Discussion Class 1
Questions
Q4) For normal scalar numbers we have that subtraction is not commutative(𝑖. 𝑒. 𝑥 − 𝑦 ≠ 𝑦 − 𝑥). Is
this also true for vectors? If yes then prove it. If No then give a counter example
Yes (it is not commutative). 𝑏⃗ − 𝑎 = −(𝑎 − 𝑏⃗) which is opposite in direction to 𝑎 − 𝑏⃗ by definition
of the negative and hence not the same vector.
Q6) Describe two vectors 𝑎 and 𝑏⃗ such that
a) 𝑎 + 𝑏⃗ = 𝑐 and 𝑎 + 𝑏 = 𝑐
𝑎 parallel to 𝑏⃗
b) 𝑎 + 𝑏⃗ = 𝑎 − 𝑏⃗
𝑎 can be any vector. 𝑏⃗ = 0
c) 𝑎 + 𝑏⃗ = 𝑐 and 𝑎2 + 𝑏 2 = 𝑐 2
𝑎 perpendicular to 𝑏⃗
Problems
P1) What are the x and y components of a vector 𝑎 in the xy plane if its direction is 250°
counterclockwise from the positive direction of the x axis and its magnitude is 7.3 m?
𝑎 = 7.3 cos 250° 𝑖̂ + 7.3 sin 250° 𝑗̂
= −2.5𝑖̂ − 6.9𝑗̂
P3) The x component of vector 𝐴 is -25.0 m and the y component is +40.0 m. Write this vector in
direction-magnitude notation.
𝐴 = −25.0𝑖̂ + 40.0𝑗̂
𝐴 = √252 + 402 = 47.17
𝜃 = tan−1 (
40
) = 122°
−25
𝐴 = 47.17 𝑚 𝑎𝑡 122° 𝑡𝑜 + 𝑥 𝑎𝑥𝑖𝑠
P5) A ship sets out to sail to appoint 120 km due north, but an unexpected storm blows the ship to a
point 100 km due east of its starting point. How far and in what direction must it now sail to reach its
original destination?
𝑟 = −100𝑖̂ + 120𝑗̂
𝑟 = √1002 + 1202 = 156.2
120
𝜃 = tan−1 (
) = 50.19° 𝑁 𝑜𝑓 𝑊
100
Discussion Class 2
Questions
Q8) If 𝑎 ∙ 𝑏⃗ = 𝑎 ∙ 𝑐 then must 𝑏⃗ = 𝑐? If yes then prove it. If No then give a counter example
No. Let 𝑎 = 1𝑖̂ + 1𝑗̂, 𝑏⃗ = 1𝑖̂ and 𝑐 = 1𝑗̂ then 𝑎 ∙ 𝑏⃗ = 1 and 𝑎 ∙ 𝑐 = 1 but 𝑏⃗ ≠ 𝑐
Problems
P10) Find the unit vector notation for the resultant vector 𝑟, which is the sum of the two vectors
𝑐 = 7.4𝑖̂ − 3.8𝑗̂ − 6.1𝑘̂ and 𝑑 = 4.4𝑖̂ − 2.0𝑗̂ + 3.3𝑘̂
𝑟 = (7.4 + 4.4)𝑖̂ + (−3.8 − 2.0)𝑗̂ + (−6.1 + 3.3)𝑘̂ = 11.8𝑖̂ − 5.8𝑗̂ − 2.8𝑘̂
P30) Consider the following two vectors:
𝑎 = (4.0 𝑚)𝑖̂ − (3.0 𝑚)𝑗̂ 𝑎𝑛𝑑 𝑏⃗ = (6.0 𝑚)𝑖̂ + (8.0 𝑚)𝑗̂
a) Write both vectors in direction-magnitude notation
3
𝑎 = √42 + 32 = 5
