Pure 10.3 - Vector addition and scalar multiplication

Pure 10.3 - Vector addition and scalar multiplication

Vector addition combines directed movements, while scalar multiplication changes a vector's length and sometimes its direction. You will connect triangle and parallelogram diagrams to component calculations, combine scaled vectors, and establish when two vectors are parallel. The diagram chooses the operation; the components carry it out.

Addition as a Resultant Journey

Can two nonzero journeys add to no movement at all? Walk three units right, then three left. You have travelled six units, but the direct vector from start to finish is the zero vector. Vector addition keeps the direction information that a sum of distances loses.

You already know

A vector can be translated without changing it. Reversing an arrow reverses every component, so the reverse of a\mathbf a is −a-\mathbf a.

Following one movement and then another gives their resultant: the single vector from the first start to the final finish.

[DIAGRAM: asset_slug: p1_10_3_add_vectors_and_multiply_by_scalars_understand_geometric_interpretations__diagram_01; asset_name: Triangle and parallelogram addition; caption: Both constructions give the same resultant; description: A head-to-tail triangle with a followed by b and resultant a plus b, alongside the same two vectors drawn from a common tail with a parallelogram diagonal a plus b]
Both constructions give the same resultant

The two pictures use the same movements. They differ only in where copies of the arrows are placed.

Triangle and parallelogram laws

Triangle law: put the tail of b\mathbf b at the head of a\mathbf a; the vector from the first tail to the final head is a+b\mathbf a+\mathbf b.

Parallelogram law: draw a\mathbf a and b\mathbf b from a common tail and complete the parallelogram. Its diagonal from that tail is a+b\mathbf a+\mathbf b.

Doing the two journeys in the reverse order reaches the same final point, so a+b=b+a\mathbf a+\mathbf b=\mathbf b+\mathbf a. This is why both constructions agree. In particular, a+(−a)=0\mathbf a+(-\mathbf a)=\mathbf0: the opening puzzle has an exact algebraic answer.

Worked example 1: following a triangle route

Question. In triangle ABCABC, AB→=a\overrightarrow{AB}=\mathbf a and BC→=b\overrightarrow{BC}=\mathbf b. Find AC→\overrightarrow{AC}.

Plan. Follow the connected route from AA through BB to CC.

AC→=AB→+BC→head-to-tail route=a+bsubstitute the given vectors\begin{aligned} \overrightarrow{AC}&=\overrightarrow{AB}+\overrightarrow{BC} &&\text{head-to-tail route}\\ &=\mathbf a+\mathbf b &&\text{substitute the given vectors} \end{aligned}

Why this step? The finish of ABAB is the start of BCBC, so adding those directed journeys gives ACAC.

Answer. AC→=a+b\overrightarrow{AC}=\mathbf a+\mathbf b.

Check. Reversing the complete route gives CA→=−a−b\overrightarrow{CA}=-\mathbf a-\mathbf b.

Change only the direction of the given second arrow: it is now CBCB, rather than BCBC.

Worked example 2: reversing an arrow before adding

Question. In triangle ABCABC, AB→=a\overrightarrow{AB}=\mathbf a and CB→=b\overrightarrow{CB}=\mathbf b. Find AC→\overrightarrow{AC}.

Plan. Reverse CBCB so that the two pieces form a connected route.

BC→=−breverse the supplied arrowAC→=AB→+BC→follow A to B to C=a−bsubstitute with its sign\begin{aligned} \overrightarrow{BC}&=-\mathbf b &&\text{reverse the supplied arrow}\\ \overrightarrow{AC}&=\overrightarrow{AB}+\overrightarrow{BC} &&\text{follow }A\text{ to }B\text{ to }C\\ &=\mathbf a-\mathbf b &&\text{substitute with its sign} \end{aligned}

Why this step? Adding ABAB to CBCB directly would try to join two arrows whose heads both end at BB.

Answer. AC→=a−b\overrightarrow{AC}=\mathbf a-\mathbf b.

Check. Adding CB→=b\overrightarrow{CB}=\mathbf b to this answer gives AB→=a\overrightarrow{AB}=\mathbf a.

The letters on the arrow matter as much as the vector's coefficients.

Watch out

Wrong: AC→=AB→+CB→\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{CB}. Right: AC→=AB→−CB→\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{CB}. Reverse the second arrow to make the route connect.

Use a new triangle and a scaled second arrow.

Adding Components Algebraically

A movement of three units right followed by five left produces a net horizontal change of minus two. The vertical and third-direction changes combine independently in exactly the same way.

