AP Precalculus: Vectors
A vector-valued function gives a vector for every input. The most important use is planar motion: a particle with parametric equations x(t), y(t) has a position vector p(t) = 〈x(t), y(t)〉 = x(t)i + y(t)j. The signs tell you the location: x(t) > 0 means to the right of the origin, y(t) > 0 means above it. The magnitude of the position vector, |p(t)| = √(x(t)² + y(t)²), is the distance of the particle from the origin at time t. A velocity vector v(t) = 〈vx(t), vy(t)〉 describes how the particle is moving at time t: the sign of vx says whether it is moving right (+) or left (−), and the sign of vy says whether it is moving up (+) or down (−). The speed is the magnitude of the velocity, |v(t)| = √(vx² + vy²), which is a scalar and is never negative. Do not mix up the two magnitudes: |p(t)| measures how far the particle is from the origin, while |v(t)| measures how fast it is moving. Include units, such as meters and seconds, whenever the problem gives them.
1. Vectors: Scalars, Component Form, and Magnitude
A scalar is a quantity described by a single number: 40 miles per hour, 12 liters, 25°C. A vector needs two pieces of information, a magnitude (how much) and a direction (which way). “40 miles per hour” is a scalar speed, but “40 miles per hour due north” is a vector velocity. Force, displacement, and wind are all vectors. A vector is drawn as an arrow: the starting point is the tail, the end with the arrowhead is the head, and the length of the arrow is the magnitude. Only the length and direction matter, so a vector can be slid anywhere without changing it. To compute with vectors, write them in component form v = 〈a, b〉, where a is the horizontal change and b is the vertical change from tail to head. If the tail is (x1, y1) and the head is (x2, y2), then v = 〈x2 − x1, y2 − y1〉: always head minus tail. The magnitude (also called length or norm) is written |v| and comes from the Pythagorean theorem: |v| = √(a² + b²).
Worked Example: Find the component form and magnitude of the vector with tail (1, 1) and head (6, 4).
Head minus tail: 〈6 − 1, 4 − 1〉 = 〈5, 3〉. Then |v| = √(5² + 3²) = √(25 + 9) = √34 ≈ 5.83.
v = 〈5, 3〉, |v| = √34
The distance between two points is the magnitude of the vector joining them, so this also says the distance from (1, 1) to (6, 4) is √34.
1. Which of these is a vector quantity?
Explanation: A vector has both magnitude and direction. “40 mph due north” gives both; the other three are single numbers (scalars).
2. What is the component form of the vector with tail (−2, 5) and head (4, 1)?
Explanation: Head minus tail: 〈4 − (−2), 1 − 5〉 = 〈6, −4〉. Reversing the subtraction gives the opposite vector 〈−6, 4〉.
3. Find |v| for v = 〈−5, 12〉.
Explanation: |v| = √((−5)² + 12²) = √(25 + 144) = √169 = 13. Squaring removes the negative sign.
4. What is the distance between A(2, −1) and B(−4, 7)?
Explanation: AB = 〈−4 − 2, 7 − (−1)〉 = 〈−6, 8〉, and |AB| = √(36 + 64) = √100 = 10.
5. The vector 〈3, k〉 has magnitude 5 and k > 0. Find k.
Explanation: √(9 + k²) = 5 gives 9 + k² = 25, so k² = 16 and k = ±4. Since k > 0, k = 4.
2. Equal, Opposite, and Parallel Vectors
Two vectors are equal when they have the same magnitude and the same direction, no matter where they are drawn. In component form this is simple: 〈a, b〉 = 〈c, d〉 exactly when a = c and b = d. So two arrows in different places are equal as long as head minus tail gives the same components. The negative of a vector reverses its direction and keeps its length: −〈a, b〉 = 〈−a, −b〉. The zero vector 〈0, 0〉 has no length and no direction. Vectors that point in the same or exactly opposite directions are parallel; algebraically, u is parallel to v when u = k·v for some nonzero scalar k, so the components are proportional. Setting two component forms equal is a quick way to solve for unknown numbers, because it produces one equation for each component.
Worked Example: Vector p goes from (0, 0) to (3, −2). Vector q goes from (1, 4) to (4, 2). Are they equal? Write −p. Then find x and y if 〈2x − 1, y + 4〉 = p.
p = 〈3, −2〉. For q: 〈4 − 1, 2 − 4〉 = 〈3, −2〉. The components match, so q = p. The negative is −p = 〈−3, 2〉.
