Mv1–Mv3 - Vectors, components, magnitude and direction

Mv1–Mv3 - Vectors, components, magnitude and direction

Use the language and geometry of vectors to describe movements, combine arrows and calculate lengths and directions. You will move between column vectors, unit-vector notation and magnitude-direction form, with signs that agree with your sketch.

Vectors and their components

Two map instructions say “move 6 m6\,\mathrm m west, then 8 m8\,\mathrm m north”. Could one straight instruction replace them? It would need a length and a direction; “14 m14\,\mathrm m” alone would send you too far and tell you nothing about where to go.

You already know

Cartesian coordinates use perpendicular axes. Pythagoras gives a right triangle's hypotenuse, and trigonometric ratios connect its lengths and angles. A negative coordinate means the opposite direction along that axis.

Choose east as positive horizontal and north as positive vertical. The two instructions combine into the arrow (−6,8)(-6,8). Their route has length 14 m14\,\mathrm m, but the straight displacement has length 10 m10\,\mathrm m, since 62+82=1026^2+8^2=10^2. Surprisingly, the final arrow is unchanged if you swap the order of the instructions, even though you visit a different intermediate point.

Scalar, vector and components

A scalar has magnitude only; a vector has magnitude and direction. Distance and speed are scalars; displacement and velocity are vectors. Cartesian components are signed amounts along perpendicular axes. With standard unit vectors i\mathbf i and j\mathbf j, our displacement is

d=−6i+8j=(−68)\mathbf d=-6\mathbf i+8\mathbf j=\begin{pmatrix}-6\\8\end{pmatrix}

The component vectors are −6i-6\mathbf i and 8j8\mathbf j; their coefficients are scalars. Bold letters distinguish vectors in print. Underline a vector letter in handwriting, or use an arrow such as AB→\overrightarrow{AB}. Reversing that arrow gives BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}.

Vector language

Equal vectors have the same magnitude and direction, hence equal corresponding components. Non-zero parallel vectors are scalar multiples; the multiplier can be negative. Points are collinear when they lie on one straight line. A position vector starts at a chosen origin; a unit vector has magnitude 11. The zero vector has magnitude zero and no direction.

For example, (−3,4)(-3,4) and (−6,8)(-6,8) are parallel but not equal. An arrow can slide to another starting point without changing its vector. A position vector cannot do that while still describing the same point from the same origin. We will use parallel vectors to prove collinearity in the next lesson.

Key idea

A signed component can be negative; a modulus or magnitude is a non-negative scalar. Equal lengths alone do not make two vectors equal.

The distinction is small in notation and large in meaning.

Adding, subtracting and scaling vectors

Vector addition joins arrows head to tail. Its resultant runs from the first start to the final finish, so it records the combined movement without remembering the route.

Head-to-tail addition of two vectors
The blue arrow (3,1)(3,1), followed by the purple arrow (−1,3)(-1,3), gives the green resultant (2,4)(2,4). Translating the purple arrow preserves its vector.

The diagram also explains why addition is component by component. Subtracting a vector means adding its reverse. If two arrows share a tail, a−b\mathbf a-\mathbf b goes from the head of b\mathbf b to the head of a\mathbf a.

Combining and scaling vectors

For a=(a1,a2)\mathbf a=(a_1,a_2), b=(b1,b2)\mathbf b=(b_1,b_2) and real kk:

a±b=(a1±b1,a2±b2),ka=(ka1,ka2)\mathbf a\pm\mathbf b=(a_1\pm b_1,a_2\pm b_2),\qquad k\mathbf a=(ka_1,ka_2)

Scaling changes magnitude by ∣k∣|k|. For non-zero vectors, positive kk keeps direction and negative kk reverses it; k=0k=0 gives 0\mathbf0.

Not in the formula sheet: learn it.

MEI also uses general vectors whose components are unknown. The route round a triangle remains valid without coordinates: if OA→=p\overrightarrow{OA}=\mathbf p and OB→=q\overrightarrow{OB}=\mathbf q, then AB→=q−p\overrightarrow{AB}=\mathbf q-\mathbf p.

Worked example 1: a combination of general vectors

Question. In triangle OABOAB, OA→=p\overrightarrow{OA}=\mathbf p and OB→=q\overrightarrow{OB}=\mathbf q. Find 2AB→+OA→2\overrightarrow{AB}+\overrightarrow{OA}. Then evaluate it for p=(2,−1)\mathbf p=(2,-1), q=(5,3)\mathbf q=(5,3).

Plan. Express each route using p\mathbf p and q\mathbf q, simplify, then insert components.

AB→=q−pfinish minus start2AB→+OA→=2(q−p)+p=2q−pcollect vectors=2(5,3)−(2,−1)=(8,7)use the components\begin{aligned} \overrightarrow{AB}&=\mathbf q-\mathbf p &&\text{finish minus start}\\ 2\overrightarrow{AB}+\overrightarrow{OA}&=2(\mathbf q-\mathbf p)+\mathbf p=2\mathbf q-\mathbf p &&\text{collect vectors}\\ &=2(5,3)-(2,-1)=(8,7) &&\text{use the components} \end{aligned}

Why this step? Expressing the route first prevents the arrow ABAB being confused with the position of BB.

Answer. 2q−p2\mathbf q-\mathbf p, or 8i+7j8\mathbf i+7\mathbf j for the given vectors.

Check. AB→=(3,4)\overrightarrow{AB}=(3,4), and 2(3,4)+(2,−1)=(8,7)2(3,4)+(2,-1)=(8,7).

Change only the final plus sign to a minus sign. Keep the route and both vectors unchanged.

Worked example 2: reversing one contribution

Question. For the same triangle and components, find 2AB→−OA→2\overrightarrow{AB}-\overrightarrow{OA}.

Plan. Subtract the entire vector p\mathbf p after doubling AB→\overrightarrow{AB}.

2AB→−OA→=2(q−p)−psubstitute the routes=2q−3pcollect coefficients=(10,6)−(6,−3)=(4,9)subtract both components\begin{aligned} 2\overrightarrow{AB}-\overrightarrow{OA}&=2(\mathbf q-\mathbf p)-\mathbf p &&\text{substitute the routes}\\ &=2\mathbf q-3\mathbf p &&\text{collect coefficients}\\ &=(10,6)-(6,-3)=(4,9) &&\text{subtract both components} \end{aligned}

Why this step? Changing the sign of OA→\overrightarrow{OA} reverses the whole vector, including its negative vertical component.

Answer. 2q−3p=4i+9j2\mathbf q-3\mathbf p=4\mathbf i+9\mathbf j.

Check. 2(3,4)−(2,−1)=(4,9)2(3,4)-(2,-1)=(4,9).

For a second numerical check on the diagram's arrows, a+b=(2,4)\mathbf a+\mathbf b=(2,4), a−b=(4,−2)\mathbf a-\mathbf b=(4,-2) and 2a−3b=(9,−7)2\mathbf a-3\mathbf b=(9,-7). Each operation treats the two components separately.

Watch out

Wrong: (6,8)−(2,−1)=(4,7)(6,8)-(2,-1)=(4,7). Right: the vertical component is 8−(−1)=98-(-1)=9, so the result is (4,9)(4,9).

Use the same route with fresh component values.

Continue this lesson.

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