Vector component form (2D)
WS1 $$ \vec{v} = \langle a, b\rangle = a\mathbf{i} + b\mathbf{j} = \begin{bmatrix} a \\ b \end{bmatrix} $$
In words. A 2D vector v has two equivalent notations: angle-bracket components, basis-vector form using i and j, or a column matrix. The first component is the x-component (how far east); the second is the y-component (how far north).
Show more
Derivation
Once a vector's tail is anchored at the origin, its tip's (x, y) coordinates ARE its components. The basis vectors i = ⟨1, 0⟩ and j = ⟨0, 1⟩ point east and north with unit length, so any vector a*i + b*j is just 'go a east, then b north' — the same instruction as ⟨a, b⟩.
Example
If v points from the origin to (3, 4), then v = ⟨3, 4⟩ = 3i + 4j. Going 3 units east and 4 units north lands you at the tip.
Picture
Arrow from origin to point (a, b). The right-triangle legs are a (horizontal) and b (vertical); the hypotenuse is the vector itself.
Magnitude of a 2D vector
WS1 $$ |\vec{v}| = \sqrt{a^2 + b^2} $$
In words. The magnitude (length) of vector ⟨a, b⟩ equals the square root of the sum of squared components. It is the Pythagorean distance from the tail to the tip.
Domain / range. Always non-negative; equals 0 iff v is the zero vector.
Show more
Derivation
The components a and b are the legs of a right triangle whose hypotenuse is the vector. By the Pythagorean theorem, a² + b² = |v|². Take the positive square root.
Example
For v = ⟨3, 4⟩: |v| = sqrt(9 + 16) = sqrt(25) = 5 (the classic 3-4-5 Pythagorean triple).
Picture
Right triangle with horizontal leg a, vertical leg b, hypotenuse |v|. Pythagoras applied directly.
Angle from positive x-axis (standard math angle)
WS1 $$ \theta = \arctan\!\left(\dfrac{b}{a}\right)\quad\text{(with quadrant adjustment)} $$
In words. For vector ⟨a, b⟩, the angle theta measured counterclockwise from the positive x-axis has tangent equal to b/a — BUT you must sketch first and add 180° if the vector lies in Q2 or Q3, because arctan only outputs angles in (−90°, 90°).
Domain / range. Defined for nonzero vectors; result in [0°, 360°) or (−180°, 180°] depending on convention.
Show more
Derivation
Cosine and sine of theta give the horizontal and vertical legs of a unit vector in that direction; the ratio sin/cos = tan equals b/a. The calculator's arctan returns only the principal value, so a sketch tells you which quadrant the vector actually occupies, and you correct by +180° if needed.
Example
For t = ⟨−1, 3⟩ (Q2): arctan(3/−1) = arctan(−3) ≈ −71.57°. Vector is in Q2, so add 180°: theta ≈ 108.43°.
Picture
Counterclockwise arc from +x-axis to the vector. Always sketch the quadrant first before trusting the calculator.
Polar to rectangular (2D)
WS1 $$ \vec{v} = \langle r\cos\theta,\, r\sin\theta\rangle $$
In words. A vector with magnitude r at angle theta (measured counterclockwise from the positive x-axis) has rectangular components ⟨r·cos(theta), r·sin(theta)⟩. Magnitude times direction.
Domain / range. r >= 0; theta any real angle (typically [0°, 360°) or [0, 2π)).
Show more
Derivation
The unit vector at angle theta is ⟨cos(theta), sin(theta)⟩ — a point on the unit circle. Scaling this by r stretches it to length r without changing its direction. So v = r·⟨cos(theta), sin(theta)⟩.
Example
For r = 10, theta = 45°: v = ⟨10·cos 45°, 10·sin 45°⟩ = ⟨5√2, 5√2⟩ ≈ ⟨7.07, 7.07⟩.
Picture
Point on a circle of radius r at angle theta from +x. The (x, y) coordinates of that point are the components.
Universal bearing formula (clockwise from North)
WS1 $$ \vec{v} = \langle r\sin B,\, r\cos B\rangle $$
In words. For a vector with magnitude r and bearing B (measured clockwise from due North), the x-component (east) is r·sin(B) and the y-component (north) is r·cos(B). Sine and cosine swap roles relative to the standard math convention.
Domain / range. r >= 0; B any real angle, typically [0°, 360°).
Show more
Derivation
Bearing B = 0° means due North (+y); B = 90° means due East (+x). The bearing B and the math angle theta from +x satisfy theta = 90° − B. Using cos(90° − B) = sin(B) and sin(90° − B) = cos(B): x = r·cos(theta) = r·sin(B), y = r·sin(theta) = r·cos(B).
Example
For r = 10 knots, bearing 45° (NE): v = ⟨10·sin 45°, 10·cos 45°⟩ = ⟨5√2, 5√2⟩ ≈ ⟨7.07, 7.07⟩. Equal east and north components, as NE demands.
Picture
Compass rose with N at top. Bearing measured clockwise. Vector points r units in that direction.
Compass direction N α° E
WS1 $$ \vec{v} = \langle +r\sin\alpha,\, +r\cos\alpha\rangle $$
In words. A vector of magnitude r at compass direction 'N alpha degrees E' means: start pointing North, rotate alpha degrees toward East. The x-component is +r·sin(alpha); the y-component is +r·cos(alpha). Both positive because the vector lies in the NE quadrant (Q1).
Domain / range. r >= 0; alpha in [0°, 90°].
Show more
Derivation
Alpha is measured from the y-axis (north), so swap the trig functions relative to standard. Since rotation is toward east (+x), both components are positive.
Example
v = (55 km/h, N15°E): x = 55·sin 15° ≈ 14.24, y = 55·cos 15° ≈ 53.13. Mostly north with a small lean east.
Picture
Compass rose. Arrow leans alpha degrees from due North toward East. Tip lands in Q1.
Compass direction N α° W
WS1 $$ \vec{v} = \langle -r\sin\alpha,\, +r\cos\alpha\rangle $$
In words. A vector of magnitude r at compass direction 'N alpha degrees W' means: start pointing North, rotate alpha degrees toward West. The x-component is −r·sin(alpha) (negative because west); the y-component is +r·cos(alpha) (positive because north).
Domain / range. r >= 0; alpha in [0°, 90°].
Show more
Derivation
Alpha is measured from north toward west. Same magnitudes as NαE, but x flips sign because we're going west, not east.
Example
v = (25 km/h, N30°W): x = −25·sin 30° = −12.5, y = 25·cos 30° = 25√3/2 ≈ 21.65.
Picture
Compass rose. Arrow leans alpha degrees from due North toward West. Tip lands in Q2.
Compass direction S α° E
WS1 $$ \vec{v} = \langle +r\sin\alpha,\, -r\cos\alpha\rangle $$
In words. A vector of magnitude r at compass direction 'S alpha degrees E' means: start pointing South, rotate alpha degrees toward East. The x-component is +r·sin(alpha) (positive — east); the y-component is −r·cos(alpha) (negative — south).
Domain / range. r >= 0; alpha in [0°, 90°].
Show more
Derivation
Alpha measured from south toward east. The 'south' reference makes y negative; the 'east' rotation makes x positive.
Example
v = (70 m/s, S40°E): x = 70·sin 40° ≈ 45.00, y = −70·cos 40° ≈ −53.62.
Picture
Compass rose. Arrow leans alpha degrees from due South toward East. Tip lands in Q4.
Compass direction S α° W
WS1 $$ \vec{v} = \langle -r\sin\alpha,\, -r\cos\alpha\rangle $$
In words. A vector of magnitude r at compass direction 'S alpha degrees W' means: start pointing South, rotate alpha degrees toward West. Both components are negative — the vector lies in Q3.
Domain / range. r >= 0; alpha in [0°, 90°].
Show more
Derivation
Same logic as the other compass directions, but both reference directions (south and west) give negative components.
Example
v = (80 m/s, SW) is the same as S45°W: x = −80·sin 45° = −40√2 ≈ −56.57, y = −80·cos 45° = −40√2 ≈ −56.57.
Picture
Compass rose. Arrow leans alpha degrees from due South toward West. Tip lands in Q3.
