Vectors

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1. Vectors and their components

Discover

You walk from P=(1,2)P = (1, 2) to Q=(4,6)Q = (4, 6). How far do you go to the right, and how far up?

  1. 3 to the right and 4 up
  2. 3 to the left and 4 down

Key idea

A vector has a length and a direction, but no fixed place. We write it by its components.

From P=(x1,y1)P = (x_1, y_1) to Q=(x2,y2)Q = (x_2, y_2): PQ→=⟨x2−x1, y2−y1⟩\overrightarrow{PQ} = \langle x_2 - x_1,\ y_2 - y_1 \rangle (end minus start). Two vectors with the same components are equal, wherever they start.

Worked example

Find PQ→\overrightarrow{PQ} for P=(5,−1)P = (5, -1) and Q=(2,3)Q = (2, 3).

2−5=−32 - 5 = -3
End minus start, horizontally.
3−(−1)=43 - (-1) = 4
End minus start, vertically.
PQ→=⟨−3,4⟩\overrightarrow{PQ} = \langle -3, 4 \rangle
Write the components.

Your turn, guided

Your turn: P=(2,7)P = (2, 7) and Q=(6,1)Q = (6, 1).

Horizontal: 44
6 − 2.
Vertical: −6-6, so PQ→=⟨4,−6⟩\overrightarrow{PQ} = \langle 4, -6 \rangle
1 − 7.

Practice ★

  1. P=(2,4)P = (2, 4) and Q=(4,−6)Q = (4, -6). Find the components of PQ→\overrightarrow{PQ}.
  2. P=(5,1)P = (5, 1) and Q=(7,−2)Q = (7, -2). Find the components of PQ→\overrightarrow{PQ}.
  3. P=(2,3)P = (2, 3) and Q=(9,−8)Q = (9, -8). Find the components of PQ→\overrightarrow{PQ}.

Watch out!

Practice ★★

  1. The vector ⟨2,1⟩\langle 2, 1 \rangle starts at the point (1,1)(1, 1). Tap the point where it ends.
  2. The vector ⟨2,3⟩\langle 2, 3 \rangle starts at the point (1,−3)(1, -3). Tap the point where it ends.
  3. The vector ⟨3,1⟩\langle 3, 1 \rangle starts at the point (−2,−1)(-2, -1). Tap the point where it ends.
  1. P=(3,3)P = (3, 3) and Q=(8,8)Q = (8, 8). Which vector is PQ→\overrightarrow{PQ}?
    1. ⟨5,5⟩\langle 5, 5 \rangle
    2. ⟨−5,−5⟩\langle -5, -5 \rangle
    3. ⟨11,11⟩\langle 11, 11 \rangle
  2. P=(2,2)P = (2, 2) and Q=(8,6)Q = (8, 6). Which vector is PQ→\overrightarrow{PQ}?
    1. ⟨6,4⟩\langle 6, 4 \rangle
    2. ⟨−6,−4⟩\langle -6, -4 \rangle
    3. ⟨10,8⟩\langle 10, 8 \rangle

Exit ticket

Three questions to check you have got it.

  1. P=(2,4)P = (2, 4) and Q=(4,−6)Q = (4, -6). Find the components of PQ→\overrightarrow{PQ}.
  2. P=(5,1)P = (5, 1) and Q=(7,−2)Q = (7, -2). Find the components of PQ→\overrightarrow{PQ}.
  3. P=(2,3)P = (2, 3) and Q=(9,−8)Q = (9, -8). Find the components of PQ→\overrightarrow{PQ}.

Summary card

Vectors and components

  • PQ→=⟨x2−x1, y2−y1⟩\overrightarrow{PQ} = \langle x_2 - x_1,\ y_2 - y_1 \rangle: end minus start.
  • Same components, same vector, wherever it starts.

Watch out: start minus end gives the opposite vector.

2. Magnitude and direction of a vector

Warm-up

Three quick questions from earlier lessons.

Discover

How long is the vector ⟨3,4⟩\langle 3, 4 \rangle?

  1. 5
  2. 7

Key idea

A vector's magnitude is its length; its direction angle is measured counterclockwise from the positive xx-axis.

