Vectors
Describe directed motion with components and combine vectors geometrically and algebraically.
Learning goals
- Find a vector from its endpoints and calculate magnitude.
- Add, subtract, and scale vectors.
- Interpret a resultant using a head-to-tail diagram.
Directed segments and components
A journey has both a size and a direction. To describe its displacement with coordinates, compare the endpoint with the starting point in each direction. Moving the whole arrow without turning or stretching it preserves the vector. That is why vectors can describe the same displacement from different starting locations.
A vector has magnitude and direction. Translating an arrow without turning or resizing it gives an equivalent vector. In standard position its initial point is the origin.
\[\overrightarrow{PQ}=\langle q_1-p_1,q_2-p_2\rangle\]Two vectors are equal exactly when corresponding components agree. The zero vector is \(\langle0,0\rangle\); it has magnitude zero and no unique direction.
Magnitude
Length without direction
The magnitude of a Euclidean vector is the square root of the sum of the squares of its components in an orthonormal coordinate system. It is the vector’s length and is a nonnegative scalar.
The components are signed changes along perpendicular axes, so the Pythagorean theorem combines their squares to obtain that length. Vectors can have the same magnitude while pointing in different directions.
The components tell us how much a displacement changes in each coordinate direction. Its magnitude answers another question: how long is the displacement itself? A right-triangle picture connects these two descriptions.
\[\|\mathbf v\|=\sqrt{v_1^2+v_2^2}\]From two endpoints
A robot moves from P to Q on a coordinate grid with equal units on both axes. Find its displacement vector and the straight-line distance traveled.
\(P=(4,-7),\ Q=(-1,5)\)Show worked solution
Subtract initial coordinates from terminal coordinates. Reversing the endpoints changes direction but not length.
Scalar multiplication
Changing the size of a displacement
A scalar is an ordinary real number used here to multiply every vector component. Its absolute value changes the vector’s length by that factor; its sign determines whether the direction is retained or reversed.
Multiplying by zero removes any nonzero displacement.
A scalar changes a vector through stretching, shrinking, or reversal. Before calculating components, predict what the sign of the scalar will do to the direction. Then compare the new magnitude with the old one; the absolute size of the scalar controls the length change.
\[k\langle v_1,v_2\rangle=\langle kv_1,kv_2\rangle,\qquad\|k\mathbf v\|=|k|\|\mathbf v\|\]A positive scalar preserves direction, a negative scalar reverses it, and zero produces the zero vector.
Double a vector
A robot repeats the same displacement twice. Find its total displacement and describe how its length and direction compare with the original.
\(\mathbf v=\langle-2,5\rangle\)Show worked solution
The arrow becomes twice as long and keeps the same direction.
Separate direction from length
A unit vector has magnitude one. For any nonzero vector \(\mathbf v\), the vector \(\widehat{\mathbf v}=\mathbf v/\|\mathbf v\|\) has the same direction and unit magnitude, since \(\|\mathbf v/\|\mathbf v\|\|=1\). Multiplying this unit vector by a desired nonnegative length constructs a vector with that length and direction. The zero vector cannot be normalized because it has no direction and division by zero is undefined.
A prescribed displacement
A drone must move \(15\) m in the direction \(\langle-3,4\rangle\). Find its displacement components, and its endpoint if it starts at \((8,-2)\).
Show worked solution
The direction vector has length \(5\), so its unit vector is \(\langle-3/5,4/5\rangle\). Multiplying by \(15\) gives \(\langle-9,12\rangle\) m. Add the displacement to the starting point: \((8,-2)+(-9,12)=(-1,10)\). The endpoint is a position, while the vector records change in position.
Dividing by the sum of components does not normalize a vector. For example, the component sum of ⟨−3,4⟩ is 1, but its magnitude is 5. Use the square-root-of-squares length.
Addition and the resultant
Think of vector addition as completing one displacement and then the next. The resultant goes straight from the initial point to the final point. This picture explains the component rule and keeps the order of the journey distinct from its final effect.
\[\mathbf u+\mathbf v=\langle u_1+v_1,u_2+v_2\rangle\]Place the second arrow’s tail at the first arrow’s head. The resultant joins the first tail to the last head. Do not rotate either arrow.
Subtraction and combinations
Addition joined two displacements. For subtraction, first imagine reversing the vector being subtracted, then use the same addition idea. A sketch is especially useful here: the difference can describe the arrow from one vector endpoint to the other. Check its direction before trusting the signs in your components.
\[\mathbf w-\mathbf v=\mathbf w+(-\mathbf v)\]Subtract component by component
Two routes from the same starting point have displacements v and w below. Find the displacement from the endpoint of v to the endpoint of w.
\(\mathbf v=\langle-2,5\rangle,\quad\mathbf w=\langle3,4\rangle\)Show worked solution
A linear combination
Use v = ⟨−2, 5⟩ and w = ⟨3, 4⟩. A robot follows v once and w twice; calculate the total displacement.
\(\mathbf v+2\mathbf w\)Show worked solution
Properties and common mistakes
Component calculations should agree with the motion represented by the arrows. Use a round trip or two different routes to test the rules for combining displacements. This also helps distinguish a vector from its length or from the particular location where it is drawn.