Equal components give 2x − 1 = 3, so x = 2, and y + 4 = −2, so y = −6.
q = p, −p = 〈−3, 2〉, x = 2, y = −6
1. What is the negative of 〈−4, 7〉?
Explanation: Negating a vector changes the sign of both components: −〈−4, 7〉 = 〈4, −7〉.
2. Which vector equals the vector from (0, 0) to (3, 5)?
Explanation: The target is 〈3, 5〉. From (2, 1) to (5, 6) gives 〈3, 5〉. The second is 〈−3, −5〉, the third is 〈5, 3〉, and the last is 〈5, 3〉.
3. If 〈2x + 1, y − 3〉 = 〈7, 5〉, what are x and y?
Explanation: 2x + 1 = 7 gives x = 3, and y − 3 = 5 gives y = 8.
4. What is |−〈6, −8〉|?
Explanation: −〈6, −8〉 = 〈−6, 8〉, whose magnitude is √(36 + 64) = 10. A negative vector has the same length as the original, and a magnitude is never negative.
5. Which vector is parallel to 〈2, −3〉?
Explanation: 〈−6, 9〉 = −3·〈2, −3〉, so it is parallel (pointing the opposite way). The other choices are not scalar multiples.
3. Adding Vectors: Geometric and Algebraic
There are two equivalent ways to add vectors. Geometrically (head to tail): draw the first vector, slide the second so its tail touches the head of the first, and then the sum is the vector from the tail of the first to the head of the second. This is the triangle method. If both vectors start at the same point, complete the parallelogram and the sum is its diagonal from that point, the parallelogram method. Algebraically: add the matching components, 〈a, b〉 + 〈c, d〉 = 〈a + c, b + d〉. Vector addition is commutative (a + b = b + a) and associative. Adding the zero vector changes nothing, and adding a vector to its negative gives the zero vector. A useful shortcut for points: AB + BC = AC, because going from A to B and then B to C is the same as going from A to C. Be careful with magnitudes: |a + b| is usually not |a| + |b|, which is only true when the vectors point the same way.
Worked Example: Let a = 〈3, 1〉 and b = 〈1, 3〉. Find a + b and compare |a + b| with |a| + |b|.
Add components: a + b = 〈3 + 1, 1 + 3〉 = 〈4, 4〉. Then |a + b| = √(16 + 16) = 4√2 ≈ 5.66.
But |a| = |b| = √10 ≈ 3.16, so |a| + |b| ≈ 6.32. The triangle inequality says |a + b| ≤ |a| + |b|, and here it is strictly smaller because the vectors point in different directions.
a + b = 〈4, 4〉
1. Find 〈4, −2〉 + 〈−7, 5〉.
Explanation: Add components: 4 + (−7) = −3 and −2 + 5 = 3.
2. Given A(1, 2), B(4, 6), and C(9, 3), what is AB + BC?
Explanation: AB + BC = AC = 〈9 − 1, 3 − 2〉 = 〈8, 1〉. You can also add AB = 〈3, 4〉 and BC = 〈5, −3〉.
3. If a = 〈6, 0〉 and b = 〈0, −8〉, what is |a + b|?
Explanation: a + b = 〈6, −8〉, so |a + b| = √(36 + 64) = 10. The sum of the magnitudes (14) is not the magnitude of the sum.
4. If a = 〈3, −1〉 and a + b = 〈5, 4〉, find b.
Explanation: b = (a + b) − a = 〈5 − 3, 4 − (−1)〉 = 〈2, 5〉.
5. For a = 〈1, 0〉 and b = 〈0, 1〉, what is |a + b|?
Explanation: a + b = 〈1, 1〉, so |a + b| = √2 ≈ 1.41, which is less than |a| + |b| = 2.
4. Force and Wind Problems: Resultants with the Law of Cosines and Sines
Many real problems give vectors by magnitude and the angle between them instead of components. Two forces pulling on the same object, or a plane’s own velocity combined with a wind, can be replaced by one resultant vector, their sum. Sketch the two vectors from the same point and complete the parallelogram. Cut it along the diagonal and you get a triangle whose two known sides are the original vectors. If the angle between the vectors when their tails touch is θ, then the angle of the triangle opposite the resultant is the supplement 180° − θ, because the head-to-tail copy of the second vector turns the interior angle into its supplement. Magnitude comes from the law of cosines: R² = a² + b² − 2ab·cos(180° − θ), which is the same as R² = a² + b² + 2ab·cos θ. Direction comes from the law of sines: sin φ / (opposite side) = sin(180° − θ) / R, where φ is the angle between the resultant and one of the forces. When the vectors are perpendicular, skip the laws and use the Pythagorean theorem. Always check that the answer is reasonable: the resultant must be shorter than the sum of the two lengths and longer than their difference.