∣⟨a,b⟩∣=a2+b2|\langle a, b \rangle| = \sqrt{a^2 + b^2}. For the direction angle θ\theta: tan⁡θ=ba\tan\theta = \dfrac{b}{a}. The calculator's tan⁡−1\tan^{-1} gives an angle between −90∘-90^\circ and 90∘90^\circ; if the vector points left (a<0a < 0), add 180∘180^\circ.

Worked example

Find the magnitude and direction of ⟨−2,2⟩\langle -2, 2 \rangle.

∣v⃗∣=4+4=22|\vec{v}| = \sqrt{4 + 4} = 2\sqrt{2}
Pythagoras.
tan⁡−1(−1)=−45∘\tan^{-1}(-1) = -45^\circ
tan θ = 2 / (−2) = −1; the calculator gives −45°.
θ=135∘\theta = 135^\circ
The vector points left and up (quadrant II): add 180°.

Your turn, guided

Your turn: ⟨5,−12⟩\langle 5, -12 \rangle.

∣v⃗∣=13|\vec{v}| = 13
√(25 + 144).
θ≈292.6∘\theta \approx 292.6^\circ
Right and down: quadrant IV, so tan⁻¹(−12/5) ≈ −67.4°, or 292.6°.

Practice ★

  1. Find the exact magnitude of v⃗=⟨6,−5⟩\vec{v} = \langle 6, -5 \rangle.
  2. Find the exact magnitude of v⃗=⟨7,−7⟩\vec{v} = \langle 7, -7 \rangle.
  3. Find the exact magnitude of v⃗=⟨3,−7⟩\vec{v} = \langle 3, -7 \rangle.
  4. Find the exact magnitude of v⃗=⟨8,−2⟩\vec{v} = \langle 8, -2 \rangle.

Watch out!

Practice ★★

  1. Find the direction angle of ⟨2,3⟩\langle 2, 3 \rangle, in degrees, to 1 decimal place.
  2. Find the direction angle of ⟨8,9⟩\langle 8, 9 \rangle, in degrees, to 1 decimal place.
  1. Find the direction angle of ⟨−2,3⟩\langle -2, 3 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.
  2. Find the direction angle of ⟨−1,7⟩\langle -1, 7 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.

Exit ticket

Three questions to check you have got it.

  1. Find the direction angle of ⟨−2,3⟩\langle -2, 3 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.
  2. Find the direction angle of ⟨−1,7⟩\langle -1, 7 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.
  3. Find the direction angle of ⟨−8,9⟩\langle -8, 9 \rangle, between 0∘0^\circ and 360∘360^\circ, to 1 decimal place.

Summary card

Magnitude and direction

  • ∣⟨a,b⟩∣=a2+b2|\langle a, b \rangle| = \sqrt{a^2 + b^2}.
  • tan⁡θ=ba\tan\theta = \dfrac{b}{a}; sketch the vector to pick the quadrant.
  • Pointing left: add 180∘180^\circ to the calculator's angle.

Watch out: the magnitude is not a+ba + b.

3. Add, subtract and scale vectors

Warm-up

Three quick questions from earlier lessons.

Discover

You walk ⟨3,1⟩\langle 3, 1 \rangle, then ⟨1,2⟩\langle 1, 2 \rangle. Where are you, compared with your start?

  1. ⟨4,3⟩\langle 4, 3 \rangle
  2. ⟨2,−1⟩\langle 2, -1 \rangle

Key idea

Vectors combine component by component.

⟨a,b⟩+⟨c,d⟩=⟨a+c, b+d⟩\langle a, b \rangle + \langle c, d \rangle = \langle a + c,\ b + d \rangle;  ⟨a,b⟩−⟨c,d⟩=⟨a−c, b−d⟩\ \langle a, b \rangle - \langle c, d \rangle = \langle a - c,\ b - d \rangle;  k⟨a,b⟩=⟨ka, kb⟩\ k\langle a, b \rangle = \langle ka,\ kb \rangle. Geometrically, u⃗+v⃗\vec{u} + \vec{v} goes tip to tail. In general ∣u⃗+v⃗∣≠∣u⃗∣+∣v⃗∣|\vec{u} + \vec{v}| \ne |\vec{u}| + |\vec{v}|.