Bearing ↔ math angle conversion
WS1 $$ \theta = 90° - B \quad\Leftrightarrow\quad B = 90° - \theta $$
In words. The math angle theta (counterclockwise from +x) and the bearing B (clockwise from +y) for the same vector are complementary in a flipped sense: theta = 90° − B.
Domain / range. Any angle pair.
Show more
Derivation
Both measure direction, but from different reference axes and with opposite rotational senses. Starting at +x and rotating CCW by 90° lands on +y; equivalently, starting at +y and rotating CW by 90° lands on +x. So the two systems differ by a 90° shift and a sign flip on rotation.
Example
A vector at math angle 60° (60° CCW from east) has bearing B = 90° − 60° = 30° (30° CW from north — pointing northeast).
Picture
Two overlapping protractors: one measures CCW from east (math); the other measures CW from north (bearing). They sum to 90° for the same vector.
Vector addition (component-wise / tip-to-tail)
WS2 $$ \vec{v_1} + \vec{v_2} = \langle a_1 + a_2,\, b_1 + b_2\rangle $$
In words. To add two vectors, add their components separately. Geometrically: place v2's tail at v1's tip; the sum is the arrow from v1's tail to v2's tip (tip-to-tail). Equivalent: build a parallelogram with v1 and v2 as adjacent sides; the diagonal from the shared corner is the sum.
Domain / range. Always defined; commutative and associative.
Show more
Derivation
A vector is a displacement. Walking v1 then v2 is the same total trip as the straight line from start to end. Coordinates are bookkeeping for displacement, so the components add. Parallelogram method is just doing it two ways: v1+v2 vs v2+v1.
Example
⟨−5, 1⟩ + ⟨2, 6⟩ = ⟨−3, 7⟩. Add x-components (−5+2=−3) and y-components (1+6=7) separately.
Picture
Two arrows placed tip-to-tail forming an open chain. The resultant is the straight arrow closing the chain.
Vector subtraction
WS2 $$ \vec{v} - \vec{w} = \vec{v} + (-\vec{w}) = \langle a_v - a_w,\, b_v - b_w\rangle $$
In words. To subtract vectors, subtract their components. Equivalent: add v to the opposite of w. Geometrically: v − w points from the tip of w to the tip of v.
Domain / range. Always defined.
Show more
Derivation
Subtraction is addition of the opposite. The opposite of w is −w = ⟨−$a_w$, −$b_w$⟩ — same magnitude, reversed direction. Then use component-wise addition.
Example
If v = ⟨1, 2⟩ and w = ⟨−3, 4⟩, then v − w = ⟨1−(−3), 2−4⟩ = ⟨4, −2⟩, and w − v = −(v − w) = ⟨−4, 2⟩.
Picture
Place v and w with tails together. The vector from w's tip to v's tip is v − w.
Opposite (negation) of a vector
WS2 $$ -\vec{v} = \langle -a, -b\rangle $$
In words. The opposite of v has the same magnitude as v but points in the reverse direction. Negate every component.
Domain / range. Always defined.
Show more
Derivation
Scalar multiplication by −1 reverses direction while preserving magnitude: |−v| = |−1|·|v| = |v|, and the direction is flipped (rotated 180°).
Example
If v = ⟨3, −2⟩, then −v = ⟨−3, 2⟩.
Picture
An arrow and its 180°-rotated twin, head-to-head.
Magnitude of a vector difference
WS2 $$ |\vec{v} - \vec{w}| = \sqrt{(a_v - a_w)^2 + (b_v - b_w)^2} $$
In words. The magnitude of v − w is the distance between the tips of v and w (when both share a tail). You must subtract the vectors FIRST, then take the magnitude; |v − w| is generally NOT equal to |v| − |w|.
Domain / range. Always non-negative.
Show more
Derivation
v − w is itself a vector; its magnitude is computed by Pythagoras on its components. The triangle inequality guarantees |v − w| ≠ ||v| − |w|| unless v and w are parallel.
Example
$v_2$ = ⟨4, 1⟩, $v_4$ = ⟨−4, −2⟩. $v_2$ − $v_4$ = ⟨8, 3⟩, so |$v_2$ − $v_4$| = sqrt(64 + 9) = sqrt(73) ≈ 8.544. (Compare: |$v_2$| − |$v_4$| = sqrt(17) − sqrt(20) ≈ −0.35 — totally different!)
Picture
Triangle with sides v, w, and v − w. The third side closes the triangle from the tip of w to the tip of v.
Ground velocity equation (wind/current)
WS2 $$ \vec{v}_{\text{ground}} = \vec{v}_{\text{heading}} + \vec{v}_{\text{wind/current}} $$
In words. The actual velocity of a vehicle relative to the ground equals its velocity relative to the air/water (heading) PLUS the air/water's velocity relative to the ground (wind or current). Solve for $v_h$eading when you know the desired ground velocity: $v_h$eading = $v_g$round − $v_w$ind.
Domain / range. All velocity vectors in the same frame.
Show more
Derivation
Relative velocities add. The vehicle moves at $v_h$eading through the medium; the medium moves at $v_w$ind through the ground; combine to get ground motion. This is vector addition of two displacements that happen simultaneously.
Example
Plane heads South at 400 mph; wind blows West at 20 mph. $v_g$round = ⟨0, −400⟩ + ⟨−20, 0⟩ = ⟨−20, −400⟩. Speed ≈ 400.5 mph at S2.86°W.
Picture
Two arrows tip-to-tail: the heading vector and the wind vector. The resultant is the actual ground path.
Scalar multiplication
WS3 $$ c\vec{v} = \langle ca,\, cb\rangle $$
In words. Multiplying a vector by a scalar c scales every component by c. The magnitude scales by |c|; if c > 0 the direction is preserved; if c < 0 the direction is reversed; if c = 0 you get the zero vector.
Domain / range. c any real scalar; v any vector.
Show more
Derivation
Component-wise multiplication preserves the direction (when c > 0) because each component scales by the same factor. The magnitude scales because |cv| = sqrt((ca)² + (cb)²) = |c|·sqrt(a² + b²) = |c|·|v|.
Example
If v = ⟨3, −2⟩, then 3v = ⟨9, −6⟩, (1/2)v = ⟨3/2, −1⟩, −v = ⟨−3, 2⟩.
Picture
An arrow stretched (|c| > 1), shrunk (0 < |c| < 1), or flipped (c < 0). Always along the same line through the origin.
Linear combination
WS3 $$ \vec{v} = a\vec{x} + b\vec{y} $$
In words. A linear combination of vectors x and y with scalar coefficients a and b. Any 2D vector v can be written uniquely as a linear combination of two non-parallel vectors x and y — they form a basis for the plane. Solve by equating components to get a system of two equations in two unknowns.
Domain / range. Requires x and y to be linearly independent (not parallel) for a unique solution.
Show more
Derivation
Set ax + by = v and equate component-by-component: a·$x_1$ + b·$y_1$ = $v_1$ and a·$x_2$ + b·$y_2$ = $v_2$. A 2x2 linear system. Unique solution iff the determinant $x_1$·$y_2$ − $x_2$·$y_1$ is nonzero (vectors not parallel).
Example
Express v = ⟨5, −3⟩ as a·x + b·y with x = ⟨−4, 1⟩, y = ⟨3, −1⟩. System: −4a + 3b = 5, a − b = −3. Solve: a = 4, b = 7. Check: 4⟨−4,1⟩ + 7⟨3,−1⟩ = ⟨−16+21, 4−7⟩ = ⟨5, −3⟩. ✓
Picture
A grid of parallelograms with sides x and y covering the plane; v lands at the (a, b) cell of that grid.
Midpoint Connector Theorem (via vectors)
WS3 $$ M,\,N \text{ midpoints of } AB,\,AC \;\Longrightarrow\; \vec{MN} = \tfrac{1}{2}\vec{BC} $$
In words. In a triangle, the segment joining the midpoints of two sides is parallel to the third side and exactly half its length. Vector proof reduces to a one-line calculation: MN = (1/2)·BC.
Domain / range. Any triangle.
Show more
Derivation
Let u = AQ and v = QB so that AB = 2u (since A is the midpoint) and QR = 2v (since R is the midpoint). Then AB = u + v and PR = AB + BR... in the standard setup: MN = MA + AC + CN... pure vector arithmetic shows MN = (1/2)(BC), equivalently the third side scaled by 1/2.