Worked Example: Forces of 60 N and 80 N act on an object with 50° between them. Find the magnitude of the resultant and the angle it makes with the 60 N force.
Magnitude. The angle opposite R is 130°, so R² = 60² + 80² − 2(60)(80)cos 130° = 3600 + 6400 + 6170.9 ≈ 16170.9, giving R ≈ 127.2 N.
Direction. Let φ be the angle opposite the 80 N side, which is the angle between R and the 60 N force. By the law of sines, sin φ / 80 = sin 130° / 127.2, so sin φ ≈ 0.4818 and φ ≈ 28.8°.
R ≈ 127.2 N at 28.8° from the 60 N force
1. Forces of 5 N and 12 N act at right angles. What is the magnitude of the resultant?
Explanation: For perpendicular forces, R = √(5² + 12²) = √169 = 13 N. Adding the magnitudes (17) ignores the angle.
2. Forces of 30 N and 40 N have 60° between them. What is the magnitude of the resultant?
Explanation: R² = 30² + 40² + 2(30)(40)cos 60° = 900 + 1600 + 1200 = 3700, so R ≈ 60.8 N. The 36.1 N choice comes from using the wrong sign on the cosine term.
3. For the same 30 N and 40 N forces (60° apart), what angle does the resultant make with the 40 N force?
Explanation: sin φ / 30 = sin 120° / 60.83 gives sin φ ≈ 0.4271, so φ ≈ 25.3°. The resultant lies closer to the larger force. The angle to the 30 N force is 60° − 25.3° = 34.7°.
4. A plane flies at 300 mph heading due east. A wind of 40 mph blows due north. What is the plane’s speed relative to the ground?
Explanation: The two velocities are perpendicular, so the ground speed is √(300² + 40²) = √91600 ≈ 302.6 mph.
5. Forces of 50 N and 70 N act with 70° between them. What is the magnitude of the resultant?
Explanation: R² = 50² + 70² + 2(50)(70)cos 70° = 7400 + 2394 ≈ 9794, so R ≈ 99.0 N. Using the interior angle 70° in the minus form of the law of cosines would give 70.7 N, which is wrong.
5. Scalar Multiplication
Multiplying a vector by a scalar k scales its length and, if k is negative, flips its direction: k〈a, b〉 = 〈ka, kb〉. The length changes by the factor |k|, so |kv| = |k|·|v|. If k > 1 the vector stretches; if 0 < k < 1 it shrinks; if k < 0 it also points the opposite way; and if k = 0 it becomes the zero vector. A vector and any scalar multiple of it are parallel. This gives a powerful way to build a vector with a prescribed direction and length: first make a vector of length 1 in the wanted direction by dividing by the magnitude, then multiply by the desired length. To get a vector opposite to v with length L, use −L·v/|v|. Scalar multiplication distributes over addition: k(a + b) = ka + kb, and (k + m)a = ka + ma.
Worked Example: For a = 〈2, 1〉, find 2a and −1.5a, and find |−1.5a|. Then find a vector of length 10 that points opposite to 〈3, 4〉.
2a = 〈4, 2〉, −1.5a = 〈−3, −1.5〉, and |−1.5a| = 1.5|a| = 1.5√5 ≈ 3.35.
For the second part, |〈3, 4〉| = 5, so the unit vector in the same direction is 〈3/5, 4/5〉. Opposite with length 10 means multiplying by −10: 〈−6, −8〉. Check: √(36 + 64) = 10.
−10·〈3/5, 4/5〉 = 〈−6, −8〉
1. Find −2〈3, −4〉.
Explanation: Multiply each component by −2: 〈−6, 8〉.
2. What is |−3〈5, 12〉|?
Explanation: |−3〈5, 12〉| = 3·13 = 39. The scalar enters as its absolute value, so the result is never negative.