Worked example

u⃗=⟨2,−1⟩\vec{u} = \langle 2, -1 \rangle and v⃗=⟨−3,4⟩\vec{v} = \langle -3, 4 \rangle. Find 3u⃗+v⃗3\vec{u} + \vec{v}.

3u⃗=⟨6,−3⟩3\vec{u} = \langle 6, -3 \rangle
Scale u.
3u⃗+v⃗=⟨6−3, −3+4⟩=⟨3,1⟩3\vec{u} + \vec{v} = \langle 6 - 3,\ -3 + 4 \rangle = \langle 3, 1 \rangle
Add component by component.

Your turn, guided

Your turn: u⃗=⟨4,1⟩\vec{u} = \langle 4, 1 \rangle, v⃗=⟨1,−2⟩\vec{v} = \langle 1, -2 \rangle. Find u⃗−2v⃗\vec{u} - 2\vec{v}.

Horizontal: 4−2=24 - 2 = 2
2v = ⟨2, −4⟩.
Vertical: 55, so u⃗−2v⃗=⟨2,5⟩\vec{u} - 2\vec{v} = \langle 2, 5 \rangle
1 − (−4).

Practice ★

  1. u⃗=⟨4,6⟩\vec{u} = \langle 4, 6 \rangle and v⃗=⟨1,−4⟩\vec{v} = \langle 1, -4 \rangle. Find 4u⃗−v⃗ 4\vec{u} - \vec{v}.
  2. u⃗=⟨3,1⟩\vec{u} = \langle 3, 1 \rangle and v⃗=⟨4,−6⟩\vec{v} = \langle 4, -6 \rangle. Find 4u⃗−v⃗ 4\vec{u} - \vec{v}.
  3. u⃗=⟨5,2⟩\vec{u} = \langle 5, 2 \rangle and v⃗=⟨3,−5⟩\vec{v} = \langle 3, -5 \rangle. Find 2u⃗−v⃗ 2\vec{u} - \vec{v}.

Watch out!

Practice ★★

  1. u⃗=⟨4,1⟩\vec{u} = \langle 4, 1 \rangle and v⃗=⟨5,6⟩\vec{v} = \langle 5, 6 \rangle. Find the exact value of ∣u⃗+v⃗∣|\vec{u} + \vec{v}|.
  2. u⃗=⟨5,5⟩\vec{u} = \langle 5, 5 \rangle and v⃗=⟨5,3⟩\vec{v} = \langle 5, 3 \rangle. Find the exact value of ∣u⃗+v⃗∣|\vec{u} + \vec{v}|.
  3. u⃗=⟨1,4⟩\vec{u} = \langle 1, 4 \rangle and v⃗=⟨2,2⟩\vec{v} = \langle 2, 2 \rangle. Find the exact value of ∣u⃗+v⃗∣|\vec{u} + \vec{v}|.
  1. u⃗=⟨2,3⟩\vec{u} = \langle 2, 3 \rangle and v⃗=⟨−3,1⟩\vec{v} = \langle -3, 1 \rangle, both starting at the origin. Tap the end point of u⃗+v⃗\vec{u} + \vec{v}.
  2. u⃗=⟨4,1⟩\vec{u} = \langle 4, 1 \rangle and v⃗=⟨−2,2⟩\vec{v} = \langle -2, 2 \rangle, both starting at the origin. Tap the end point of u⃗+v⃗\vec{u} + \vec{v}.

Exit ticket

Three questions to check you have got it.

  1. u⃗=⟨4,1⟩\vec{u} = \langle 4, 1 \rangle and v⃗=⟨5,6⟩\vec{v} = \langle 5, 6 \rangle. Find the exact value of ∣u⃗+v⃗∣|\vec{u} + \vec{v}|.
  2. u⃗=⟨5,5⟩\vec{u} = \langle 5, 5 \rangle and v⃗=⟨5,3⟩\vec{v} = \langle 5, 3 \rangle. Find the exact value of ∣u⃗+v⃗∣|\vec{u} + \vec{v}|.
  3. u⃗=⟨1,4⟩\vec{u} = \langle 1, 4 \rangle and v⃗=⟨2,2⟩\vec{v} = \langle 2, 2 \rangle. Find the exact value of ∣u⃗+v⃗∣|\vec{u} + \vec{v}|.

