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Physics / Chapter 3: Vector / Section 3-1

Vectors and Scalars — Question 5

Section 3-1 · Physics pages 320–321

Question 5 A direction angle needs a reference and a turning sense A map fixes north upward and east to the right. Bearings are measured clockwise from north and reduced to 0◦≤β < 360◦. The following records describe displacement vectors of magnitude 8.0 m: P is 30◦north of east; Q has bearing 60◦; R is 30◦east of north. S lists angle 30◦but omits its reference and turning sense. x y P, Q: 8 m R: 8 m 30◦ S: 8 m, 30◦ reference omitted East is positive x and north positive y. Bearing is measured clockwise from north. (a) Which of P, Q and R are equal vectors? Give the bearing of each, using angle geometry only. (b) Show why S cannot be uniquely reconstructed by giving two distinct directions compatible with its bare angle field. (c) Give the bearing and a cardinal-direction phrase for a vector of the same magnitude opposite to P. Does “opposite” change the magnitude’s sign? (d) A compass arrow of length 1 cm on a map and a displacement arrow of length 1 cm need not represent the same quantity. Explain the distinct roles of a direction reference, an angle and a displacement vector.
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Question 5 - Solution A direction angle needs a reference and a turning sense x y P, Q: 8 m R: 8 m 30◦ S: 8 m, 30◦ reference omitted (a) Express all directions in one convention. βP = 90◦−30◦= 60◦= βQ, βR = 30◦,⃗ P =⃗Q̸ =⃗R Convert to the same bearing convention before comparing. Equal magnitude and equal bearing make P and Q equal; R has a different bearing. (b) An angle without its reference does not identify a direction. θS = 30◦from east or θS = 90◦−30◦= 60◦from east These two interpretations of S are distinct. The missing reference axis or turning sense prevents unique reconstruction even though its 8 m magnitude is known. (c) Reverse the direction, preserving the nonnegative magnitude. β−P = (60◦+ 180◦) mod 360◦= 240◦, | −⃗P| = 8 m The opposite direction is 30◦south of west. Reversal changes direction by half a turn, leaving the nonnegative magnitude unchanged. (d) A drawing symbol is not automatically a measured vector. d = kℓpaper, [k] = m/cm; a compass symbol supplies orientation only A compass arrow establishes a reference direction; its printed length can be arbitrary. An angle describes orientation relative to a convention. A displacement additionally requires a calibrated physical length.

Question 5

A direction angle needs a reference and a turning sense

A map fixes north upward and east to the right. Bearings are measured clockwise from north and reduced to 0∘≤β<360∘0^\circ\leq\beta<360^\circ. The following records describe displacement vectors of magnitude 8.0 m: P is 30∘30^\circ north of east; Q has bearing 60∘60^\circ; R is 30∘30^\circ east of north. S lists angle 30∘30^\circ but omits its reference and turning sense.

Diagram shown in the original SVG worksheet.

  1. Which of P, Q and R are equal vectors? Give the bearing of each, using angle geometry only.

  2. Show why S cannot be uniquely reconstructed by giving two distinct directions compatible with its bare angle field.

  3. Give the bearing and a cardinal-direction phrase for a vector of the same magnitude opposite to P. Does “opposite” change the magnitude’s sign?

  4. A compass arrow of length 1 cm on a map and a displacement arrow of length 1 cm need not represent the same quantity. Explain the distinct roles of a direction reference, an angle and a displacement vector.

Question 5 - Solution

A direction angle needs a reference and a turning sense

Diagram shown in the original SVG worksheet.

(a) Express all directions in one convention.

βP=90∘−30∘=60∘=βQ,βR=30∘,P→=Q→≠R→\beta_P=90^\circ-30^\circ=60^\circ=\beta_Q,\quad\beta_R=30^\circ,\quad\boxed{\vec P=\vec Q\ne\vec R} Convert to the same bearing convention before comparing. Equal magnitude and equal bearing make P and Q equal; R has a different bearing.

(b) An angle without its reference does not identify a direction.

θS=30∘ from eastorθS=90∘−30∘=60∘ from east\theta_S=30^\circ\text{ from east}\quad\text{or}\quad\theta_S=90^\circ-30^\circ=60^\circ\text{ from east} These two interpretations of S are distinct. The missing reference axis or turning sense prevents unique reconstruction even though its 8 m magnitude is known.

(c) Reverse the direction, preserving the nonnegative magnitude.

β−P=(60∘+180∘)mod⁡360∘=240∘,|−P→|=8m\beta_{-P}=(60^\circ+180^\circ)\bmod360^\circ=240^\circ,\quad|-\vec P|=8\,\mathrm m The opposite direction is 30∘30^\circ south of west. Reversal changes direction by half a turn, leaving the nonnegative magnitude unchanged.

(d) A drawing symbol is not automatically a measured vector.

d=kℓpaper,[k]=m/cm;a compass symbol supplies orientation onlyd=k\ell_{\mathrm{paper}},\qquad[k]=\mathrm{m/cm};\quad\text{a compass symbol supplies orientation only} A compass arrow establishes a reference direction; its printed length can be arbitrary. An angle describes orientation relative to a convention. A displacement additionally requires a calibrated physical length.

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