J1, J2, J3 - Vector components and operations

J1, J2, J3 - Vector components and operations

Describe directed quantities in two and three dimensions, combine their effects, and move between components and magnitude with a clearly stated direction. J1 supplies the notation, J2 connects an arrow to its length and angles, and J3 explains what addition and multiplication by a scalar do geometrically.

Components and vector notation

If you walk four metres east and three metres south, have you moved seven metres from your starting point? Your route has length seven metres, but your final displacement is a different quantity: it records the direct change of position, including direction.

You already know

Coordinates locate points on perpendicular axes. Pythagoras connects perpendicular lengths, and sine and cosine connect a right triangle's sides to its angles.

We will give those coordinate changes their own notation so that a movement is distinct from its endpoint.

Vector and scalar

A vector has magnitude and direction. A scalar has magnitude only. Displacement and velocity are vectors; distance and speed are scalars. Equal vectors have the same magnitude and direction even if their arrows start at different points.

Think of a vector as a movement instruction. Choose perpendicular axes with equal distance units. The unit vectors i\mathbf i and j\mathbf j have length one in the positive horizontal and vertical directions. Four units right and three down is

a=4i−3j=[4−3]\mathbf a=4\mathbf i-3\mathbf j=\begin{bmatrix}4\\-3\end{bmatrix}

The entries are signed components, rather than two positive lengths. The arrow AB→\overrightarrow{AB} starts at AA and ends at BB; reversing it gives BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}. These notation and reversal rules are not in the formula booklet: learn them.

In space, add k\mathbf k, the unit vector along the third perpendicular axis. Thus 2i−j+5k=[2−15]2\mathbf i-\mathbf j+5\mathbf k=\begin{bmatrix}2\\-1\\5\end{bmatrix}. Use the order x,y,zx,y,z consistently in a right-handed Cartesian coordinate system. Printed bold letters distinguish vectors from scalars; in handwriting an arrow, underline or overline is acceptable. AQA uses square brackets for columns, but round brackets are equally acceptable. The magnitude of a\mathbf a is written ∣a∣|\mathbf a|. In applications, r,s,v,a\mathbf r,\mathbf s,\mathbf v,\mathbf a commonly denote position, displacement, velocity and acceleration vectors respectively.

Worked example 1: translating a graph

Question. A graph is translated two units left and five units up. Write the translation vector and find the image of (3,−1)(3,-1).

Plan. Give each movement a sign, then add those components to the point's coordinates.

t=[−25]=−2i+5jleft is negative(3,−1)⟼(3−2,−1+5)same movement at every point=(1,4)\begin{aligned} \mathbf t&=\begin{bmatrix}-2\\5\end{bmatrix}=-2\mathbf i+5\mathbf j && \text{left is negative}\\ (3,-1)&\longmapsto(3-2,-1+5) && \text{same movement at every point}\\ &=(1,4) \end{aligned}

Why this step? A translation describes the change of position, so the components are added to the existing coordinates.

Answer. The vector is [−25]\begin{bmatrix}-2\\5\end{bmatrix}, and the image is (1,4)(1,4). The graph keeps its shape and size.

Check. From (3,−1)(3,-1) to (1,4)(1,4), the horizontal change is −2-2 and the vertical change is 55.

The next example changes one feature: the movement now has a third component. The first two instructions keep exactly the same meaning.

Worked example 2: translating a point in space

Question. A point P=(3,−1,0)P=(3,-1,0) is translated two units left, five units in the positive yy direction and four units in the positive zz direction. Find its image and give the translation in both vector notations.

Plan. Put the three signed changes in coordinate order and add them to PP.

t=[−254]=−2i+5j+4korder x,y,zP′=(3−2,−1+5,0+4)add corresponding entries=(1,4,4)\begin{aligned} \mathbf t&=\begin{bmatrix}-2\\5\\4\end{bmatrix}=-2\mathbf i+5\mathbf j+4\mathbf k && \text{order }x,y,z\\ P'&=(3-2,-1+5,0+4) && \text{add corresponding entries}\\ &=(1,4,4) \end{aligned}

Why this step? The third component changes depth or height along the chosen zz axis; it does not alter the first two components.

Answer. P′=(1,4,4)P'=(1,4,4), with translation −2i+5j+4k-2\mathbf i+5\mathbf j+4\mathbf k.

Check. Subtracting the coordinates of PP from those of P′P' returns (−2,5,4)(-2,5,4).

A zero component is still useful information: it says there is no movement along that axis.

Watch out

Wrong: “Three units left means component 33.” A component records a signed change, so left along the chosen horizontal axis is −3-3. Right: three left and four up is [−34]\begin{bmatrix}-3\\4\end{bmatrix}.

Try the same conversion with one stationary coordinate.

Addition and scalar multiplication

Two successive movements combine by adding their signed changes, coordinate by coordinate. For u=3i+j\mathbf u=3\mathbf i+\mathbf j and v=−i+3j\mathbf v=-\mathbf i+3\mathbf j, the combined movement is 2i+4j2\mathbf i+4\mathbf j.