Summary card

Vector operations

  • Add, subtract, scale: component by component.
  • u⃗+v⃗\vec{u} + \vec{v}: tip to tail.
  • ∣u⃗+v⃗∣|\vec{u} + \vec{v}|: add first, then take the length.

Watch out: ∣u⃗+v⃗∣≠∣u⃗∣+∣v⃗∣|\vec{u} + \vec{v}| \ne |\vec{u}| + |\vec{v}| unless they point the same way.

4. Components from magnitude and direction; unit vectors

Warm-up

Three quick questions from earlier lessons.

Discover

A vector has length 10 and direction 30∘30^\circ. Its horizontal component is the side next to the 30∘30^\circ angle. Which is it?

  1. 10cos⁡30∘10\cos 30^\circ
  2. 10sin⁡30∘10\sin 30^\circ

Key idea

Magnitude and direction give the components, and a unit vector (length 1) gives just the direction.

A vector with magnitude rr and direction θ\theta is ⟨rcos⁡θ, rsin⁡θ⟩\langle r\cos\theta,\ r\sin\theta \rangle. The unit vector in the direction of v⃗\vec{v} is v⃗∣v⃗∣\dfrac{\vec{v}}{|\vec{v}|}.

Worked example

Write the vector with magnitude 6 and direction 120∘120^\circ.

6cos⁡120∘=−36\cos 120^\circ = -3
cos 120° = −1/2.
6sin⁡120∘=336\sin 120^\circ = 3\sqrt{3}
sin 120° = √3/2.
v⃗=⟨−3,33⟩\vec{v} = \langle -3, 3\sqrt{3} \rangle
Components.

Your turn, guided

Your turn: the unit vector in the direction of ⟨6,8⟩\langle 6, 8 \rangle.

∣v⃗∣=10|\vec{v}| = 10
√(36 + 64).
v⃗∣v⃗∣=⟨35,45⟩\dfrac{\vec{v}}{|\vec{v}|} = \left\langle \dfrac{3}{5}, \dfrac{4}{5} \right\rangle
6 ÷ 10: type the horizontal component.

Practice ★

  1. A vector has magnitude 12 and direction angle 120∘120^\circ. Find its exact components.Not to scale
  2. A vector has magnitude 3 and direction angle 120∘120^\circ. Find its exact components.Not to scale
  3. A vector has magnitude 3 and direction angle 60∘60^\circ. Find its exact components.Not to scale

Watch out!

Practice ★★

  1. Find the horizontal component of the unit vector in the direction of ⟨3,4⟩\langle 3, 4 \rangle.
  2. Find the horizontal component of the unit vector in the direction of ⟨15,8⟩\langle 15, 8 \rangle.
  1. A ball is kicked at 51 meters per second, at 35∘35^\circ above the ground. Find the horizontal component of its velocity, in meters per second, to 1 decimal place.Not to scale
  2. A ball is kicked at 40 meters per second, at 40∘40^\circ above the ground. Find the horizontal component of its velocity, in meters per second, to 1 decimal place.Not to scale
  3. A ball is kicked at 40 meters per second, at 70∘70^\circ above the ground. Find the horizontal component of its velocity, in meters per second, to 1 decimal place.Not to scale

Exit ticket

Three questions to check you have got it.

  1. A ball is kicked at 51 meters per second, at 35∘35^\circ above the ground. Find the horizontal component of its velocity, in meters per second, to 1 decimal place.Not to scale
  2. A ball is kicked at 40 meters per second, at 40∘40^\circ above the ground. Find the horizontal component of its velocity, in meters per second, to 1 decimal place.Not to scale
  3. A ball is kicked at 40 meters per second, at 70∘70^\circ above the ground. Find the horizontal component of its velocity, in meters per second, to 1 decimal place.Not to scale

Summary card

From magnitude and direction

  • ⟨rcos⁡θ, rsin⁡θ⟩\langle r\cos\theta,\ r\sin\theta \rangle.
  • Unit vector: v⃗∣v⃗∣\dfrac{\vec{v}}{|\vec{v}|}, length 1.