3. Which vector has length 10 and points in the direction opposite to 〈3, 4〉?
Explanation: The unit vector is 〈3/5, 4/5〉; times −10 gives 〈−6, −8〉, whose length is 10. 〈−3, −4〉 only has length 5.
4. For what value of k is k〈2, −5〉 = 〈−6, 15〉?
Explanation: 2k = −6 gives k = −3, and the second component checks: −5(−3) = 15.
5. Which vector has the same direction as 〈5, 12〉 and length 26?
Explanation: |〈5, 12〉| = 13, so multiply by 26/13 = 2 to get 〈10, 24〉. Its length is √(100 + 576) = 26.
6. Subtracting Vectors
Subtracting a vector means adding its negative: a − b = a + (−b). In components, subtract the matching entries: 〈a, b〉 − 〈c, d〉 = 〈a − c, b − d〉. Geometrically there are two pictures. Triangle form: draw a, then draw −b starting at the head of a; the arrow from the tail of a to the head of −b is a − b. Tails together: if a and b start at the same point, a − b is the arrow that goes from the head of b to the head of a. Memorize that phrase, “to the first, from the second,”, because it is the quickest way to draw a difference and it is exactly why a vector between two points is head minus tail. Subtraction is not commutative: b − a = −(a − b), which has the same length and the opposite direction. Combining operations follows ordinary algebra: 2a − 3b means scale first, then subtract.
Worked Example: Let p = 〈6, 2〉 and q = 〈2, 4〉. Find p − q, q − p, and |p − q|.
p − q = 〈4, −2〉 and q − p = 〈−4, 2〉 (the opposite vector). The length is |p − q| = √(16 + 4) = √20 = 2√5 ≈ 4.47.
p − q = 〈4, −2〉
1. Find 〈7, 2〉 − 〈3, −4〉.
Explanation: Subtract components: 7 − 3 = 4 and 2 − (−4) = 6.
2. If a = 〈5, 1〉 and b = 〈2, 3〉, find 2a − 3b.
Explanation: 2a = 〈10, 2〉 and 3b = 〈6, 9〉. The difference is 〈4, −7〉.
3. If a = 〈9, 5〉 and b = 〈3, −3〉, find |a − b|.
Explanation: a − b = 〈6, 8〉, so |a − b| = √(36 + 64) = 10.
4. When a and b are drawn from the same tail, the vector b − a goes
Explanation: The difference b − a points to b (the first vector named) from a (the second). Reversing the order gives a − b, which is the opposite arrow.
5. If a − b = 〈1, −2〉 and a = 〈4, 3〉, find b.
Explanation: b = a − (a − b) = 〈4 − 1, 3 − (−2)〉 = 〈3, 5〉.
7. The Scalar (Dot) Product
The dot product (or scalar product) of a = 〈a1, a2〉 and b = 〈b1, b2〉 is a · b = a1b1 + a2b2. Multiply matching components and add. The result is a single number, not a vector, which is why it is called the scalar product, and why it is different from scalar multiplication (a number times a vector gives a vector). Four properties are worth knowing: (1) a · b = b · a; (2) a · a = |a|²; (3) a · (b + c) = a · b + a · c; (4) (ka) · b = k(a · b). The second one is a handy way to get a magnitude: |a| = √(a · a). Because the dot product is a sum of products, it can be expanded like ordinary multiplication, so (a + b) · (c + d) = a · c + a · d + b · c + b · d. Do the dot product of vectors only; a dot product with a number such as 3 · a is not defined (that is scalar multiplication).
Worked Example: Let a = 〈3, 1〉, b = 〈−1, 3〉, and c = 〈2, −2〉. Find a · b, a · (b + c), and a · a.
a · b = (3)(−1) + (1)(3) = 0. Next b + c = 〈1, 1〉, so a · (b + c) = 3 + 1 = 4. You can check the distributive property: a · b + a · c = 0 + (6 − 2) = 4. Finally a · a = 9 + 1 = 10 = |a|².
a · b = 0, a · (b + c) = 4, a · a = 10
1. Find 〈3, −2〉 · 〈4, 5〉.
Explanation: (3)(4) + (−2)(5) = 12 − 10 = 2. The result is a number, not a vector.
2. Let a = 〈2, −1〉, b = 〈−3, 4〉, c = 〈1, 2〉. Find a · (b + c).
Explanation: b + c = 〈−2, 6〉, so a · (b + c) = (2)(−2) + (−1)(6) = −4 − 6 = −10.