- Vector addition is commutative and associative: \(\mathbf u+\mathbf v=\mathbf v+\mathbf u\).
- Scalars distribute: \(k(\mathbf u+\mathbf v)=k\mathbf u+k\mathbf v\).
- The opposite vector satisfies \(\mathbf v+(-\mathbf v)=\mathbf0\).
Add components, not magnitudes. Two nonzero vectors can cancel. A negative component describes direction; magnitude is never negative.
Practice
For each question, decide whether you are being asked for a vector, a length, or an endpoint. Those answers describe different things. Keep a quick sketch beside the calculation so the signs and direction remain meaningful.
Equivalent arrows
Show worked solution
They represent the same vector.
A round trip
Show worked solution
Net displacement
Show worked solution
Distance traveled is \(4+3+1=8\) units. It differs from the straight-line displacement.
Find an unknown
Find the missing displacement vector u that makes this journey end at the stated final displacement.
\(\mathbf u+\langle2,-3\rangle=\langle7,1\rangle\)Show worked solution
A resultant must satisfy the whole situation
Can the boat travel straight across?
Take east and north as positive directions. A current flows east at 3 m/s. A boat moves at speed 5 m/s relative to the water. Find its heading components relative to the water if its ground motion is due north. How long does it take to cross a 100 m wide river?
Show worked solution
Ground velocity equals boat-relative-to-water velocity plus current. To cancel the east component, choose boat velocity \(\langle-3,v_y\rangle\). Its speed condition gives \(9+v_y^2=25\). Choose \(v_y=4\) for northward travel. Ground velocity is \(\langle0,4\rangle\) m/s, so the time is \(100/4=25\) seconds. If the current speed were at least the boat’s speed, a due-north crossing with positive north component would be impossible: all or more of the speed would be needed to cancel drift.
Additional practice: mixed exercises
Component operations
Given \(\mathbf v=\langle-2,5\rangle\) and \(\mathbf w=\langle3,4\rangle\), calculate each vector.
- \(2\mathbf v\)
- \(\mathbf w-\mathbf v\)
- \(\mathbf v+2\mathbf w\)
Show worked solution
- Multiply each component by \(2\): \(\langle-4,10\rangle\).
- Subtract corresponding components: \(\langle3+2,4-5\rangle=\langle5,-1\rangle\).
- \(\langle-2,5\rangle+\langle6,8\rangle=\langle4,13\rangle\).
Component arithmetic
Calculate each component separately and interpret scaling geometrically.
- Add \(\mathbf a=\langle8,13\rangle\) and \(\mathbf b=\langle26,7\rangle\).
- Subtract \(\mathbf k=\langle4,5\rangle\) from \(\mathbf v=\langle12,2\rangle\).
- Multiply \(\mathbf m=\langle7,3\rangle\) by \(3\).
Show worked solution
- \(\mathbf a+\mathbf b=\langle8+26,13+7\rangle=\langle34,20\rangle\). This is the resultant of placing the vectors head to tail.
- \(\mathbf v-\mathbf k=\langle12-4,2-5\rangle=\langle8,-3\rangle\). Reverse the subtracted vector, then add.
- \(3\mathbf m=\langle21,9\rangle\). Its length is three times as large and its direction is unchanged.
Takeaways and connections
- Keep component order consistent.
- Magnitude comes from the Pythagorean theorem.
- Geometric and component methods describe the same operations.
Reasoning, connections and deeper practice
Components describe a displacement while magnitude describes its length. The following tasks connect those two representations and use vectors to establish a geometric fact, rather than merely calculating an arrow.
Application · A current changes the resultant
A boat moves relative to the water at \(\langle0,4\rangle\) km/h while a current moves at \(\langle3,0\rangle\) km/h. Find its ground velocity, ground speed, and displacement after \(30\) minutes. Why is adding the two speeds wrong?
Hint
Velocities add as vectors. Convert time to hours before multiplying.
Show worked solution
Ground velocity is \(\langle3,4\rangle\) km/h, so speed is \(\sqrt{3^2+4^2}=5\) km/h. In \(1/2\) hour the displacement is \(\langle1.5,2\rangle\) km, with magnitude \(2.5\) km. Adding speeds would give \(7\), which assumes both motions point in the same direction; these motions are perpendicular.
Advanced / Honors · Diagonals characterize a parallelogram
Let \(A,B,C,D\) be distinct vertices in order of a nondegenerate quadrilateral, with position vectors \(a,b,c,d\). Prove it is a parallelogram if and only if \(a+c=b+d\). Interpret the equality using diagonal midpoints.
Hint
Rearrange the equation into equal opposite-side displacements, and divide it by two for the midpoint interpretation.
Show worked solution
If \(a+c=b+d\), then \(b-a=c-d\) and \(d-a=c-b\), so both opposite-side pairs have equal directed displacements and are parallel with equal lengths. Hence the quadrilateral is a parallelogram. Conversely, in a parallelogram write \(b=a+u,d=a+v,c=a+u+v\); then both sums equal \(2a+u+v\). Dividing by two gives \((a+c)/2=(b+d)/2\), saying the diagonals bisect one another. Nondegeneracy excludes four collinear vertices.