Head-to-tail addition of u=(3,1) and v=(-1,3), with resultant (2,4)

Draw u\mathbf u, then place the tail of an equal copy of v\mathbf v at its head. The resultant joins the first tail to the last head. The dashed copy shows that translating an arrow without rotating or resizing it preserves its vector. Starting both vectors at the same point gives the parallelogram construction: its diagonal is the same resultant. Reversing the order reaches the same endpoint, so u+v=v+u\mathbf u+\mathbf v=\mathbf v+\mathbf u.

Subtracting v\mathbf v means adding its reverse, −v-\mathbf v. Multiplication by a scalar λ\lambda multiplies every component by λ\lambda. A positive scalar keeps direction; a negative scalar reverses it. Multiplication by zero gives the zero vector, which has magnitude zero and no direction.

Operations and their geometry

For vectors in the same coordinate system,

[a1a2a3]+[b1b2b3]=[a1+b1a2+b2a3+b3]λ[a1a2a3]=[λa1λa2λa3]∣λa∣=∣λ∣ ∣a∣\begin{aligned} \begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}+\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}&=\begin{bmatrix}a_1+b_1\\a_2+b_2\\a_3+b_3\end{bmatrix}\\ \lambda\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}&=\begin{bmatrix}\lambda a_1\\\lambda a_2\\\lambda a_3\end{bmatrix}\\ |\lambda\mathbf a|&=|\lambda|\,|\mathbf a| \end{aligned}

Omit the third row in two dimensions. Not in the formula booklet: learn it.

For example, [6−4]=−2[−32]\begin{bmatrix}6\\-4\end{bmatrix}=-2\begin{bmatrix}-3\\2\end{bmatrix}: the arrows are parallel, twice as long and oppositely directed. In general, two nonzero vectors are parallel if one is a scalar multiple of the other. The same scalar must work in every component. This criterion is not in the formula booklet: learn it. Parallel arrows drawn in unrelated places do not establish that all their endpoints lie on one line.

Worked example 3: subtracting a vector in space

Question. Given a=[1−23]\mathbf a=\begin{bmatrix}1\\-2\\3\end{bmatrix} and b=[−214]\mathbf b=\begin{bmatrix}-2\\1\\4\end{bmatrix}, find 2a−b2\mathbf a-\mathbf b.

Plan. Multiply every component of a\mathbf a by two, then subtract each component of b\mathbf b.

2a−b=[2−46]−[−214]scale before subtracting=[2−(−2)−4−16−4]retain each sign=[4−52]\begin{aligned} 2\mathbf a-\mathbf b&=\begin{bmatrix}2\\-4\\6\end{bmatrix}-\begin{bmatrix}-2\\1\\4\end{bmatrix} && \text{scale before subtracting}\\ &=\begin{bmatrix}2-(-2)\\-4-1\\6-4\end{bmatrix} && \text{retain each sign}\\ &=\begin{bmatrix}4\\-5\\2\end{bmatrix} \end{aligned}

Why this step? Subtracting a negative horizontal component adds a positive change.

Answer. 2a−b=4i−5j+2k2\mathbf a-\mathbf b=4\mathbf i-5\mathbf j+2\mathbf k.

Check. Adding b\mathbf b to the answer returns (2,−4,6)=2a(2,-4,6)=2\mathbf a.

Now change only the operation before b\mathbf b, from subtraction to addition. Every component of b\mathbf b must respond to that change.

Worked example 4: adding the same vector

Question. For the same a=[1−23]\mathbf a=\begin{bmatrix}1\\-2\\3\end{bmatrix} and b=[−214]\mathbf b=\begin{bmatrix}-2\\1\\4\end{bmatrix}, find 2a+b2\mathbf a+\mathbf b.

Plan. Use the same doubled vector, then add the corresponding entries.

2a+b=[2−46]+[−214]addition in every row=[0−310]\begin{aligned} 2\mathbf a+\mathbf b&=\begin{bmatrix}2\\-4\\6\end{bmatrix}+\begin{bmatrix}-2\\1\\4\end{bmatrix} && \text{addition in every row}\\ &=\begin{bmatrix}0\\-3\\10\end{bmatrix} \end{aligned}

Why this step? The operation applies to the whole vector, including its negative entries.

Answer. 2a+b=−3j+10k2\mathbf a+\mathbf b=-3\mathbf j+10\mathbf k.

Check. The difference between this answer and the previous one is (−4,2,8)=2b(-4,2,8)=2\mathbf b, as changing −b-\mathbf b to +b+\mathbf b requires.

A resultant may have a zero component even though both added vectors have nonzero components there: they cancel.

Watch out

Wrong: [2−46]−[−214]=[0−52]\begin{bmatrix}2\\-4\\6\end{bmatrix}-\begin{bmatrix}-2\\1\\4\end{bmatrix}=\begin{bmatrix}0\\-5\\2\end{bmatrix}. The first subtraction is 2−(−2)=42-(-2)=4, not 2−22-2. Right: [4−52]\begin{bmatrix}4\\-5\\2\end{bmatrix}.

Keep that sign check when the vectors change.

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