P4 7.1–P4 7.3 - Vectors in two and three dimensions

P4 7.1–P4 7.3 - Vectors in two and three dimensions

Describe vectors using two or three components, find their magnitudes and unit directions, and combine them algebraically. Connect component calculations to arrows, journeys and parallel directions so that the algebra has a geometric meaning.

Components and basis vectors

Can two arrows drawn in different places represent the same vector? Yes: a vector records a size and direction, rather than a fixed starting point.

Moving 33 units right and 22 units down gives the vector (3−2)\binom{3}{-2}. Moving that arrow elsewhere does not change either component. Reversing it does: the reverse journey is (−32)\binom{-3}{2}.

Vector and scalar

A vector has magnitude and direction. A scalar is a number used to describe a quantity without a direction, or to multiply a vector. A vector's components are its signed changes along perpendicular coordinate axes.

The basis vectors i\mathbf i, j\mathbf j and k\mathbf k have length 11 along the positive x-, y- and z-directions. In three dimensions,

i=(100),j=(010),k=(001).\mathbf i=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad \mathbf j=\begin{pmatrix}0\\1\\0\end{pmatrix},\quad \mathbf k=\begin{pmatrix}0\\0\\1\end{pmatrix}.

Thus ai+bj+cka\mathbf i+b\mathbf j+c\mathbf k and (abc)\begin{pmatrix}a\\b\\c\end{pmatrix} describe the same vector. Two vectors are equal exactly when their corresponding components are equal.

[DIAGRAM: asset_name: A vector can be translated; asset_slug: p4-vector-components; caption: Both arrows have components three and negative two, despite different starting points.; description: A coordinate grid shows the vector from zero zero to three negative two and a translated copy from one four to four two. Horizontal and vertical component steps are labelled plus three and negative two for both arrows.]
Both arrows have components three and negative two, despite different starting points.

Worked example 1: two components

Question. Write a=3i−2j\mathbf a=3\mathbf i-2\mathbf j as a column vector and describe the reverse vector.

Plan. Read the coefficient of each basis vector, then reverse both component changes.

a=(3−2),−a=(−32).\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}, \qquad -\mathbf a=\begin{pmatrix}-3\\2\end{pmatrix}.

Answer. a\mathbf a moves 33 units in the positive x-direction and 22 in the negative y-direction; −a-\mathbf a reverses this.

Check. Following a\mathbf a then −a-\mathbf a returns to the start: the total changes are 3−3=03-3=0 and −2+2=0-2+2=0.

Add only a z-component. The first two changes stay the same.

Worked example 2: a third component

Question. Write b=3i−2j+6k\mathbf b=3\mathbf i-2\mathbf j+6\mathbf k as a column vector and describe the reverse vector.

Plan. Include the z-change as the third entry and reverse all three entries for −b-\mathbf b.

b=(3−26),−b=(−32−6).\mathbf b=\begin{pmatrix}3\\-2\\6\end{pmatrix}, \qquad -\mathbf b=\begin{pmatrix}-3\\2\\-6\end{pmatrix}.

Answer. The additional change is 66 units in the positive z-direction; the reverse has z-change −6-6.

Check. Each component of b+(−b)\mathbf b+(-\mathbf b) is zero, so the reverse still undoes the whole journey.

A missing component is zero, rather than an unknown value.

Watch out

3i+6k3\mathbf i+6\mathbf k means (306)\begin{pmatrix}3\\0\\6\end{pmatrix}. The middle entry cannot be omitted from a three-dimensional column vector.

Use the same component order here.

Magnitude and unit vectors

Magnitude measures an arrow's length, while a unit vector keeps its direction and reduces its length to one.

For a two-dimensional vector, perpendicular components form a right-angled triangle. Pythagoras gives ∣a∣=a12+a22|\mathbf a|=\sqrt{a_1^2+a_2^2}. In three dimensions, first find the length of the projection in the xy-plane, then combine it with the perpendicular z-component. This gives one more squared component.

Magnitude and unit direction

For a=(a1a2a3)\mathbf a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix},

∣a∣=a12+a22+a32,a^=a∣a∣(a≠0).|\mathbf a|=\sqrt{a_1^2+a_2^2+a_3^2}, \qquad \widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}\quad(\mathbf a\ne\mathbf0).

The hat denotes a unit vector in the direction of a\mathbf a. These results are not printed in the P1–P4 formula booklet: learn them.

The zero vector has magnitude zero and no direction. Dividing it by its magnitude is undefined, so it has no unit vector in its direction.

Worked example 3: normalising a vector

Question. Find the magnitude of a=2i+3j+6k\mathbf a=2\mathbf i+3\mathbf j+6\mathbf k, then a unit vector in its direction.

Plan. Use Pythagoras for the length, then divide every component by that length.

∣a∣=22+32+62=7,a^=17(236).|\mathbf a|=\sqrt{2^2+3^2+6^2}=7, \qquad \widehat{\mathbf a}=\frac17\begin{pmatrix}2\\3\\6\end{pmatrix}.

Answer. Magnitude 77; unit vector 27i+37j+67k\frac27\mathbf i+\frac37\mathbf j+\frac67\mathbf k.

Check. Its squared magnitude is (4+9+36)/49=1(4+9+36)/49=1; multiplying it by 77 recovers a\mathbf a.

Change only the direction to its reverse. Squaring the components preserves the length.

Worked example 4: the opposite direction

Question. Find the magnitude of b=−2i−3j−6k\mathbf b=-2\mathbf i-3\mathbf j-6\mathbf k, then a unit vector in its direction.

Plan. Calculate the positive magnitude, retaining the signs in the unit vector.

∣b∣=(−2)2+(−3)2+(−6)2=7,b^=17(−2−3−6).|\mathbf b|=\sqrt{(-2)^2+(-3)^2+(-6)^2}=7, \qquad \widehat{\mathbf b}=\frac17\begin{pmatrix}-2\\-3\\-6\end{pmatrix}.

Answer. Magnitude 77; the unit vector is the negative of that in Example 3.

Check. Its squared magnitude is again 11, and its negative components point in the same direction as b\mathbf b.

A magnitude cannot become negative when an arrow is reversed.

Watch out

Divide the vector by its magnitude, not by the sum of its components. Also, a12+a22+a32\sqrt{a_1^2+a_2^2+a_3^2} is one square root around the whole sum.

Keep the signs in the vector and squares in the magnitude calculation.

Continue this lesson.

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