Watch out: horizontal goes with cosine, vertical with sine.

5. The dot product and the angle between vectors

Warm-up

Three quick questions from earlier lessons.

Discover

⟨2,1⟩\langle 2, 1 \rangle and ⟨−1,2⟩\langle -1, 2 \rangle look perpendicular. Multiply matching components and add: 2⋅(−1)+1⋅22 \cdot (-1) + 1 \cdot 2. What do you get?

  1. 00
  2. ⟨−2,2⟩\langle -2, 2 \rangle

Key idea

The dot product of two vectors is a number.

⟨a,b⟩⋅⟨c,d⟩=ac+bd\langle a, b \rangle \cdot \langle c, d \rangle = ac + bd. Also u⃗⋅v⃗=∣u⃗∣ ∣v⃗∣cos⁡θ\vec{u} \cdot \vec{v} = |\vec{u}|\,|\vec{v}|\cos\theta, where θ\theta is the angle between them, so cos⁡θ=u⃗⋅v⃗∣u⃗∣ ∣v⃗∣\cos\theta = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}. Two non-zero vectors are perpendicular exactly when u⃗⋅v⃗=0\vec{u} \cdot \vec{v} = 0.

Worked example

Find the angle between u⃗=⟨1,2⟩\vec{u} = \langle 1, 2 \rangle and v⃗=⟨3,1⟩\vec{v} = \langle 3, 1 \rangle.

u⃗⋅v⃗=3+2=5\vec{u} \cdot \vec{v} = 3 + 2 = 5
Multiply and add.
∣u⃗∣ ∣v⃗∣=5 10=52|\vec{u}|\,|\vec{v}| = \sqrt{5}\,\sqrt{10} = 5\sqrt{2}
√5 and √10.
θ=45∘\theta = 45^\circ
cos θ = 5 / (5√2) = √2/2.

Your turn, guided

Your turn: ⟨4,−1⟩⋅⟨2,3⟩\langle 4, -1 \rangle \cdot \langle 2, 3 \rangle.

First products: 88
4 · 2.
⟨4,−1⟩⋅⟨2,3⟩=5\langle 4, -1 \rangle \cdot \langle 2, 3 \rangle = 5
8 + (−1)(3).

Practice ★

  1. Find ⟨5,2⟩⋅⟨8,−9⟩\langle 5, 2 \rangle \cdot \langle 8, -9 \rangle.
  2. Find ⟨9,8⟩⋅⟨8,−2⟩\langle 9, 8 \rangle \cdot \langle 8, -2 \rangle.
  3. Find ⟨3,4⟩⋅⟨7,−7⟩\langle 3, 4 \rangle \cdot \langle 7, -7 \rangle.
  4. Find ⟨7,5⟩⋅⟨5,−1⟩\langle 7, 5 \rangle \cdot \langle 5, -1 \rangle.

Watch out!

Practice ★★

  1. Are ⟨6,4⟩\langle 6, 4 \rangle and ⟨−8,12⟩\langle -8, 12 \rangle perpendicular?
    1. Yes: their dot product is 0
    2. No: neither of them is the zero vector
    3. No: their components do not add up to 0
  2. Are ⟨2,1⟩\langle 2, 1 \rangle and ⟨−2,4⟩\langle -2, 4 \rangle perpendicular?
    1. Yes: their dot product is 0
    2. No: neither of them is the zero vector
    3. No: their components do not add up to 0
  1. Find the angle between ⟨4,4⟩\langle 4, 4 \rangle and ⟨4,−3⟩\langle 4, -3 \rangle, in degrees, to 1 decimal place.
  2. Find the angle between ⟨3,4⟩\langle 3, 4 \rangle and ⟨5,−4⟩\langle 5, -4 \rangle, in degrees, to 1 decimal place.

Exit ticket

Three questions to check you have got it.

