MAT1503 Assignment with Complete Solution 2023
QUESTION 1
QUESTION 1.1
⃗→ = 〈1,3, −2〉 𝑎𝑛𝑑 𝑣→ = 〈−5,3,2〉
𝑢
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑙𝑔𝑙𝑒 = 𝜃
𝑢
⃗→ ∙ 𝑣→
cos(𝜃) =
‖𝑢
⃗→‖‖𝑣→‖
⃗→ ∙ 𝑣→ = 〈1,3, −2〉 ∙ 〈−5,3,2〉
𝑢
𝑢
⃗→ ∙ 𝑣→ = −5 + 9 − 4 = 0
𝑆𝑖𝑛𝑐𝑒 𝑢
⃗→ ∙ 𝑣→ = 0 𝑡ℎ𝑒𝑛 𝑢
⃗→ 𝑎𝑛𝑑 𝑣→ 𝑎𝑟𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
QUESTION 1.2
⃗→ = 〈1, −2,4〉 𝑎𝑛𝑑 𝑣→ = 〈5,3,7〉
𝑢
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑙𝑔𝑙𝑒 = 𝜃
𝑢
⃗→ ∙ 𝑣→
cos(𝜃) =
‖𝑢
⃗→‖‖𝑣→‖
⃗→ ∙ 𝑣→ = 〈1, −2,4〉 ∙ 〈5,3,7〉
𝑢
= 5 − 6 + 28
= 27
⃗→‖ = √(1)2 + (−2)2 + (4)2 = √21
‖𝑢
‖𝑣→‖ = √(5)2 + (3)2 + (2)2 = √38
27
cos(𝜃) =
√21 ∙ √38
, 27
𝜃 = cos−1 ( )
√21 ∙ √38
𝜃 = 17.1° 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑎𝑛 𝑎𝑐𝑢𝑡𝑒
QUESTION 2
QUESTION 2.1
𝑎→ ∙ 𝑢
⃗→
𝑝𝑟𝑜𝑎 𝑢
⃗→ = 𝑎→
|𝑎→|2
𝑎→ = 〈−1, −3〉
|𝑎→| = √(−1)2 + (−3)2 = √10
⃗→ = 〈−1,3〉
𝑢
〈−1, −3〉 ∙ 〈−1,3〉
𝑝𝑟𝑜𝑎 𝑢
⃗→ = 2 〈−1, −3〉
(√10)
1−9
= 〈−1, −3〉
10
8
=− 〈−1, −3〉
10
4
= − 〈−1, −3〉
5
4 12
=〈 , 〉
5 5
4 12
⃗→ = 〈 , 〉
𝑝𝑟𝑜𝑎 𝑢
5 5
QUESTION 2.2
𝑎→ ∙ 𝑢
⃗→
𝑝𝑟𝑜𝑎 𝑢
⃗→ = 𝑎→
|𝑎→|2
QUESTION 1
QUESTION 1.1
⃗→ = 〈1,3, −2〉 𝑎𝑛𝑑 𝑣→ = 〈−5,3,2〉
𝑢
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑙𝑔𝑙𝑒 = 𝜃
𝑢
⃗→ ∙ 𝑣→
cos(𝜃) =
‖𝑢
⃗→‖‖𝑣→‖
⃗→ ∙ 𝑣→ = 〈1,3, −2〉 ∙ 〈−5,3,2〉
𝑢
𝑢
⃗→ ∙ 𝑣→ = −5 + 9 − 4 = 0
𝑆𝑖𝑛𝑐𝑒 𝑢
⃗→ ∙ 𝑣→ = 0 𝑡ℎ𝑒𝑛 𝑢
⃗→ 𝑎𝑛𝑑 𝑣→ 𝑎𝑟𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
QUESTION 1.2
⃗→ = 〈1, −2,4〉 𝑎𝑛𝑑 𝑣→ = 〈5,3,7〉
𝑢
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑙𝑔𝑙𝑒 = 𝜃
𝑢
⃗→ ∙ 𝑣→
cos(𝜃) =
‖𝑢
⃗→‖‖𝑣→‖
⃗→ ∙ 𝑣→ = 〈1, −2,4〉 ∙ 〈5,3,7〉
𝑢
= 5 − 6 + 28
= 27
⃗→‖ = √(1)2 + (−2)2 + (4)2 = √21
‖𝑢
‖𝑣→‖ = √(5)2 + (3)2 + (2)2 = √38
27
cos(𝜃) =
√21 ∙ √38
, 27
𝜃 = cos−1 ( )
√21 ∙ √38
𝜃 = 17.1° 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑎𝑛 𝑎𝑐𝑢𝑡𝑒
QUESTION 2
QUESTION 2.1
𝑎→ ∙ 𝑢
⃗→
𝑝𝑟𝑜𝑎 𝑢
⃗→ = 𝑎→
|𝑎→|2
𝑎→ = 〈−1, −3〉
|𝑎→| = √(−1)2 + (−3)2 = √10
⃗→ = 〈−1,3〉
𝑢
〈−1, −3〉 ∙ 〈−1,3〉
𝑝𝑟𝑜𝑎 𝑢
⃗→ = 2 〈−1, −3〉
(√10)
1−9
= 〈−1, −3〉
10
8
=− 〈−1, −3〉
10
4
= − 〈−1, −3〉
5
4 12
=〈 , 〉
5 5
4 12
⃗→ = 〈 , 〉
𝑝𝑟𝑜𝑎 𝑢
5 5
QUESTION 2.2
𝑎→ ∙ 𝑢
⃗→
𝑝𝑟𝑜𝑎 𝑢
⃗→ = 𝑎→
|𝑎→|2