Pure 10.2 - Magnitude, direction and component form

Pure 10.2 - Magnitude, direction and component form

Components tell you how to move along the axes; magnitude and direction tell you the length and orientation of the same movement. You will convert between these descriptions, choose the correct quadrant for an angle, and use a unit vector to build a vector of a required length. The methods work in space as well as on a flat diagram.

Components Mean Movement

Does moving three units right and four units up put you seven units from where you started? The route along the axes has length seven, but the direct arrow is the hypotenuse of a right-angled triangle. Its length is five.

You already know

Pythagoras gives c2=a2+b2c^2=a^2+b^2 in a right-angled triangle. Vector components are signed changes along perpendicular coordinate axes.

The vector 3i+4j3\mathbf i+4\mathbf j records both changes. Its magnitude records just the direct length.

[DIAGRAM: asset_slug: p1_10_2_calculate_magnitude_and_direction_and_convert_between_component_and_magnitude_direction_forms__diagram_01; asset_name: Components and the direct vector length; caption: The component route has length seven but the vector has magnitude five; description: A right-angled component triangle from the origin to (3, 4), with horizontal component 3i, vertical component 4j, diagonal magnitude 5 and an angle measured anticlockwise from the positive x-axis]
The component route has length seven but the vector has magnitude five

A magnitude is always non-negative, even when some components are negative.

Magnitude

The magnitude ∣a∣\lvert\mathbf a\rvert of a vector a\mathbf a is its length. It is a scalar. Only the zero vector has magnitude zero; its direction is undefined.

Pythagoras gives the two-dimensional result. In three dimensions, first find the length of the projection in the xyxy-plane, then use Pythagoras again with the perpendicular zz-component. Thus the three squared components add.

Magnitude from components

For real components relative to perpendicular unit axes,

∣xi+yj∣=x2+y2\lvert x\mathbf i+y\mathbf j\rvert=\sqrt{x^2+y^2} ∣xi+yj+zk∣=x2+y2+z2\lvert x\mathbf i+y\mathbf j+z\mathbf k\rvert=\sqrt{x^2+y^2+z^2}

Not in the formula booklet: learn it.

There is a neat spatial version of the opening puzzle: three perpendicular one-unit steps have total route length three, but their resultant has magnitude 1+1+1=3\sqrt{1+1+1}=\sqrt3. Even with every component positive, adding component lengths is not the magnitude rule.

Key idea

Square the signed components, add the squares, then take the non-negative square root.

Test that distinction on a new triangle.

From Components to Magnitude and Direction

To find a direction, decide which quadrant the signed components describe before using inverse trigonometry. Here we report anticlockwise directions in 0∘≤θ<360∘0^\circ\le\theta<360^\circ, unless a question specifies another convention. For instance, −6i+8j-6\mathbf i+8\mathbf j points left and up, so its anticlockwise direction from positive xx must lie between 90∘90^\circ and 180∘180^\circ.

Direction from components

For a nonzero planar vector xi+yjx\mathbf i+y\mathbf j, an angle θ\theta from the positive xx-axis satisfies

tan⁡θ=yx(x≠0)\tan\theta=\frac{y}{x}\qquad(x\ne0)

Use the signs of both components to choose the quadrant. When x=0x=0, read the vertical direction directly.

Not in the formula booklet: learn it.

A reference angle from tan⁡−1(∣y/x∣)\tan^{-1}(|y/x|) is acute. The required direction may not be acute.

Worked example 1: a quadrant-II direction

Question. Find the magnitude and direction of a=−6i+8j\mathbf a=-6\mathbf i+8\mathbf j. Measure direction anticlockwise from positive xx, to one decimal place.

Plan. Use Pythagoras for length and subtract the acute reference angle from 180∘180^\circ.

∣a∣=(−6)2+82=10sum of squaresα=tan⁡−1 ⁣(86)=53.130…∘acute reference angleθ=180∘−α=126.869…∘quadrant II\begin{aligned} \lvert\mathbf a\rvert&=\sqrt{(-6)^2+8^2}=10 &&\text{sum of squares}\\ \alpha&=\tan^{-1}\!\left(\frac86\right)=53.130\ldots^\circ &&\text{acute reference angle}\\ \theta&=180^\circ-\alpha=126.869\ldots^\circ &&\text{quadrant II} \end{aligned}

Why this step? A negative horizontal component and positive vertical component put the arrow in quadrant II, so 53.1∘53.1^\circ itself cannot be the answer.