Example
Triangle PQR. A = midpoint PQ, B = midpoint QR. Let u = AQ, v = QB. Then AB = u + v, and PR = 2u + 2v = 2(u+v) = 2·AB. So PR is parallel to AB and twice as long — equivalently AB is parallel to and half of PR.
Picture
Triangle with three midpoints connected to form a smaller similar triangle (1:2 scale). The midpoint segment is parallel to the opposite side.
Parallelogram diagonals bisect each other
WS3 $$ E = \text{midpoint of } \overline{AC} = \text{midpoint of } \overline{BD} $$
In words. In parallelogram ABCD, the diagonals AC and BD cross at a point E that is the midpoint of BOTH diagonals. Equivalently: AE = (1/2)·AC and BE = (1/2)·BD.
Domain / range. Any parallelogram.
Show more
Derivation
In parallelogram ABCD, AC = AB + BC = AB + AD (opposite sides equal as vectors). Setting AE = (1/2)·AC and verifying BE = (1/2)·BD yields the same point.
Example
Parallelogram with AB = 3v and AD = 2u. Then AC = 2u + 3v and BD = 2u − 3v. Midpoint of AC from A: AE = u + (3/2)v. Midpoint of BD from B: BE = u − (3/2)v. Both reach the same point E.
Picture
A parallelogram with both diagonals drawn, meeting at a center point E that bisects each.
Unit vector (definition)
WS4 $$ \hat{v} = \dfrac{\vec{v}}{|\vec{v}|} $$
In words. The unit vector v-hat in the direction of v is v divided by its own magnitude. The result has length 1 and points the same way as v. The 'pure direction' of v with magnitude stripped out.
Domain / range. Defined for any nonzero vector v.
Show more
Derivation
Scalar multiplication by 1/|v| scales the magnitude by 1/|v|·|v| = 1, leaving direction unchanged.
Example
If v = ⟨3, 4⟩ with |v| = 5, then v-hat = (1/5)⟨3, 4⟩ = ⟨3/5, 4/5⟩. Check magnitude: sqrt(9/25 + 16/25) = sqrt(1) = 1. ✓
Picture
Original arrow shrunk down to length 1 along the same line.
Magnitude-direction decomposition
WS4 $$ \vec{v} = |\vec{v}|\,\hat{v} $$
In words. Every vector splits as the product of its magnitude (how long) and its unit vector (which way). This is THE foundational identity for the entire unit — every other formula sits on top of it.
Domain / range. Any nonzero vector.
Show more
Derivation
Direct from the definition v-hat = v/|v|: multiply both sides by |v|.
Example
For v = ⟨3, 4⟩: |v| = 5 and v-hat = ⟨3/5, 4/5⟩. Reconstruct: 5·⟨3/5, 4/5⟩ = ⟨3, 4⟩ = v. ✓
Picture
An arrow labeled |v| above the line and v-hat for the unit-length copy of the arrow.
Unit vector at angle θ from +x axis
WS4 $$ \hat{u}(\theta) = \langle\cos\theta,\, \sin\theta\rangle $$
In words. The unit vector at angle theta (measured CCW from the positive x-axis) has components ⟨cos(theta), sin(theta)⟩. This is the bridge between trigonometry (angles) and vectors (direction): every unit vector in 2D is a point on the unit circle.
Domain / range. theta any real angle.
Show more
Derivation
By definition, cos(theta) and sin(theta) are the x and y coordinates of the unit-circle point at angle theta. That point, viewed as a position vector from the origin, IS the unit vector at angle theta. Magnitude check: sqrt(cos² + sin²) = sqrt(1) = 1. ✓
Example
At theta = 45°: u-hat = ⟨cos 45°, sin 45°⟩ = ⟨√2/2, √2/2⟩ ≈ ⟨0.707, 0.707⟩.
Picture
Unit circle with a radius drawn to the point (cos(theta), sin(theta)) at angle theta from the positive x-axis.
Standard basis vectors (2D)
WS4 $$ \mathbf{i} = \langle 1, 0\rangle,\;\; \mathbf{j} = \langle 0, 1\rangle $$
In words. The two standard unit vectors. i points east (along +x), j points north (along +y). Every 2D vector ⟨a, b⟩ = a·i + b·j.
Show more
Derivation
These are the unit vectors at angles 0° and 90°. By definition i = ⟨cos 0°, sin 0°⟩ = ⟨1, 0⟩ and j = ⟨cos 90°, sin 90°⟩ = ⟨0, 1⟩.
Example
v = ⟨3, 4⟩ = 3i + 4j. The component 3 is 'how many i' and 4 is 'how many j' you stack to build v.
Picture
Two perpendicular unit-length arrows along the +x and +y axes.
Dot product (algebraic / component form)
WS5 $$ \vec{v}\cdot\vec{w} = v_1 w_1 + v_2 w_2 \;\;(+\; v_3 w_3\text{ in 3D}) $$
In words. The dot product of two vectors is the sum of products of corresponding components. It returns a scalar (single number), not a vector. Extends to 3D by adding the third component product.
Domain / range. Defined for two vectors of the same dimension; output is a real scalar.
Show more
Derivation
Defined component-wise. Justified by the equivalent geometric form |v|·|w|·cos(theta), which emerges from the Law of Cosines applied to the triangle with sides v, w, v−w. The cross terms in the expansion of |v−w|² are exactly the dot product.
Example
For v = ⟨−2, 6⟩ and w = ⟨3, 4⟩: v·w = (−2)(3) + (6)(4) = −6 + 24 = 18.
Picture
Two vectors with the formula written below: multiply matching components, sum the products. The result is a single number.
Dot product (geometric form)
WS5 $$ \vec{v}\cdot\vec{w} = |\vec{v}|\,|\vec{w}|\,\cos\theta $$
In words. The dot product also equals the product of the magnitudes times the cosine of the angle between the vectors. Combined with the algebraic form, this lets you recover the angle from purely algebraic data (and vice versa).
Domain / range. theta in [0°, 180°]: the geometric angle between v and w when placed tail-to-tail.
Show more
Derivation
From the Law of Cosines applied to the triangle with sides v, w, v−w: |v−w|² = |v|² + |w|² − 2·|v|·|w|·cos(theta). Expanding |v−w|² in components: ($v_1$−$w_1$)² + ($v_2$−$w_2$)² = $v_1$² + $w_1$² + $v_2$² + $w_2$² − 2($v_1$·$w_1$ + $v_2$·$w_2$). The squared-magnitude terms cancel, leaving $v_1$·$w_1$ + $v_2$·$w_2$ = |v|·|w|·cos(theta).
Example
v = ⟨3, 0⟩, w = ⟨0, 4⟩: v·w = 0 (perpendicular). Algebraic check: 0 = 3·4·cos 90° = 0. ✓ The dot product being zero IS the perpendicularity.
Picture
Two vectors with the angle theta between them; the dot product encodes how much they 'agree' in direction.
Angle between two vectors
WS5 $$ \cos\theta = \dfrac{\vec{v}\cdot\vec{w}}{|\vec{v}|\,|\vec{w}|} $$
In words. The cosine of the angle between two vectors equals their dot product divided by the product of their magnitudes. Take arccos to recover theta. Works in any dimension. Range of theta is [0°, 180°].
Domain / range. Defined for nonzero vectors; theta in [0°, 180°].
Show more
Derivation
Solve the geometric dot product formula v·w = |v|·|w|·cos(theta) for cos(theta). Arccos handles both acute and obtuse angles correctly (unlike arcsin from the cross product, which can't distinguish).
Example
v = ⟨3, 2⟩, w = ⟨4, 1⟩. v·w = 14. |v| = √13, |w| = √17. cos(theta) = 14/√221 ≈ 0.9417. theta = arccos(0.9417) ≈ 19.65°.
Picture
Two vectors meeting at a point with arc showing theta; arccos lookup table or unit circle for evaluating.
Perpendicularity test (dot product)
WS5 $$ \vec{v}\perp\vec{w} \;\Longleftrightarrow\; \vec{v}\cdot\vec{w} = 0 $$
In words. Two nonzero vectors are perpendicular if and only if their dot product is zero. This is the single most useful application of the dot product: a one-line algebraic test for orthogonality.