  1. Find ⟨5,2⟩⋅⟨8,−9⟩\langle 5, 2 \rangle \cdot \langle 8, -9 \rangle.
  2. Find ⟨9,8⟩⋅⟨8,−2⟩\langle 9, 8 \rangle \cdot \langle 8, -2 \rangle.
  3. Find ⟨3,4⟩⋅⟨7,−7⟩\langle 3, 4 \rangle \cdot \langle 7, -7 \rangle.

Summary card

The dot product

  • ⟨a,b⟩⋅⟨c,d⟩=ac+bd\langle a, b \rangle \cdot \langle c, d \rangle = ac + bd, a number.
  • cos⁡θ=u⃗⋅v⃗∣u⃗∣ ∣v⃗∣\cos\theta = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}.
  • u⃗⋅v⃗=0\vec{u} \cdot \vec{v} = 0: perpendicular.

Watch out: multiply matching components, not crossed ones.

6. Vectors in context: velocity and force

Warm-up

Three quick questions from earlier lessons.

Discover

A swimmer swims north at 3 km/h across a river flowing east at 4 km/h. How fast does she move over the ground?

  1. 5 km/h
  2. 7 km/h

Key idea

Velocities and forces are vectors: when several act together, add them.

The resultant of several vectors is their sum. Its magnitude is the speed (or total force), and its direction angle is where the object actually goes. Write each vector in components first.

Worked example

Forces of 30 N due east and 40 N due north pull on a ring. Find the resultant.

F⃗=⟨30,40⟩\vec{F} = \langle 30, 40 \rangle
Components.
∣F⃗∣=50|\vec{F}| = 50 N
√(900 + 1600).
θ=tan⁡−143≈53.1∘\theta = \tan^{-1}\dfrac{4}{3} \approx 53.1^\circ north of east
tan θ = 40/30.

Your turn, guided

Your turn: a plane flies east at 200 mph with a wind from the south at 50 mph.

Ground velocity ⟨200,50⟩\langle 200, 50 \rangle
East and north components.
Speed =42500≈206.2= \sqrt{42500} \approx 206.2 mph
√(40000 + 2500) ≈ 206.2: type the exact value.

Practice ★

  1. A plane flies due east at 101 miles per hour through the air, and the wind blows due north at 32 miles per hour. Find its speed over the ground, to 1 decimal place.Not to scale
  2. A plane flies due east at 236 miles per hour through the air, and the wind blows due north at 60 miles per hour. Find its speed over the ground, to 1 decimal place.Not to scale
  3. A plane flies due east at 174 miles per hour through the air, and the wind blows due north at 17 miles per hour. Find its speed over the ground, to 1 decimal place.Not to scale

Watch out!

Practice ★★

  1. Two people pull a box: one with 36 newtons due east, the other with 59 newtons due north. At what angle north of east does the box start to move? Give degrees to 1 decimal place.Not to scale
  2. Two people pull a box: one with 27 newtons due east, the other with 23 newtons due north. At what angle north of east does the box start to move? Give degrees to 1 decimal place.Not to scale
  3. Two people pull a box: one with 88 newtons due east, the other with 72 newtons due north. At what angle north of east does the box start to move? Give degrees to 1 decimal place.Not to scale

Exit ticket

Three questions to check you have got it.

  1. A plane flies due east at 101 miles per hour through the air, and the wind blows due north at 32 miles per hour. Find its speed over the ground, to 1 decimal place.Not to scale
  2. A plane flies due east at 236 miles per hour through the air, and the wind blows due north at 60 miles per hour. Find its speed over the ground, to 1 decimal place.Not to scale
  3. A plane flies due east at 174 miles per hour through the air, and the wind blows due north at 17 miles per hour. Find its speed over the ground, to 1 decimal place.Not to scale

Summary card

Vectors in context

  1. Write each velocity or force in components.
  2. Add them: the resultant.
  3. Magnitude = speed or total force; tan⁡θ\tan\theta = direction.

Watch out: speeds in different directions do not simply add.