Answer. Magnitude 1010, direction 126.9∘126.9^\circ.

Check. 10cos⁡θ=−610\cos\theta=-6 and 10sin⁡θ=810\sin\theta=8, using the unrounded angle.

Change only the vertical sign in the second example. The magnitude is unchanged, but the direction moves to quadrant III.

Worked example 2: a quadrant-III direction

Question. Find the magnitude and anticlockwise direction from positive xx of b=−6i−8j\mathbf b=-6\mathbf i-8\mathbf j, to one decimal place.

Plan. Keep the same reference triangle and add its acute angle to 180∘180^\circ.

∣b∣=(−6)2+(−8)2=10signs disappear on squaringα=tan⁡−1 ⁣(86)=53.130…∘same reference angleθ=180∘+α=233.130…∘quadrant III\begin{aligned} \lvert\mathbf b\rvert&=\sqrt{(-6)^2+(-8)^2}=10 &&\text{signs disappear on squaring}\\ \alpha&=\tan^{-1}\!\left(\frac86\right)=53.130\ldots^\circ &&\text{same reference angle}\\ \theta&=180^\circ+\alpha=233.130\ldots^\circ &&\text{quadrant III} \end{aligned}

Why this step? Both components are negative, so the arrow points left and down.

Answer. Magnitude 1010, direction 233.1∘233.1^\circ.

Check. Both the cosine and sine of 233.130…∘233.130\ldots^\circ are negative, giving the required component signs.

On an axis there is no acute reference triangle. A nonzero upward vector has direction 90∘90^\circ, a downward one 270∘270^\circ, a leftward one 180∘180^\circ, and a rightward one 0∘0^\circ. A quadrant-IV vector such as 5i−12j5\mathbf i-12\mathbf j has direction 360∘−tan⁡−1(12/5)=292.6∘360^\circ-\tan^{-1}(12/5)=292.6^\circ, to one decimal place.

Watch out

Wrong: “tan⁡−1(8/(−6))=−53.1∘\tan^{-1}(8/(-6))=-53.1^\circ, so −6i+8j-6\mathbf i+8\mathbf j points at −53.1∘-53.1^\circ.” That angle points right and down. Right: use the quadrant to obtain 126.9∘126.9^\circ.

In this check both components are negative again, but the triangle changes.

From Magnitude and Direction to Components

Now start with the arrow's length and direction. A vector of length ten at 120∘120^\circ has negative horizontal component 10cos⁡120∘=−510\cos120^\circ=-5 and positive vertical component 10sin⁡120∘=5310\sin120^\circ=5\sqrt3.

[DIAGRAM: asset_slug: p1_10_2_calculate_magnitude_and_direction_and_convert_between_component_and_magnitude_direction_forms__diagram_02; asset_name: Resolving a known magnitude and direction; caption: Cosine and sine supply the component sizes and their signs; description: A general vector with components R cos theta and R sin theta, alongside a quadrant-II vector of magnitude 10 at 120 degrees with components minus 5 and 5 root 3]
Cosine and sine supply the component sizes and their signs

The choice of sine or cosine depends on the axis from which the angle is measured.

Components from a direction angle

If R>0R>0 and θ\theta is measured anticlockwise from positive xx,

a=Rcos⁡θ i+Rsin⁡θ j\mathbf a=R\cos\theta\,\mathbf i+R\sin\theta\,\mathbf j

If instead an acute angle β\beta is measured east of north, the east component is Rsin⁡βR\sin\beta and the north component is Rcos⁡βR\cos\beta.

Not in the formula booklet: learn it.

Identify the reference axis before substituting. A bearing is measured clockwise from north, so east/north components of a bearing β\beta are also Rsin⁡βR\sin\beta and Rcos⁡βR\cos\beta, with their signs supplied by the angle.

Continue this lesson.

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