Domain / range. Both vectors nonzero.
Show more
Derivation
From v·w = |v|·|w|·cos(theta): for nonzero vectors, the dot product is zero iff cos(theta) = 0 iff theta = 90°.
Example
Find a so ⟨2, 6⟩ ⊥ ⟨−2, a⟩: solve (2)(−2) + (6)(a) = 0 → −4 + 6a = 0 → a = 2/3.
Picture
Two arrows meeting at a right angle (small square symbol). The dot product is the algebraic detector for that 90° corner.
Sign of dot product → angle classification
WS5 $$ \text{sign}(\vec{v}\cdot\vec{w}) \;\leftrightarrow\; \begin{cases} + & \text{acute }(\theta<90°)\\ 0 & \text{right }(\theta=90°)\\ - & \text{obtuse }(\theta>90°)\end{cases} $$
In words. You can classify the angle between two vectors without computing it: positive dot product means acute, zero means perpendicular, negative means obtuse. (Caveat: positive can also mean parallel/same direction, theta = 0°; negative can mean anti-parallel, theta = 180°. Always check parallelism before declaring 'acute' or 'obtuse.')
Domain / range. Both vectors nonzero.
Show more
Derivation
Since v·w = |v|·|w|·cos(theta) and magnitudes are positive, the sign of the dot product matches the sign of cos(theta). Cosine is positive in (0°, 90°), zero at 90°, negative in (90°, 180°).
Example
$v_1$ = ⟨2, 5⟩, $v_2$ = ⟨−3, 1⟩: dot product = −6 + 5 = −1 < 0 → obtuse. $v_1$ = −11i − 3j, $v_2$ = −2i − j: dot product = 22 + 3 = 25 > 0 → acute. ⟨4, 9⟩ and ⟨−18, 8⟩: dot product = −72 + 72 = 0 → right angle.
Picture
Three vector pairs with arcs labeled <90°, =90°, >90° and corresponding dot-product signs.
Parallel test for vectors
WS5 $$ \vec{v}\parallel\vec{w} \;\Longleftrightarrow\; \vec{w} = k\vec{v}\text{ for some scalar }k\neq 0 $$
In words. Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. If k > 0, they point the same direction (theta = 0°); if k < 0, anti-parallel (theta = 180°). Equivalent component-wise test in 2D: $v_1$/$w_1$ = $v_2$/$w_2$ (proportional components).
Domain / range. Both vectors nonzero.
Show more
Derivation
Scalar multiplication preserves the line of a vector. Conversely, if two vectors lie on the same line through the origin, one is a stretched/flipped version of the other. In 3D, the cleanest test is v × w = 0.
Example
⟨−4, 6⟩ and ⟨8, −12⟩: ⟨8, −12⟩ = −2·⟨−4, 6⟩, so parallel (anti-parallel, k = −2, theta = 180°). To make ⟨3, 5⟩ parallel to ⟨a+3, 20⟩: need 20/5 = 4 = (a+3)/3, so a = 9.
Picture
Two arrows on the same line through the origin, possibly of different lengths or pointing opposite ways.
Line equation as dot product (normal-vector form)
WS5 $$ \vec{a}\cdot\vec{v} = c \;\;\Leftrightarrow\;\; a_1 x + a_2 y = c $$
In words. The equation a·v = c (where v = ⟨x, y⟩ is variable and a = ⟨$a_1$, $a_2$⟩ is fixed) defines a LINE. The vector a is the NORMAL VECTOR — perpendicular to the line. Changing c shifts the line parallel to itself without changing its normal direction. Every linear equation ax + by = c hides this dot product structure: the coefficients form the normal vector.
Domain / range. a nonzero; c any real.
Show more
Derivation
Pick any two points $P_1$, $P_2$ on the line: a·$P_1$ = c and a·$P_2$ = c. Subtract: a·($P_2$ − $P_1$) = 0. Since $P_2$ − $P_1$ is a vector along the line, the equation says a ⊥ (any vector on the line). So a is the line's normal vector.
Example
For ⟨2, 1⟩·⟨x, y⟩ = 4: this is 2x + y = 4, slope −2, y-intercept 4. The vector ⟨2, 1⟩ is perpendicular to the line. In 3D this generalizes: ax + by + cz = d is a PLANE with normal ⟨a, b, c⟩.
Picture
A family of parallel lines with the same normal vector a; varying c shifts which line you have.
Dot product properties
WS5 $$ \vec{v}\cdot\vec{w} = \vec{w}\cdot\vec{v},\;\; \vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot\vec{v} + \vec{u}\cdot\vec{w},\;\; (c\vec{v})\cdot\vec{w} = c(\vec{v}\cdot\vec{w}) $$
In words. The dot product is commutative (order doesn't matter), distributive over vector addition, and compatible with scalar multiplication (the scalar can move outside).
Domain / range. Any vectors and scalars.
Show more
Derivation
All three follow directly from the component-wise definition and the corresponding properties of real-number addition and multiplication.
Example
(2v)·w = 2(v·w): if v = ⟨1, 2⟩, w = ⟨3, 4⟩, then (2v)·w = ⟨2, 4⟩·⟨3, 4⟩ = 22 = 2·(1·3 + 2·4) = 2·11 = 22. ✓
Picture
Algebraic shorthand on a whiteboard, no figure needed.
Self dot product = magnitude squared
WS5 $$ \vec{v}\cdot\vec{v} = |\vec{v}|^2 $$
In words. The dot product of a vector with itself equals the square of its magnitude. Convenient identity used inside the projection formula and elsewhere.
Domain / range. Any vector v.
Show more
Derivation
Algebraic: v·v = $v_1$² + $v_2$² (+ $v_3$²) = |v|². Geometric: v·v = |v|·|v|·cos(0°) = |v|².
Example
v = ⟨3, 4⟩: v·v = 9 + 16 = 25 = 5² = |v|². So |v| = sqrt(v·v) = 5.
Picture
Same vector dotted with itself — angle 0°, cosine 1, product is the square of length.
Projection (length × direction form)
WS6 $$ \text{proj}_w v = (|\vec{v}|\cos\theta)\,\hat{w} $$
In words. The projection of v onto w equals the scalar (|v|·cos(theta)) — which is the signed length of the shadow — multiplied by the unit vector w-hat — which gives it direction. 'Length-times-direction' decomposition of the projection.
Domain / range. w nonzero; theta = angle between v and w.
Show more
Derivation
Drop a perpendicular from the tip of v to the line of w. In the resulting right triangle, the adjacent side (along w) has length |v|·cos(theta). The direction along w is given by w-hat. Multiply them.
Example
If v has magnitude 5 and makes a 60° angle with w, then |pro$j_w$ v| = 5·cos 60° = 2.5, pointing along w-hat.
Picture
Right triangle with v as hypotenuse, perpendicular dropped from v's tip to the line of w. The 'shadow' lies on w's line.
Projection formula (algebraic / dot-product form)
WS6 $$ \text{proj}_w v = \left(\dfrac{\vec{v}\cdot\vec{w}}{\vec{w}\cdot\vec{w}}\right)\vec{w} $$
In words. The projection of v onto w equals a scalar coefficient (the ratio of two dot products) times the vector w itself. NO cosine, NO arccos, NO explicit magnitudes — just dot products. This is the workhorse formula. Same form in 2D and 3D.
Domain / range. w nonzero.
Show more
Derivation
Start from pro$j_w$ v = (|v|·cos(theta))·w-hat. Substitute cos(theta) = (v·w)/(|v|·|w|) and w-hat = w/|w|. The |v| factors cancel and one |w| combines with the other |w| to form |w|² = w·w in the denominator. Result: ((v·w)/(w·w))·w.
Example
For v = ⟨3, 1⟩, w = ⟨2, 6⟩: v·w = 12, w·w = 40, so pro$j_w$ v = (12/40)·⟨2, 6⟩ = (3/10)·⟨2, 6⟩ = ⟨3/5, 9/5⟩.
Picture
v projected onto the line through w; the foot of the perpendicular from v's tip lies at the tip of the projection vector.