3. For a = 〈−6, 8〉, what is a · a?
Explanation: a · a = 36 + 64 = 100 = |a|², since |a| = 10.
4. If a = 〈1, 5〉 and b = 〈3, −2〉, find (2a) · b.
Explanation: a · b = 3 − 10 = −7, and (2a) · b = 2(a · b) = −14.
5. The dot product of two vectors is
Explanation: Multiplying matching components and adding produces a single number, which is why the operation is called the scalar product.
8. Geometric Properties of the Dot Product and the Angle Between Vectors
The dot product also has a geometric meaning: a · b = |a||b| cos θ, where θ (0° ≤ θ ≤ 180°) is the angle between the vectors when they are drawn from the same point. It comes from the law of cosines applied to the triangle formed by a, b, and b − a: |b − a|² = |a|² + |b|² − 2|a||b| cos θ. Because |a||b| is positive, the sign of a · b tells you the type of angle: a · b > 0 means θ is acute, a · b < 0 means θ is obtuse, and a · b = 0 means the nonzero vectors are perpendicular (θ = 90°). For parallel vectors, a · b = ±|a||b|, with + when they point the same way and − when opposite. To find the angle, solve for it: cos θ = (a · b) / (|a||b|) and θ = arccos of that number. Compute the dot product and both magnitudes separately, then combine. Keep your calculator in degree mode if you want degrees, and check the sign: a negative dot product must give an angle above 90°.
Worked Example: Find the angle between a = 〈4, 1〉 and b = 〈1, 3〉. Then find x so that 〈x, 6〉 is perpendicular to 〈3, −2〉.
a · b = 4 + 3 = 7, |a| = √17, |b| = √10. So cos θ = 7/(√17 · √10) = 7/√170 ≈ 0.5369, and θ = arccos(0.5369) ≈ 57.5°. The dot product is positive, which matches an acute angle.
For perpendicular vectors set the dot product to 0: 3x + 6(−2) = 0, so 3x = 12 and x = 4.
θ ≈ 57.5°, x = 4
1. Find x so that 〈x, −3〉 is perpendicular to 〈6, 2〉.
Explanation: Perpendicular means the dot product is zero: 6x + (−3)(2) = 0, so 6x = 6 and x = 1.
2. What is the angle between 〈2, 0〉 and 〈0, 5〉?
Explanation: The dot product is (2)(0) + (0)(5) = 0, so the vectors are perpendicular. They point along the x-axis and the y-axis.
3. The angle between 〈3, 4〉 and 〈−4, 1〉 is
Explanation: The dot product is (3)(−4) + (4)(1) = −8, which is negative, so the angle is obtuse (between 90° and 180°).
4. Find the angle between 〈1, 2〉 and 〈3, −1〉 to the nearest tenth of a degree.
Explanation: The dot product is 3 − 2 = 1. Magnitudes are √5 and √10, so cos θ = 1/√50 ≈ 0.1414 and θ ≈ 81.9°. The dot product is positive, so the angle must be acute, not 98.1°.
5. If |a| = 4, |b| = 6, and the angle between them is 60°, what is a · b?
Explanation: a · b = |a||b| cos 60° = 4 · 6 · 0.5 = 12. The value 20.8 would come from using sine instead of cosine.
9. Unit Vectors and the i, j Form
A unit vector is any vector of length exactly 1. To check, compute the magnitude; to make one, divide a nonzero vector by its own length: u = v / |v|. This unit vector points in the same direction as v. If a component is unknown, set the magnitude equal to 1, square both sides, and solve; expect two solutions (a ± sign) unless the problem says which side. The two special unit vectors are i = 〈1, 0〉 and j = 〈0, 1〉, pointing along the positive x-axis and y-axis. Every vector can be built from them: 〈a, b〉 = ai + bj. The expression ai + bj is called unit vector form, and the magnitude formula is the same, |ai + bj| = √(a² + b²). Moving between the two notations is just reading off the coefficients. Unit vectors are the key to scaling: once you have the unit vector for a direction, multiplying by any positive length L gives a vector of length L in that direction.
Worked Example: (a) Find t so that 〈t, −3/5〉 is a unit vector. (b) Write 6i − 8j in component form and find the unit vector in its direction.
(a) Set the magnitude to 1: t² + 9/25 = 1, so t² = 16/25 and t = ±4/5.