Component addition

For vectors expressed relative to the same axes,

(x1y1z1)+(x2y2z2)=(x1+x2y1+y2z1+z2)\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}+\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}=\begin{pmatrix}x_1+x_2\\y_1+y_2\\z_1+z_2\end{pmatrix}

In two dimensions omit the third row from each vector.

Not in the formula booklet: learn it.

The unit-vector notation makes these into familiar collections of like terms.

Worked example 3: adding three-dimensional components

Question. Given a=3i−2j+4k\mathbf a=3\mathbf i-2\mathbf j+4\mathbf k and b=−5i+7j−k\mathbf b=-5\mathbf i+7\mathbf j-\mathbf k, find a+b\mathbf a+\mathbf b.

Plan. Add the coefficients of each unit vector separately.

a+b=(3−5)i+(−2+7)j+(4−1)kcollect like components=−2i+5j+3kevaluate the sums\begin{aligned} \mathbf a+\mathbf b&=(3-5)\mathbf i+(-2+7)\mathbf j+(4-1)\mathbf k &&\text{collect like components}\\ &=-2\mathbf i+5\mathbf j+3\mathbf k &&\text{evaluate the sums} \end{aligned}

Why this step? The k\mathbf k-component contributes to movement along zz, so it cannot be combined with a j\mathbf j-component.

Answer. −2i+5j+3k-2\mathbf i+5\mathbf j+3\mathbf k.

Check. Subtracting b\mathbf b from the answer gives 3i−2j+4k=a3\mathbf i-2\mathbf j+4\mathbf k=\mathbf a.

Change only the sign of the final component of b\mathbf b. Only the final component of the sum can change.

Worked example 4: changing one component

Question. Given a=3i−2j+4k\mathbf a=3\mathbf i-2\mathbf j+4\mathbf k and c=−5i+7j+k\mathbf c=-5\mathbf i+7\mathbf j+\mathbf k, find a+c\mathbf a+\mathbf c.

Plan. Repeat the component sums with the revised k\mathbf k-coefficient.

a+c=(3−5)i+(−2+7)j+(4+1)kadd corresponding components=−2i+5j+5ksimplify\begin{aligned} \mathbf a+\mathbf c&=(3-5)\mathbf i+(-2+7)\mathbf j+(4+1)\mathbf k &&\text{add corresponding components}\\ &=-2\mathbf i+5\mathbf j+5\mathbf k &&\text{simplify} \end{aligned}

Why this step? Changing −1-1 to +1+1 increases that component by two; the other components are unaffected.

Answer. −2i+5j+5k-2\mathbf i+5\mathbf j+5\mathbf k.

Check. The difference between the two resultants is 2k2\mathbf k, exactly the change to the input vector.

Vector magnitudes do not generally add. Perpendicular vectors 3i3\mathbf i and 4j4\mathbf j have magnitudes three and four, but their sum has magnitude five.

Watch out

Wrong: ∣a+b∣=∣a∣+∣b∣\lvert\mathbf a+\mathbf b\rvert=\lvert\mathbf a\rvert+\lvert\mathbf b\rvert for every pair. Right: add the vectors first, then calculate the resultant's magnitude if it is needed.

Now combine a new pair of spatial movements.

Multiplying by a Scalar

Multiplying a movement by two doubles each component, so the resulting arrow has twice the length in the same direction. Multiplying by a negative number also reverses its direction.

Scalar

A scalar is a number with no direction. Multiplying a vector by a real scalar gives a vector, rather than a scalar.

The diagrams show a stretch, a shortening and a reversal of one reference vector.

[DIAGRAM: asset_slug: p1_10_3_add_vectors_and_multiply_by_scalars_understand_geometric_interpretations__diagram_02; asset_name: Positive and negative vector scalings; caption: Positive scalars preserve direction and negative scalars reverse it; description: Three panels compare a reference vector a with 2a, one half a and minus a, with lengths and directions drawn consistently]
Positive scalars preserve direction and negative scalars reverse it

Length depends on the absolute value of the scalar.

Scaling a vector

For a real scalar λ\lambda,

λ(xyz)=(λxλyλz)∣λa∣=∣λ∣∣a∣\lambda\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\lambda x\\\lambda y\\\lambda z\end{pmatrix} \qquad \lvert\lambda\mathbf a\rvert=\lvert\lambda\rvert\lvert\mathbf a\rvert

For nonzero a\mathbf a, a positive scalar preserves direction, a negative scalar reverses it, and zero gives 0\mathbf0.

Not in the formula booklet: learn it.

In particular, 0<λ<10<\lambda<1 shortens a nonzero vector, while λ>1\lambda>1 lengthens it.

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