Projection works for all angles (sign of dot product)
WS6 $$ \text{sign}(\text{coefficient of }\vec{w}) = \text{sign}(\vec{v}\cdot\vec{w}) $$
In words. The projection formula handles acute, right, and obtuse angles uniformly. The sign of v·w determines whether pro$j_w$ v points along w (positive, acute angle), is zero (perpendicular), or points opposite to w (negative, obtuse angle). The formula has no triangle in it — purely algebraic, valid in all three cases.
Domain / range. w nonzero; theta in [0°, 180°].
Show more
Derivation
The denominator w·w = |w|² is always positive, so the sign of the coefficient (v·w)/(w·w) matches sign(v·w). For acute theta, v·w > 0 → projection along w. For theta = 90°, v·w = 0 → zero projection. For obtuse theta, v·w < 0 → projection opposite to w (the 'shadow falls on the negative side of the line').
Example
v = 3i + 4j, w = −2i − j (angle ≈ 153° obtuse). v·w = −10 < 0. pro$j_w$ v = (−10/5)·⟨−2, −1⟩ = −2·⟨−2, −1⟩ = ⟨4, 2⟩ — points along +x and +y, OPPOSITE to w = ⟨−2, −1⟩, as expected for the obtuse case.
Picture
Three side-by-side cases: acute (projection along w), perpendicular (zero projection), obtuse (projection points opposite w). All from the same formula.
When projection is zero
WS6 $$ \text{proj}_w v = \vec{0} \;\Longleftrightarrow\; \vec{v}\perp\vec{w} $$
In words. The projection of v onto w is the zero vector if and only if v and w are perpendicular. The shadow of v has zero length because v is hitting w's line edge-on.
Domain / range. w nonzero.
Show more
Derivation
The projection coefficient (v·w)/(w·w) is zero iff v·w = 0, which is the perpendicularity condition.
Example
v = ⟨1, 0⟩, w = ⟨0, 1⟩ (perpendicular). v·w = 0 → pro$j_w$ v = 0·⟨0, 1⟩ = ⟨0, 0⟩. ✓
Picture
v sticking out perpendicular to w; no shadow falls on w's line.
When projection equals the original vector
WS6 $$ \text{proj}_w v = \vec{v} \;\Longleftrightarrow\; \vec{v}\parallel\vec{w} $$
In words. The projection of v onto w equals v itself if and only if v is parallel to w. The shadow IS the vector — because v already lies along w's line.
Domain / range. w nonzero.
Show more
Derivation
If v = k·w, then pro$j_w$ v = ((k·w)·w/(w·w))·w = (k·(w·w)/(w·w))·w = k·w = v.
Example
v = ⟨6, 3⟩, w = ⟨2, 1⟩ (v = 3w). pro$j_w$ v = ((6·2+3·1)/(4+1))·⟨2, 1⟩ = (15/5)·⟨2, 1⟩ = 3·⟨2, 1⟩ = ⟨6, 3⟩ = v. ✓
Picture
v drawn along the same line as w; the projection is just v.
2D rotation matrix
WS7 $$ R(\theta) = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} $$
In words. The 2x2 matrix that rotates any 2D vector counterclockwise by theta. Apply via matrix-vector multiplication: R(theta)·v gives v rotated by theta. The columns of R are the rotated basis vectors: first column = where i lands, second column = where j lands.
Domain / range. theta any real angle.
Show more
Derivation
Any 2D linear transformation is determined by where it sends i and j. Rotation sends i = ⟨1, 0⟩ to ⟨cos(theta), sin(theta)⟩ (the unit vector at angle theta — from WS4) and sends j = ⟨0, 1⟩ to ⟨−sin(theta), cos(theta)⟩ (unit vector at angle theta+90°). Place these images as columns of the matrix.
Example
R(90°) = [[0, −1], [1, 0]]. Apply to ⟨1, 0⟩: [[0,−1],[1,0]]·⟨1,0⟩ = ⟨0, 1⟩ — i has rotated to j, exactly 90° CCW.
Picture
Original vector and its rotated image, with a curved arrow showing the angle theta of rotation.
Rotation matrix properties
WS7 $$ \det R(\theta)=1,\;\;|R(\theta)\vec{v}|=|\vec{v}|,\;\;R(\theta)^{-1}=R(-\theta)=R(\theta)^T $$
In words. Rotation matrices preserve magnitudes (isometries), preserve angles, have determinant 1 (area-preserving and orientation-preserving), and their inverse equals their transpose (which is the same as rotation by negative theta, i.e., clockwise rotation by the same angle).
Domain / range. Any theta.
Show more
Derivation
det R = cos²(theta) + sin²(theta) = 1 by the Pythagorean identity. Magnitude preservation: expand |R·v|² in components and show it collapses to $v_1$² + $v_2$² = |v|². The transpose has cos's on the diagonal and the sines with flipped signs — exactly R(−theta).
Example
R(45°)·R(−45°) = R(0°) = I (identity). The transpose of R(45°) has top-right sin(45°) instead of −sin(45°), giving R(−45°).
Picture
A square rotated and unchanged in size; arrow lengths preserved; orientations preserved.
Rotation composition (matrix product = angle sum)
WS7 $$ R(\alpha)\,R(\beta) = R(\alpha+\beta) $$
In words. Composing two rotations is equivalent to a single rotation by the sum of the angles. The matrix product encodes the trig angle-sum identities: when you multiply out R(alpha)·R(beta), the resulting entries are cos(alpha+beta), −sin(alpha+beta), sin(alpha+beta), cos(alpha+beta).
Domain / range. Any angles alpha, beta. 2D rotations commute; 3D rotations in general do not.
Show more
Derivation
Direct matrix multiplication. The (1,1) entry of R(alpha)·R(beta) is cos(alpha)·cos(beta) − sin(alpha)·sin(beta) = cos(alpha+beta) by the cosine sum identity. Similarly for the other entries.
Example
R(30°)·R(60°) = R(90°). The 90° rotation matrix [[0, −1], [1, 0]] should equal the product of the 30° and 60° matrices — verifies the angle sum identity cos(90°) = cos(30°)cos(60°) − sin(30°)sin(60°) = (√3/2)(1/2) − (1/2)(√3/2) = 0.
Picture
Two curved arrows added head-to-tail, total rotation equals the sum.
Inclined-plane tangential component (sliding force)
WS8 $$ |\vec{w}_T| = |\vec{w}|\sin\theta $$
In words. On a ramp inclined at angle theta above horizontal, the component of weight parallel to the ramp surface (the part that tries to slide the object down) equals weight times sin(theta).
Domain / range. theta in [0°, 90°].
Show more
Derivation
Decompose weight vector into components along and perpendicular to the ramp. By alternate-interior-angles, the ramp angle theta also appears in the force triangle between w (hypotenuse) and $w_N$ (adjacent). Opposite-to-theta is $w_T$, hence |$w_T$| = |w|·sin(theta).
Example
433 N object on 17° ramp: |$w_T$| = 433·sin 17° ≈ 126.59 N. Sanity check at theta = 0° (flat): $w_T$ = 0, no sliding force. ✓ At theta = 90° (vertical wall): $w_T$ = w, all weight tries to slide. ✓
Picture
Block on a ramp with the weight arrow decomposed into 'along ramp' ($w_T$, sliding) and 'into ramp' ($w_N$, normal) components.
Inclined-plane normal component (pressing force)
WS8 $$ |\vec{w}_N| = |\vec{w}|\cos\theta $$
In words. On a ramp at angle theta, the component of weight perpendicular to the ramp (pressing the object into the surface) equals weight times cos(theta).
Domain / range. theta in [0°, 90°].
Show more
Derivation
In the force triangle, $w_N$ is adjacent to theta with hypotenuse |w|, so |$w_N$| = |w|·cos(theta). Pythagorean check: $w_T$² + $w_N$² = |w|²·(sin²+cos²) = |w|². ✓
Example
415 N block on 23° ramp: |$w_N$| = 415·cos 23° ≈ 382 N (force into the plank).
Picture
Same ramp diagram as tangential; the normal component points into the ramp surface (perpendicular to it).
Inclined-plane ratio (find angle from forces)
WS8 $$ \tan\theta = \dfrac{|\vec{w}_T|}{|\vec{w}_N|} $$
In words. The ratio of tangential to normal force component equals the tangent of the ramp angle. Lets you find theta without knowing the actual weight — the weight cancels.
Domain / range. theta in [0°, 90°); $w_N$ > 0.