𝛼 = tan−1 (− 4) = −36.9°
𝑏 = √62 + 82 = 10
𝛽 = tan−1 (8) = 36.9°
6
𝑎 = 5 𝑚 𝑎𝑡 − 36.9°𝑡𝑜 + 𝑥 𝑎𝑥𝑖𝑠
𝑏⃗ = 10 𝑚 𝑎𝑡 36.9°𝑡𝑜 + 𝑥 𝑎𝑥𝑖𝑠
b) Write 𝑎 + 𝑏⃗ in direction-magnitude notation
𝑎 + 𝑏⃗ = (10 𝑚)𝑖̂ + (5.0 𝑚)𝑗̂
5
10
|𝑎 + 𝑏⃗| = √102 + 152 = 18.03 𝑚
𝜃 = tan−1 ( ) = 26.6°
𝑎 + 𝑏⃗ = 18.03 𝑚 𝑎𝑡 26.6°𝑡𝑜 + 𝑥 𝑎𝑥𝑖𝑠
c) Write 𝑏⃗ − 𝑎 in direction-magnitude notation
𝑏⃗ − 𝑎 = (2.0 𝑚)𝑖̂ − (11 𝑚)𝑗̂
|𝑏⃗ − 𝑎| = √22 + 112 = 11.2 𝑚
−11
)
2
𝜃 = tan−1 (
= −79.7°
𝑏⃗ − 𝑎 = 11.2 𝑚 𝑎𝑡 − 79.7° 𝑡𝑜 + 𝑥 𝑎𝑥𝑖𝑠
d) Write 𝑎 − 𝑏⃗ in direction-magnitude notation
𝑎 − 𝑏⃗ = −(𝑏⃗ − 𝑎)
𝑎 − 𝑏⃗ = 11.2 𝑚 𝑎𝑡 100.3° 𝑡𝑜 + 𝑥 𝑎𝑥𝑖𝑠
⃗ is added to 𝐴 the result is 6.0𝑖̂ + 1.0𝑗̂ and if 𝐵
⃗ is subtracted from 𝐴 the result is −4.0𝑖̂ +
P57) If 𝐵
7.0𝑗̂. Calculate vector 𝐴
⃗ = 6.0𝑖̂ + 1.0𝑗̂
𝐴+𝐵
⃗ = −4.0𝑖̂ + 7.0𝑗̂
𝐴−𝐵
2𝐴 = (6.0𝑖̂ + 1.0𝑗̂) + (−4.0𝑖̂ + 7.0𝑗̂) = 2.0𝑖̂ + 8.0𝑗̂
𝐴 = 1.0𝑖̂ + 4.0𝑗̂
Discussion Class 3
Problems
P34) Consider the two vectors 𝑎 = 3.0𝑖̂ + 5.0𝑗̂ and 𝑏⃗ = 2.0𝑖̂ + 4.0𝑗̂. Find the following:
a) 𝑎 × 𝑏⃗
= (12 − 10)𝑘̂ = 2𝑘̂
b) 𝑎 ∙ 𝑏⃗
= 3(2) + 5(4) = 26
c) (𝑎 + 𝑏⃗) ∙ 𝑏⃗
= (5.0𝑖̂ + 9.0𝑗̂) ∙ (2.0𝑖̂ + 4.0𝑗̂) = 10 + 36 = 46
d) The component of 𝑎 along (parallel to) the direction of 𝑏⃗
𝑎 ∙ 𝑏⃗
26
𝑎𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 =
=
= 5.81
𝑏
√22 + 42
P52) The following three displacements are all measured in meters:
𝑑1 = 4.0𝑖̂ + 5.0𝑗̂ − 6.0𝑘̂
𝑑2 = −1.0𝑖̂ + 2.0𝑗̂ + 3.0𝑘̂
𝑑3 = 4.0𝑖̂ + 3.0𝑗̂ + 2.0𝑘̂
a) What is 𝑟 = 𝑑1 − 𝑑2 + 𝑑3
= (4.0 + 1.0 + 4.0 )𝑖̂ + (5.0 − 2.0 + 3.0)𝑗̂ + (−6.0 − 3.0 + 2.0)𝑘̂
= 9.0𝑖̂ + 6.0𝑗̂ − 7.0𝑘̂
b) What is the angle between 𝑟 and the positive z axis?
𝑧 = 1.0𝑘̂
𝜃 = cos−1 (
𝑧∙𝑟
7
) = 122.9°
) = cos−1 (−
2
𝑧𝑟
√9 + 62 + 72
c) What is the component of 𝑑1 parallel to 𝑑2 ?
𝑑1 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 =
𝑑1 ∙ 𝑑2 −4 + 10 − 18
=
= −3.21
𝑑2
√12 + 22 + 32
d) What is the component of 𝑑1 perpendicular to 𝑑2 ?
𝑑1 × 𝑑2 = 27𝑖̂ − 6.0𝑗̂ + 13𝑘̂
𝑑1 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 =
|𝑑1 × 𝑑2 | √272 + 62 + 132
=
= 8.16
𝑑2
√12 + 22 + 32