(b) The component form is 〈6, −8〉 and |〈6, −8〉| = √(36 + 64) = 10. Dividing by 10 gives u = 〈3/5, −4/5〉 = (3/5)i − (4/5)j. Check: √(9/25 + 16/25) = 1.
t = ±4/5, u = 〈3/5, −4/5〉
1. Find the unit vector in the direction of 〈5, −12〉.
Explanation: |〈5, −12〉| = 13, so divide each component by 13. The −5/13, 12/13 choice is the unit vector in the opposite direction.
2. Find t > 0 so that 〈t, 1/2〉 is a unit vector.
Explanation: t² + 1/4 = 1 gives t² = 3/4, so t = √3/2 (taking the positive root).
3. Write −2i + 7j in component form.
Explanation: The coefficient of i is the x-component and the coefficient of j is the y-component.
4. What is |3i − 4j|?
Explanation: |3i − 4j| = √(9 + 16) = 5.
5. Which of these is a unit vector?
Explanation: √(9/25 + 16/25) = 1. The others have magnitudes √2/2 ≈ 0.71, √2 ≈ 1.41, and √8/3 ≈ 0.94.
10. Finding a Vector from Its Direction and Magnitude
Real situations rarely give components. A pilot reports a speed and a heading; a wind has a speed and a direction. To turn magnitude and direction into components, note that the unit vector at angle θ (measured from the positive x-axis) is 〈cos θ, sin θ〉, which is a point on the unit circle. Scaling by the magnitude gives v = |v|〈cos θ, sin θ〉 = |v|cos θ i + |v|sin θ j. In the right triangle formed by the vector and its horizontal and vertical legs, the horizontal component is the adjacent side and the vertical component is the opposite side. Going the other way, from components to direction, use tan θ = b/a, but choose the quadrant carefully: the arctangent gives an angle in quadrant I or IV only, so add 180° when a < 0. Always sketch the vector first. Directions measured as compass bearings (clockwise from north) must be converted to this standard angle before using the formula.
Worked Example: A boat moves at 10 miles per hour in a direction 30° above the positive x-axis. Write its velocity as a vector. Then find the direction angle of 〈−3, 3〉.
v = 10〈cos 30°, sin 30°〉 = 10〈√3/2, 1/2〉 = 〈5√3, 5〉 ≈ 〈8.66, 5〉.
For 〈−3, 3〉: tan θ = 3/(−3) = −1, so the reference angle is 45°. The vector is in quadrant II (x negative, y positive), so θ = 180° − 45° = 135°.
v = 〈5√3, 5〉, θ = 135°
1. A vector has magnitude 8 and direction 60°. What are its components?
Explanation: 〈8 cos 60°, 8 sin 60°〉 = 〈8(1/2), 8(√3/2)〉 = 〈4, 4√3〉.
2. A drone flies at speed 50 straight north (positive y-direction). What is its velocity vector?
Explanation: North is 90°: 〈50 cos 90°, 50 sin 90°〉 = 〈0, 50〉.
3. What is the direction angle (0° ≤ θ < 360°) of 〈−1, 1〉?
Explanation: The vector lies in quadrant II. The reference angle is 45°, so θ = 180° − 45° = 135°.
4. A wind blows at 20 mph in direction 150°. What is the wind vector?
Explanation: cos 150° = −√3/2 and sin 150° = 1/2, so the vector is 〈−10√3, 10〉.
5. Which vector has magnitude 6 and direction 225°?
Explanation: cos 225° = sin 225° = −√2/2, and 6 · (−√2/2) = −3√2 for both components. The last choice has magnitude 6√2.
11. Vector-Valued Functions: Position, Velocity, and Speed
A vector-valued function gives a vector for every input. The most important use is planar motion: a particle with parametric equations x(t), y(t) has a position vector p(t) = 〈x(t), y(t)〉 = x(t)i + y(t)j. The signs tell you the location: x(t) > 0 means to the right of the origin, y(t) > 0 means above it. The magnitude of the position vector, |p(t)| = √(x(t)² + y(t)²), is the distance of the particle from the origin at time t. A velocity vector v(t) = 〈vx(t), vy(t)〉 describes how the particle is moving at time t: the sign of vx says whether it is moving right (+) or left (−), and the sign of vy says whether it is moving up (+) or down (−). The speed is the magnitude of the velocity, |v(t)| = √(vx² + vy²), which is a scalar and is never negative. Do not mix up the two magnitudes: |p(t)| measures how far the particle is from the origin, while |v(t)| measures how fast it is moving. Include units, such as meters and seconds, whenever the problem gives them.