Show more
Derivation
Divide |$w_T$| = |w|·sin(theta) by |$w_N$| = |w|·cos(theta). The |w| cancels and sin/cos = tan.
Example
Tangential 117.6 N, normal 49 N: tan(theta) = 117.6/49 = 2.4. theta = arctan(2.4) ≈ 67.38°.
Picture
Right triangle with legs labeled |$w_T$| (opposite) and |$w_N$| (adjacent); theta at the corner; tan = opp/adj.
Equilibrium of forces
WS8 $$ \sum_i \vec{F}_i = \vec{0} $$
In words. An object is in equilibrium (not accelerating) when the vector sum of all forces on it equals the zero vector. In 2D this single vector equation becomes TWO scalar equations: sum of x-components = 0, sum of y-components = 0. With two unknowns, the system is solvable.
Domain / range. Any system of force vectors.
Show more
Derivation
Newton's first law: an object at rest stays at rest if net force is zero. The vector equation breaks into one scalar equation per dimension.
Example
20 kg mass hanging from cords at 30° and 45° above horizontal. Weight = 196 N. x-equation: |$T_1$|·cos 150° + |$T_2$|·cos 45° = 0. y-equation: |$T_1$|·sin 150° + |$T_2$|·sin 45° = 196. Solve: |$T_2$| ≈ 175.73 N, |$T_1$| ≈ 143.5 N.
Picture
An object with multiple force arrows (tensions, weight) that form a closed polygon when laid tip-to-tail.
Weight vector
WS8 $$ \vec{w} = -mg\,\mathbf{j},\quad |\vec{w}| = mg $$
In words. An object of mass m has weight vector pointing straight down (negative y) with magnitude m·g, where g ≈ 9.8 m/s² is gravitational acceleration. Units: kilograms times m/s² = Newtons.
Domain / range. m > 0; g ≈ 9.8 m/s² near Earth's surface.
Show more
Derivation
Force = mass × acceleration. Gravity gives every object near Earth's surface a downward acceleration of g, so the force is m·g pointing down.
Example
20 kg mass: |w| = 20·9.8 = 196 N; weight vector = ⟨0, −196⟩. 10 kg mass: |w| = 98 N; weight vector = ⟨0, −98⟩.
Picture
Arrow pointing straight down from an object's center, labeled mg in Newtons.
Cord tension components
WS8 $$ \vec{T} = \langle |\vec{T}|\cos\alpha,\, |\vec{T}|\sin\alpha\rangle $$
In words. A cord tension with magnitude |T|, oriented at angle alpha measured CCW from the positive x-axis, has components |T|·cos(alpha) and |T|·sin(alpha). For a cord pulling up-left at angle beta above horizontal, alpha = 180° − beta. For up-right at angle beta, alpha = beta.
Domain / range. |T| >= 0; alpha any angle (typically chosen so the cord pulls 'up' away from the object).
Show more
Derivation
A cord pulls along its length. The pulling vector has magnitude |T| and direction given by the unit vector ⟨cos(alpha), sin(alpha)⟩ at angle alpha from +x. Multiply magnitude by direction.
Example
Cord pulling up-right at 45° above horizontal with tension 100 N: T = ⟨100·cos 45°, 100·sin 45°⟩ = ⟨50√2, 50√2⟩ ≈ ⟨70.7, 70.7⟩.
Picture
Cord drawn at an angle; the tension vector along the cord has horizontal and vertical components.
3D vector component form
WS9 $$ \vec{v} = \langle v_1, v_2, v_3\rangle = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} $$
In words. A 3D vector has three components corresponding to the three coordinate axes. The new basis vector k = ⟨0, 0, 1⟩ points in the +z direction (up). Same notation conventions as 2D, just one more component.
Show more
Derivation
Natural extension: add a third axis perpendicular to x and y. k is the third unit basis vector.
Example
v = ⟨5, 3, 2⟩ = 5i + 3j + 2k. Tip at point (5, 3, 2). Draw as a 3D arrow with a dashed-line box showing the (5, 3, 2) corner.
Picture
Three perpendicular axes labeled x, y, z. A dashed rectangular box from origin to the point ($v_1$, $v_2$, $v_3$); the vector is the diagonal.
Standard basis vectors (3D)
WS9 $$ \mathbf{i}=\langle 1,0,0\rangle,\;\mathbf{j}=\langle 0,1,0\rangle,\;\mathbf{k}=\langle 0,0,1\rangle $$
In words. The three standard 3D unit vectors: i along +x (east), j along +y (north), k along +z (up). Mutually perpendicular, each of length 1.
Show more
Derivation
Definitions. Mutually perpendicular: i·j = j·k = i·k = 0. Each unit length.
Example
Cross-product identities: i × j = k, j × k = i, k × i = j (cyclic). Useful for sanity-checking cross products.
Picture
Three short arrows from origin, one along each positive axis, forming a right-handed coordinate system.
Magnitude of a 3D vector
WS9 $$ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} $$
In words. The magnitude of a 3D vector is the square root of the sum of the three squared components. Pythagoras applied twice (stacked): once for the base diagonal in the xy-plane, again for the full 3D diagonal.
Domain / range. Always non-negative; equals 0 only for the zero vector.
Show more
Derivation
The 2D projection of v in the xy-plane has length sqrt($v_1$² + $v_2$²). The full vector is the hypotenuse of a right triangle with legs (that 2D length) and $v_3$. Pythagoras again: |v| = sqrt(($v_1$² + $v_2$²) + $v_3$²) = sqrt($v_1$² + $v_2$² + $v_3$²). 'Pythagoras stacks.'
Example
v = ⟨2, 2, 1⟩: |v| = sqrt(4 + 4 + 1) = sqrt(9) = 3. (A '2-2-1' Pythagorean-style triple in 3D.) Or v = ⟨3, 4, 12⟩: |v| = sqrt(9 + 16 + 144) = sqrt(169) = 13.
Picture
A 3D box with the diagonal as the vector; first form the floor diagonal (Pythagoras in xy), then stand up to the top corner (Pythagoras with z).
Dot product (3D)
WS9 $$ \vec{v}\cdot\vec{w} = v_1 w_1 + v_2 w_2 + v_3 w_3 $$
In words. The 3D dot product is the sum of the three products of corresponding components. The geometric identity v·w = |v|·|w|·cos(theta) holds unchanged in 3D — same angle interpretation as 2D.
Domain / range. Both vectors 3D.
Show more
Derivation
Direct extension of the 2D formula. The Law-of-Cosines derivation goes through identically in 3D (just one more squared term in |v−w|² that cancels).
Example
v = ⟨3, 2, 4⟩, w = ⟨−1, 3, −2⟩: v·w = −3 + 6 − 8 = −5 (obtuse angle).
Picture
Three pairs of matching components multiplied and summed — same picture as 2D, just with three terms instead of two.
3D linear combination (change of basis)
WS9 $$ \vec{v} = a\vec{x} + b\vec{y} + c\vec{z} $$
In words. Any 3D vector can be written uniquely as a linear combination of three linearly independent basis vectors x, y, z. Component-wise equality produces a 3x3 linear system in unknowns a, b, c.
Domain / range. Requires x, y, z linearly independent — equivalent to det[x | y | z] ≠ 0.
Show more
Derivation
Equate component-wise: three scalar equations in three unknowns. Solve by substitution or matrix methods. Unique solution exists iff the determinant of the basis matrix is nonzero.
Example
v = ⟨3, −1, 4⟩, x = ⟨1, 2, 1⟩, y = ⟨0, 4, −1⟩, z = ⟨−3, −2, 3⟩. System: a − 3c = 3; 2a + 4b − 2c = −1; a − b + 3c = 4. Solve: a = 75/28, b = −23/14, c = −3/28.
Picture
Three non-coplanar vectors forming an oblique basis; v expressed as a stretch-and-combine of these three.
Direction cosines
WS9 $$ \cos\alpha = \dfrac{v_1}{|\vec{v}|},\;\cos\beta = \dfrac{v_2}{|\vec{v}|},\;\cos\gamma = \dfrac{v_3}{|\vec{v}|};\;\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 $$
In words. The angles a 3D vector makes with each coordinate axis have cosines equal to the corresponding components of the unit vector. The squared direction cosines sum to 1 — this is just the magnitude-of-unit-vector identity in disguise.
