrac{\pi}{3}.

Example 7 — Fixing a parameter by perpendicularity

Find λ\lambda if a⃗=2i^+λj^+k^\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k} and b⃗=i^−2j^+3k^\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k} are perpendicular.

Step 1 — set the dot product to zero: 2−2λ+3=02 - 2\lambda + 3 = 0.

Answer: λ=52\lambda = \dfrac{5}{2} — perpendicularity is always one linear equation via the dot product.

Example 8 — Projection

Find the projection of a⃗=2i^+3j^+2k^\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} on b⃗=i^+2j^+k^\vec{b} = \hat{i} + 2\hat{j} + \hat{k}.

Step 1 — the formula: projection =a⃗⋅b⃗∣b⃗∣=2+6+26= \dfrac{\vec{a}\cdot\vec{b}}{\vert\vec{b}\vert} = \dfrac{2 + 6 + 2}{\sqrt{6}}.

Answer: 106=563\dfrac{10}{\sqrt{6}} = \dfrac{5\sqrt{6}}{3} — divide by ∣b⃗∣\vert\vec{b}\vert, the vector being projected onto, never ∣a⃗∣\vert\vec{a}\vert.

Worked Examples — Dot-Product Identities and the Cross Product

Example 9 — Three vectors summing to zero

If a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0} with ∣a⃗∣=3\vert\vec{a}\vert = 3, ∣b⃗∣=5\vert\vec{b}\vert = 5, ∣c⃗∣=7\vert\vec{c}\vert = 7, find the angle between a⃗\vec{a} and b⃗\vec{b}.

Step 1 — isolate and square: a⃗+b⃗=−c⃗\vec{a} + \vec{b} = -\vec{c}, so ∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2=∣c⃗∣2\vert\vec{a}\vert^2 + 2\vec{a}\cdot\vec{b} + \vert\vec{b}\vert^2 = \vert\vec{c}\vert^2.

Step 2 — solve: 9+25+2a⃗⋅b⃗=499 + 25 + 2\vec{a}\cdot\vec{b} = 49, so a⃗⋅b⃗=152\vec{a}\cdot\vec{b} = \dfrac{15}{2} and cos⁡θ=15/23⋅5=12\cos\theta = \dfrac{15/2}{3\cdot 5} = \dfrac{1}{2}.

Answer: θ=π3\theta = \dfrac{\pi}{3} — squaring a vector equation with a⃗⋅a⃗=∣a⃗∣2\vec{a}\cdot\vec{a} = \vert\vec{a}\vert^2 is the standard unlock.

Example 10 — Cross product and the unit perpendicular

Find a unit vector perpendicular to both a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k} and b⃗=i^+2j^+3k^\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}.

Step 1 — cross product by determinant: a⃗×b⃗=∣i^j^k^111123∣=i^(3−2)−j^(3−1)+k^(2−1)=i^−2j^+k^\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3 - 2) - \hat{j}(3 - 1) + \hat{k}(2 - 1) = \hat{i} - 2\hat{j} + \hat{k}

Step 2 — normalise: ∣a⃗×b⃗∣=1+4+1=6\vert\vec{a}\times\vec{b}\vert = \sqrt{1 + 4 + 1} = \sqrt{6}.

Answer: ±16(i^−2j^+k^)\pm\dfrac{1}{\sqrt{6}}\left(\hat{i} - 2\hat{j} + \hat{k}\right) — both signs are perpendicular; state the ±\pm unless an orientation is given.

Example 11 — Area of a parallelogram

Find the area of the parallelogram with adjacent sides a⃗=3i^+j^+4k^\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k} and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}.

Step 1 — cross product: a⃗×b⃗=∣i^j^k^3141−11∣=5i^+j^−4k^\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 5\hat{i} + \hat{j} - 4\hat{k}

Step 2 — magnitude: 25+1+16\sqrt{25 + 1 + 16}.

Answer: 42\sqrt{42} square units.

Example 12 — Area of a triangle from vertices

Find the area of the triangle with vertices A(1,1,2)A(1, 1, 2), B(2,3,5)B(2, 3, 5) and C(1,5,5)C(1, 5, 5).

Step 1 — two side vectors from one vertex: AB→=i^+2j^+3k^\overrightarrow{AB} = \hat{i} + 2\hat{j} + 3\hat{k}, AC→=4j^+3k^\overrightarrow{AC} = 4\hat{j} + 3\hat{k}.

Step 2 — cross product: AB→×AC→=∣i^j^k^123043∣=−6i^−3j^+4k^,∣⋅∣=36+9+16=61\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = -6\hat{i} - 3\hat{j} + 4\hat{k}, \qquad \vert\cdot\vert = \sqrt{36 + 9 + 16} = \sqrt{61}

Step 3 — halve it:

Answer: 612\dfrac{\sqrt{61}}{2} square units — forgetting the 12\frac{1}{2} turns a triangle into a parallelogram.

Example 13 — Lagrange's identity in action

If ∣a⃗∣=10\vert\vec{a}\vert = 10, ∣b⃗∣=2\vert\vec{b}\vert = 2 and a⃗⋅b⃗=12\vec{a}\cdot\vec{b} = 12, find ∣a⃗×b⃗∣\vert\vec{a}\times\vec{b}\vert.

Step 1 — apply ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2\vert\vec{a}\times\vec{b}\vert^2 + (\vec{a}\cdot\vec{b})^2 = \vert\vec{a}\vert^2\vert\vec{b}\vert^2: ∣a⃗×b⃗∣2=100⋅4−144=256\vert\vec{a}\times\vec{b}\vert^2 = 100\cdot 4 - 144 = 256

Answer: ∣a⃗×b⃗∣=16\vert\vec{a}\times\vec{b}\vert = 16 — no angle needed; the identity carries sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 inside it.