Worked Example: A particle has position p(t) = 〈2t, t² − 1〉 and velocity v(t) = 〈2, 2t〉, with distance in meters and time in seconds. Find its position, its distance from the origin, its direction of motion, and its speed at t = 2.
Position: p(2) = 〈4, 3〉. Distance from the origin: |p(2)| = √(16 + 9) = 5 meters.
Direction: v(2) = 〈2, 4〉. Both components are positive, so the particle is moving right and up. Speed: |v(2)| = √(4 + 16) = √20 = 2√5 ≈ 4.47 meters per second.
p(2) = 〈4, 3〉, |p(2)| = 5, |v(2)| = 2√5
1. A particle has position p(t) = 〈t² − 4, 3t〉. Where is it at t = 3?
Explanation: x(3) = 9 − 4 = 5 and y(3) = 9, so the position is (5, 9).
2. For p(t) = 〈3t, 4t²〉, what is the particle’s distance from the origin at t = 1?
Explanation: p(1) = 〈3, 4〉, so |p(1)| = √(9 + 16) = 5.
3. A velocity vector is v(t) = 〈3 − t, 2t − 4〉. At t = 1, the particle is moving
Explanation: v(1) = 〈2, −2〉. The x-component is positive (right) and the y-component is negative (down).
4. A particle has velocity v(t) = 〈t², 3t〉. What is its speed at t = 2?
Explanation: v(2) = 〈4, 6〉, so the speed is √(16 + 36) = √52 ≈ 7.21. Do not add the components, because speed is a magnitude.
5. An object has position s(t) = 〈2 cos t, 2 sin t〉. What is its distance from the origin at any time t?
Explanation: |s(t)| = √(4 cos² t + 4 sin² t) = √4 = 2. The object travels on a circle of radius 2 centered at the origin.
Common Mistakes to Avoid
- Computing a vector between points as tail minus head. The component form is always head minus tail; reversing it gives the opposite vector.
- Adding magnitudes when adding vectors. |a + b| equals |a| + |b| only when the vectors point in the same direction; otherwise add the components first, then find the magnitude.
- Using the interior angle of the triangle with the wrong sign in the law of cosines. With the angle between the vectors when tails touch, use R² = a² + b² + 2ab cos θ (or the minus form with 180° − θ).
- Forgetting that a negative scalar reverses direction. Also, |kv| = |k||v|, so the magnitude never becomes negative.
- Confusing the dot product with scalar multiplication. a · b is a number; k·a is a vector.
- Reading the sign of the dot product backwards. Positive means an acute angle, negative means obtuse, and zero means perpendicular.
- Giving a unit vector that is not divided by the magnitude, or forgetting that making a vector of length L means L times the unit vector.
- Mixing up distance from the origin (|p(t)|, the magnitude of position) with speed (|v(t)|, the magnitude of velocity).
Frequently Asked Questions
What is the difference between a vector and a scalar?
A scalar is a single number (a size only), such as speed, mass, or temperature. A vector has both size and direction, such as velocity, force, or displacement.
How do I find the magnitude of a vector between two points?
Subtract the coordinates (head minus tail) to get the components, then use the Pythagorean theorem: |v| = √(a² + b²). That is the same as the distance formula.
How do I know if two vectors are perpendicular or parallel?
Perpendicular vectors have a dot product of 0. Parallel vectors are scalar multiples of each other, so their components are proportional, for example 〈2, −3〉 and 〈−6, 9〉.
How do I find the angle between two vectors?
Use cos θ = (a · b)/(|a||b|). Compute the dot product and the two magnitudes, divide, and take the inverse cosine. The angle is always between 0° and 180°.
How do I make a vector of a given length in a given direction?
Divide the direction vector by its own magnitude to get a unit vector, then multiply by the required length. Use a negative multiplier for the opposite direction.
What is the difference between position, velocity, and speed?
Position p(t) is a vector that locates the particle. Velocity v(t) is a vector that tells which way and how fast it is moving. Speed is the magnitude of velocity, a nonnegative number.
Where can I practice more problems like these?
The AP Precalculus Parametric Functions, Vectors, and Matrices Quiz 2 in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.