Domain / range. v nonzero; alpha, beta, gamma in [0°, 180°].
Show more
Derivation
cos(alpha) is the cosine of the angle between v and the x-axis. From the angle formula: cos(alpha) = (v·i)/(|v|·|i|) = $v_1$/|v|. The sum of squared direction cosines is ($v_1$² + $v_2$² + $v_3$²)/|v|² = |v|²/|v|² = 1.
Example
v = ⟨2, 2, 1⟩, |v| = 3. Direction cosines: 2/3, 2/3, 1/3. Check: 4/9 + 4/9 + 1/9 = 9/9 = 1. ✓
Picture
A 3D vector with three angles drawn to each coordinate axis; their cosines are the unit-vector components.
Cross product (determinant form)
WS10 $$ \vec{v}\times\vec{w} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{vmatrix} $$
In words. The cross product of two 3D vectors is computed as the symbolic determinant of a 3x3 matrix with the basis vectors i, j, k in the first row and the components of v and w in the second and third rows. Expand along the first row with sign pattern +, −, +.
Domain / range. Defined only in 3D. Output is a 3D vector.
Show more
Derivation
Set up two perpendicularity conditions: v·(v × w) = 0 and w·(v × w) = 0. Eliminate one component of the output and the result has the structure of the 2x2 sub-determinants. The determinant mnemonic packages this cleanly.
Example
v × w = ($v_2$·$w_3$ − $v_3$·$w_2$)·i − ($v_1$·$w_3$ − $v_3$·$w_1$)·j + ($v_1$·$w_2$ − $v_2$·$w_1$)·k. For v = ⟨5, 3, 1⟩, w = ⟨2, 5, 6⟩: i: (3·6 − 1·5) = 13; j: −(5·6 − 1·2) = −28; k: (5·5 − 3·2) = 19. Result: ⟨13, −28, 19⟩.
Picture
A 3x3 matrix with the top row being symbols i, j, k and the bottom two rows being the vector components. Expand with alternating signs.
Cross product (component form)
WS10 $$ \vec{v}\times\vec{w} = \langle v_2 w_3 - v_3 w_2,\, v_3 w_1 - v_1 w_3,\, v_1 w_2 - v_2 w_1\rangle $$
In words. Direct component formula for the cross product. Each output component is a 2x2 determinant of the 'other two columns' from the input vectors. Equivalent to the determinant form, just written out.
Domain / range. v, w in 3D.
Show more
Derivation
Expand the determinant formula along the first row. The i-component comes from the minor of column 1 (uses y- and z-columns). The j-component is the negated minor of column 2 (uses x- and z-columns). The k-component is the minor of column 3 (uses x- and y-columns).
Example
⟨1, 2, 1⟩ × ⟨−2, −1, 3⟩: i: 2·3 − 1·(−1) = 7; j: 1·(−2) − 1·3 = −5; k: 1·(−1) − 2·(−2) = 3. Result: ⟨7, −5, 3⟩.
Picture
Three boxed terms, each a 2x2 sub-determinant of the input matrix, with the middle term negated.
Cross product magnitude = parallelogram area
WS10 $$ |\vec{v}\times\vec{w}| = |\vec{v}|\,|\vec{w}|\sin\theta = \text{area of the parallelogram spanned by }\vec{v},\vec{w} $$
In words. The magnitude of the cross product equals the product of the input magnitudes times the sine of the angle between them — which is exactly the area of the parallelogram with sides v and w. This is the geometric companion to the dot product (which extracts cosine).
Domain / range. theta in [0°, 180°]; the magnitude is always non-negative.
Show more
Derivation
For vectors in the xy-plane, the cross product is ($v_1$·$w_2$ − $v_2$·$w_1$)·k. The bounding-rectangle-minus-triangles calculation in WS10 page 10.4 verifies that |$v_1$·$w_2$ − $v_2$·$w_1$| equals the parallelogram's area. The general 3D case reduces to this by viewing the spanned plane.
Example
v = ⟨3, 0, 0⟩ and w = ⟨0, 4, 0⟩ (perpendicular). v × w = ⟨0, 0, 12⟩. |v × w| = 12, which equals the rectangle area 3 × 4 = 12. ✓
Picture
A parallelogram with sides v and w (in 3D); the area equals |v × w|, and the cross product vector sticks straight out of the parallelogram's plane.
Right-hand rule for cross product direction
WS10 $$ \widehat{\vec{v}\times\vec{w}} = \text{thumb when right-hand fingers curl from }\vec{v}\text{ to }\vec{w} $$
In words. Point your right hand's fingers in the direction of v. Curl them toward w (through the shorter angle). Your thumb points in the direction of v × w. The result is always perpendicular to both v and w; the right-hand rule disambiguates which of the two perpendicular directions you get.
Domain / range. Defined whenever v × w is nonzero.
Show more
Derivation
Convention. Equivalent statements: (v, w, v×w) forms a right-handed coordinate system; det[v | w | v×w] > 0 when nonzero; matches the standard i × j = k orientation.
Example
i × j = k. Point right hand along +x (i), curl toward +y (j); thumb points along +z (k). ✓. Reverse: j × i = −k.
Picture
A right hand with fingers curling from v to w and thumb sticking up, labeled as the direction of v × w.
Cross product anti-commutativity
WS10 $$ \vec{w}\times\vec{v} = -(\vec{v}\times\vec{w}) $$
In words. Swapping the order of the cross product flips the result. The right-hand rule's thumb points the opposite way when you curl from w to v instead of v to w. This is the defining feature that distinguishes cross product from addition, multiplication, and dot product (all of which are commutative).
Domain / range. Any 3D v, w.
Show more
Derivation
Swapping the second and third rows of the determinant changes its sign (basic determinant property). The component formula confirms: v × w has $v_2$·$w_3$ − $v_3$·$w_2$, while w × v has $w_2$·$v_3$ − $w_3$·$v_2$ = −($v_2$·$w_3$ − $v_3$·$w_2$).
Example
If v × w = ⟨13, −28, 19⟩, then w × v = ⟨−13, 28, −19⟩.
Picture
Two cross-product vectors pointing in exactly opposite directions, like a thumb flipped.
Cross product perpendicular to inputs
WS10 $$ (\vec{v}\times\vec{w})\cdot\vec{v} = 0 = (\vec{v}\times\vec{w})\cdot\vec{w} $$
In words. By construction the cross product v × w is perpendicular to BOTH v and w. Dotting the cross product with either input always gives zero. Useful as a self-check after computing a cross product.
Domain / range. Any 3D v, w.
Show more
Derivation
Algebraic verification: (v×w)·v = ($v_2$·$w_3$ − $v_3$·$w_2$)·$v_1$ + ($v_3$·$w_1$ − $v_1$·$w_3$)·$v_2$ + ($v_1$·$w_2$ − $v_2$·$w_1$)·$v_3$. Expand and watch every term cancel in pairs. Same for w.
Example
For p × q = ⟨1, 1, 1⟩ from p = ⟨−1, 0, 1⟩, q = ⟨0, −1, 1⟩: (p×q)·p = −1 + 0 + 1 = 0 ✓ and (p×q)·q = 0 − 1 + 1 = 0 ✓.
Picture
Cross-product arrow sticking straight up from the plane of v and w, forming right angles with both.
Cross product = 0 iff parallel
WS10 $$ \vec{v}\times\vec{w} = \vec{0} \;\Longleftrightarrow\; \vec{v}\parallel\vec{w}\text{ (or either vector is zero)} $$
In words. Two nonzero 3D vectors are parallel if and only if their cross product is the zero vector. The 'parallel detector' in 3D — analog of the dot product's perpendicularity detector. Also: v × v = 0 always.
Domain / range. Both vectors 3D.
Show more
Derivation
|v × w| = |v|·|w|·sin(theta). For nonzero vectors, sin(theta) = 0 iff theta = 0° or 180° iff parallel (or anti-parallel). Geometrically: the parallelogram spanned by parallel vectors has zero area.
Example
v × v = 0 for any v (the parallelogram degenerates). v = ⟨1, 2, 3⟩, w = ⟨2, 4, 6⟩ = 2v: cross product is ⟨2·6 − 3·4, 3·2 − 1·6, 1·4 − 2·2⟩ = ⟨0, 0, 0⟩. ✓
Picture
Two arrows on the same line; the 'parallelogram' is degenerate (just a line segment with zero area).
