MAT1503 ASSIGNMENT 5 2021

MAT1503


ASSIGNMENT 5



QUESTION 1


QUESTION 1.1


𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑙𝑖𝑛𝑒: 𝑟 = 𝑟𝑜 + 𝑡𝑣

𝑤ℎ𝑒𝑟𝑒 ∶ 𝑟 = 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑝𝑜𝑖𝑛𝑡 = (𝑥, 𝑦, 𝑧)

𝑟𝑜 = 𝑝𝑜𝑖𝑛𝑡 𝑡ℎ𝑎𝑡 𝑙𝑖𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = (−6, −7,0)

𝑣 = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 = 〈−1,2,4〉

𝑡 = 𝑠𝑐𝑎𝑙𝑎𝑟(𝑎𝑛𝑦 𝑣𝑎𝑙𝑢𝑒)

𝑟 = 𝑟𝑜 + 𝑡𝑣

(𝑥, 𝑦, 𝑧) = (−6, −7,0) + 𝑡(−1,2,4)



QUESTION 1.2


𝑟 = 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑝𝑜𝑖𝑛𝑡 = (𝑥, 𝑦, 𝑧)

𝑟𝑜 = (−6, −7,0)

⃗ 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 (−1,2,4)
𝑣 = −(−1,2,4) 𝑠𝑖𝑛𝑐𝑒 𝑈

𝑣 = (1, −2, −4)

𝑟 = 𝑟𝑜 + 𝑡𝑣

(𝑥, 𝑦, 𝑧) = (−6, −7,0) + 𝑡(1, −2, −4)

, QUESTION 2


⃗ = 〈3, −1, −2〉 , 𝑣 = 〈−1,0,2〉 , 𝑤
𝑢 ⃗⃗ = 〈−6,1,4〉

QUESTION 2.1


⃗⃗ = 〈−1,0,2〉 + 3〈−6,1,4〉
𝑣 + 3𝑤

= 〈−1,0,2〉 + 〈−18,3,12〉

= 〈−19,3,14〉



QUESTION 2.2


3𝑢
⃗ − 2𝑣 = 3〈3, −1, −2〉 − 2〈−1,0,2〉

= 〈9, −3, −6〉 − 〈−2,0, −4〉

= 〈11, −3, −2〉



QUESTION 2.3


⃗ ) = −[〈−1,0,2〉 + 3〈3, −1, −2〉]
−(𝑣 + 3𝑢

= −[〈−1,0,2〉 + 〈9, −3, −6〉]

= −〈8, −3 − 4〉

= 〈−8,34〉

, QUESTION 3


QUESTION 3.1


𝐶




𝐴 𝐵



⃗⃗⃗⃗⃗
𝐴𝐵 = 𝐵 − 𝐴

⃗⃗⃗⃗⃗ = (−3,5) − (1,3) = (−4,2)
𝐴𝐵



⃗⃗⃗⃗⃗ = 𝐶 − 𝐴
𝐴𝐶

⃗⃗⃗⃗⃗ = 2𝐴 − 𝐴
𝐴𝐶

⃗⃗⃗⃗⃗
𝐴𝐶 = 𝐴

⃗⃗⃗⃗⃗
𝐴𝐶 = (1,3)

𝑁𝑜𝑤 𝑙𝑒𝑡𝑠 𝑖𝑚𝑎𝑔𝑖𝑛𝑒 𝑤𝑒 𝑎𝑟𝑒 𝑖𝑛 3𝐷 𝑠𝑝𝑎𝑐𝑒 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑧𝑒𝑟𝑜

⃗⃗⃗⃗⃗ = (−4,2,0)
𝐴𝐵

⃗⃗⃗⃗⃗ = (1,3,0)
𝐴𝐶

1
⃗⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = ‖𝐴𝐵 𝐴𝐶 ‖
2



𝑖 𝑗 𝑘
⃗⃗⃗⃗⃗
𝐴𝐵 × ⃗⃗⃗⃗⃗
𝐴𝐶 = |−4 2 0|
1 3 0
2 0 −4 0 −4 2
= 𝑖| |−𝑗| |+𝑘| |
3 0 1 0 1 3



⃗⃗⃗⃗⃗ × 𝐴𝐶
𝐴𝐵 ⃗⃗⃗⃗⃗ = 0𝑖 + 0𝑗 − 14𝑘 = (0,0, −14)