Answers: Vectors

1. Vectors and their components

A

  1. −10-10
  2. −3-3
  3. −11-11

B

  1. (3,2)(3, 2)
  2. (3,0)(3, 0)
  3. (1,0)(1, 0)

C

  1. ⟨5,5⟩\langle 5, 5 \rangle
  2. ⟨6,4⟩\langle 6, 4 \rangle

D

  1. −10-10
  2. −3-3
  3. −11-11

2. Magnitude and direction of a vector

A

  1. 61\sqrt{61}
  2. 727 \sqrt{2}
  3. 58\sqrt{58}
  4. 2172 \sqrt{17}

B

  1. 180tan⁡−1(32)π\frac{180 \tan^{-1}{\left(\frac{3}{2} \right)}}{\pi}
  2. 180tan⁡−1(98)π\frac{180 \tan^{-1}{\left(\frac{9}{8} \right)}}{\pi}

C

  1. −180tan⁡−1(32)π+180- \frac{180 \tan^{-1}{\left(\frac{3}{2} \right)}}{\pi} + 180
  2. −180tan⁡−1(7)π+180- \frac{180 \tan^{-1}{\left(7 \right)}}{\pi} + 180

D

  1. −180tan⁡−1(32)π+180- \frac{180 \tan^{-1}{\left(\frac{3}{2} \right)}}{\pi} + 180
  2. −180tan⁡−1(7)π+180- \frac{180 \tan^{-1}{\left(7 \right)}}{\pi} + 180
  3. −180tan⁡−1(98)π+180- \frac{180 \tan^{-1}{\left(\frac{9}{8} \right)}}{\pi} + 180

3. Add, subtract and scale vectors

A

  1. 2828
  2. 1010
  3. 99

B

  1. 130\sqrt{130}
  2. 2412 \sqrt{41}
  3. 353 \sqrt{5}

C

  1. (−1,4)(-1, 4)
  2. (2,3)(2, 3)

D

  1. 130\sqrt{130}
  2. 2412 \sqrt{41}
  3. 353 \sqrt{5}

4. Components from magnitude and direction; unit vectors

A

  1. 636 \sqrt{3}
  2. 332\frac{3 \sqrt{3}}{2}
  3. 332\frac{3 \sqrt{3}}{2}

B

  1. 35\frac{3}{5}
  2. 1517\frac{15}{17}

C

  1. 51cos⁡(7π36)51 \cos{\left(\frac{7 \pi}{36} \right)}
  2. 40cos⁡(2π9)40 \cos{\left(\frac{2 \pi}{9} \right)}
  3. 40cos⁡(7π18)40 \cos{\left(\frac{7 \pi}{18} \right)}

D

  1. 51cos⁡(7π36)51 \cos{\left(\frac{7 \pi}{36} \right)}
  2. 40cos⁡(2π9)40 \cos{\left(\frac{2 \pi}{9} \right)}
  3. 40cos⁡(7π18)40 \cos{\left(\frac{7 \pi}{18} \right)}

5. The dot product and the angle between vectors

A

  1. 2222
  2. 5656
  3. −7-7
  4. 3030

B

  1. Yes: their dot product is 0
  2. Yes: their dot product is 0

C

  1. 180cos⁡−1(210)π\frac{180 \cos^{-1}{\left(\frac{\sqrt{2}}{10} \right)}}{\pi}
  2. 180cos⁡−1(−41205)π\frac{180 \cos^{-1}{\left(- \frac{\sqrt{41}}{205} \right)}}{\pi}

D

  1. 2222
  2. 5656
  3. −7-7

6. Vectors in context: velocity and force

A

  1. 54495 \sqrt{449}
  2. 437064 \sqrt{3706}
  3. 30565\sqrt{30565}

B

  1. 180tan⁡−1(5936)π\frac{180 \tan^{-1}{\left(\frac{59}{36} \right)}}{\pi}
  2. 180tan⁡−1(2327)π\frac{180 \tan^{-1}{\left(\frac{23}{27} \right)}}{\pi}
  3. 180tan⁡−1(911)π\frac{180 \tan^{-1}{\left(\frac{9}{11} \right)}}{\pi}

C

  1. 54495 \sqrt{449}
  2. 437064 \sqrt{3706}
  3. 30565\sqrt{30565}