Triangle area (from two side vectors)
WS10 $$ \text{Area}_{\triangle} = \tfrac{1}{2}|\vec{v}\times\vec{w}| $$
In words. The area of the triangle formed by two vectors v and w (originating at a common vertex) is half the magnitude of their cross product. A triangle is exactly half of the parallelogram spanned by its two sides.
Domain / range. v, w nonzero and not parallel.
Show more
Derivation
The parallelogram with sides v and w has area |v × w|. Cut it along its diagonal — get two congruent triangles. Each has area (1/2)·|v × w|.
Example
Triangle with sides ⟨3, 0, 0⟩ and ⟨5, 4, 0⟩: cross product = ⟨0, 0, 12⟩, magnitude 12, triangle area = 6. Sanity via 2D shoelace on vertices (0,0), (3,0), (5,4): (1/2)·|3·4 − 0·5| = 6. ✓
Picture
A parallelogram split by a diagonal into two congruent triangles; the triangle is half of the parallelogram.
Always use dot product (not cross) for angles
WS10 $$ \theta = \arccos\!\left(\dfrac{\vec{v}\cdot\vec{w}}{|\vec{v}|\,|\vec{w}|}\right)\;\;\text{(handles all }[0°,180°]) $$
In words. To find the angle between two vectors, ALWAYS use the dot product formula with arccos. The cross-product formula sin(theta) = |v×w|/(|v|·|w|) seems analogous, but arcsin only outputs angles in [−90°, 90°] and cannot distinguish acute from obtuse — you can get an answer 180° off the truth.
Domain / range. theta in [0°, 180°] for the angle between vectors.
Show more
Derivation
arccos has range [0°, 180°], which exactly matches the natural range of angles between two vectors — so arccos is one-to-one and gives the correct angle. arcsin has range [−90°, 90°]; if the true angle is obtuse (90°−180°), arcsin returns its supplementary acute angle instead. Workaround: use cross-product sine for the reference angle, then check sign of dot product to determine acute vs. obtuse.
Example
m = ⟨1,1,1⟩, w = ⟨−2,3,−6⟩. Dot: m·w = −5; |m| = √3, |w| = 7. cos(theta) = −5/(7√3) ≈ −0.412; theta ≈ 114.4° (obtuse, correct). Cross: |m × w| = √122 ≈ 11.045; sin(theta) ≈ 0.911; arcsin gives 65.6° (WRONG — that's the supplementary acute angle). 114.4° + 65.6° = 180° confirms they're supplementary. Always trust the dot product.
Picture
Two side-by-side angles summing to 180° — one obtuse (correct), one acute (the wrong arcsin answer). The dot product picks the right one.
Scalar triple product
WS11 $$ \vec{u}\cdot(\vec{v}\times\vec{w}) = \det\begin{bmatrix}u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{bmatrix} $$
In words. The scalar triple product takes three 3D vectors and produces a scalar equal to the determinant of the 3x3 matrix with the vectors as rows. Its absolute value equals the volume of the parallelepiped spanned by the three vectors.
Domain / range. u, v, w in 3D.
Show more
Derivation
Start with |v × w| = area of base parallelogram. The triple product u·(v×w) = |u|·|v×w|·cos(theta), where theta is the angle between u and the cross product. |u|·cos(theta) is the height of the parallelepiped above the base. Area times height = volume.
Example
u = ⟨1, 0, 0⟩, v = ⟨0, 1, 0⟩, w = ⟨0, 0, 1⟩ (standard basis): triple product = 1 (unit cube volume). For arbitrary vectors, expand the 3x3 determinant.
Picture
A parallelepiped (slanted box) with edges u, v, w; the triple product's absolute value is its volume.
Parallelepiped volume
WS11 $$ V_{\text{parallelepiped}} = |\vec{u}\cdot(\vec{v}\times\vec{w})| $$
In words. The volume of the parallelepiped with edges u, v, w is the absolute value of the scalar triple product.
Domain / range. u, v, w in 3D, linearly independent (else volume = 0).
Show more
Derivation
Base area = |v × w|. Height = component of u perpendicular to base = |u|·|cos(angle between u and v×w)|. Volume = base·height = |v × w|·|u|·|cos| = |u·(v × w)|.
Example
Standard unit cube: u, v, w = i, j, k. Triple product = 1. Volume = 1. For u = ⟨2, 0, 0⟩, v = ⟨0, 3, 0⟩, w = ⟨0, 0, 4⟩: triple product = 24. Volume = 24.
Picture
A 3D slanted box (parallelepiped); the triple product's absolute value is the box's volume.
Tetrahedron volume (1/6 triple product)
WS11 $$ V_{\text{tetra}} = \tfrac{1}{6}|\vec{u}\cdot(\vec{v}\times\vec{w})| $$
In words. The volume of a tetrahedron with three edge vectors u, v, w meeting at one vertex equals one-sixth of the absolute value of the scalar triple product. A tetrahedron is 1/6 of the corresponding parallelepiped.
Domain / range. u, v, w in 3D, linearly independent.
Show more
Derivation
The parallelepiped spanned by u, v, w can be sliced into 6 congruent tetrahedra (each sharing the common vertex). Triangular prism = 1/2 of parallelepiped; pyramid (tetrahedron) = 1/3 of that prism. Combine: (1/2)·(1/3) = 1/6.
Example
Tetrahedron with vertices $P_1$=(1,4,−2), $P_2$=(2,0,−1), $P_3$=(3,5,3), $P_4$=(4,−2,2). Edge vectors from $P_1$: ⟨1,−4,1⟩, ⟨2,1,5⟩, ⟨3,−6,4⟩. Triple product = −9. Volume = |−9|/6 = 3/2 cubic units.
Picture
A pyramid (tetrahedron) inside the corresponding parallelepiped; the tetrahedron occupies 1/6 of the box.
Fourth vertex of a parallelogram
WS11 $$ D = A + C - B \quad\text{(if AC is the diagonal)} $$
In words. Given three vertices A, B, C of a parallelogram, the fourth vertex D is found using the fact that the diagonals bisect each other. If A and C are opposite (diagonal AC), then midpoint of AC equals midpoint of BD, so D = A + C − B. Three different pairings of 'diagonal' give three possible fourth vertices.
Domain / range. Any three points in 2D or 3D not all collinear.
Show more
Derivation
Diagonals of a parallelogram bisect each other (vector property). If A and C are opposite vertices, midpoint of AC = (A+C)/2 = midpoint of BD = (B+D)/2, so D = A + C − B.
Example
A = (1, −2, 5), B = (2, 3, 6), C = (4, 1, −2). Three possible fourth vertices: A+C−B = (3, −4, −3); A+B−C = (−1, 0, 13); B+C−A = (5, 6, −1).
Picture
Three given points; the missing fourth vertex completes the parallelogram. Three different choices depending on which pair is treated as the diagonal.
Dot vs. cross duality (geometric companion)
WS10 $$ \vec{v}\cdot\vec{w} = |\vec{v}|\,|\vec{w}|\cos\theta;\quad |\vec{v}\times\vec{w}| = |\vec{v}|\,|\vec{w}|\sin\theta $$
In words. Dot and cross product are the geometric companions of cosine and sine. Dot extracts the cosine (and is scalar, defined in all dimensions, commutative, zero iff perpendicular). Cross extracts the sine and a perpendicular direction (and is a 3D-only vector, anti-commutative, zero iff parallel). Together they characterize the entire geometric relationship between two vectors.
Domain / range. theta in [0°, 180°].
Show more
Derivation
Both formulas come from the same geometric construction (parallelogram-with-angle-theta). Cosine is the 'shadow length' coefficient; sine is the 'perpendicular height' coefficient.
Example
v = ⟨3, 0, 0⟩, w = ⟨0, 4, 0⟩ (perpendicular, theta = 90°). Dot = 0 (cos 90° = 0). Cross = ⟨0, 0, 12⟩, magnitude 12 (sin 90° = 1, |v|·|w| = 12).
Picture
A side-by-side table: dot extracts cosine (scalar), cross extracts sine (vector); zero iff perpendicular (dot) vs parallel (cross).