MAT1503 ASSIGNMENT 4 2025

MAT1503
ASSIGNMENT 4
2025

,QUESTION 1


𝐼𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 − 𝑥 + 3𝑦 − 2𝑧 = 6 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑦 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒
𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 , 𝑠𝑜


𝑛⃗ = (−1,3, −2)


⃗⃗⃗⃗⃗⃗⃗
𝑃0 𝑃 = (𝑥, 𝑦, 𝑧) − (0,0,0)
⃗⃗⃗⃗⃗⃗⃗
𝑃0 𝑃 = (𝑥, 𝑦, 𝑧)



𝑛⃗ ∙ ⃗⃗⃗⃗⃗⃗⃗
𝑃0 𝑃 = 0



𝑛⃗ ∙ ⃗⃗⃗⃗⃗⃗⃗
𝑃0 𝑃 = 0
(𝑥, 𝑦, 𝑧) ∙ (−1,3, −2) = 0
−𝑥 + 3𝑦 − 2𝑧 = 0


𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 ∶ −𝑥 + 3𝑦 − 2𝑧 = 0




1.2).

, |𝑎𝑥0 + 𝑏𝑦0 + 𝑐𝑧0 + 𝑑|
𝐷=
√(𝑎)2 + (𝑏)2 + (𝑐)2


𝐹𝑜𝑟 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 ∶ 3𝑥 − 𝑦 + 4𝑧 = −2 ⇒ 3𝑥 − 𝑦 + 4𝑧 + 2 = 0 𝑤𝑒 ℎ𝑎𝑣𝑒


𝑎 = 3, 𝑏 = −1, 𝑐 = 4 𝑎𝑛𝑑 𝑑 = 2


𝐴𝑛𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑃0 = (−1, −2,0) 𝑤𝑒 𝑔𝑒𝑡


𝑥0 = −1, 𝑦0 = −2 𝑎𝑛𝑑 𝑧0 = 0


|𝑎𝑥0 + 𝑏𝑦0 + 𝑐𝑧0 + 𝑑|
𝐷=
√(𝑎)2 + (𝑏)2 + (𝑐)2
|(3)(−1) + (−1)(−2) + (4)(0) + 2|
𝐷=
√(3)2 + (−1)2 + (4)2
|−3 + 2 + 0 + 2|
𝐷=
√9 + 1 + 16
|1|
𝐷=
√26
1
𝐷= 𝑢𝑛𝑖𝑡𝑠
√26




QUESTION 2


2.1).


⃗ = 〈−1,1,0, −1〉 𝑎𝑛𝑑 𝑣 = 〈1, −1,3, −2〉
𝑢


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

, 𝑢
⃗ ∙ 𝑣 = −1 − 1 + 0 + 2
𝑢
⃗ ∙𝑣 =0


𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑑𝑜𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑖𝑠 𝑧𝑒𝑟𝑜 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑠 90°




2.2).


3
𝑟 = 〈0, −1, −2, 〉
4



3 2
√ 2 2 2
‖𝑟‖ = (0) + (−1) + (−2) + ( )
4

√89
‖𝑟‖ =
4


𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟𝑠:


𝑎1 = 〈1,0,0,0〉, 𝑎2 = 〈0,1,0,0〉 , 𝑎3 = 〈0,0,1,0〉 𝑎𝑛𝑑 𝑎4 = 〈0,0,0,1〉



𝑎1 ‖ = √(1)2 + (0)2 + (0)2 + (0)2 = 1
‖⃗⃗⃗⃗

⃗⃗⃗⃗2 ‖ = √(0)2 + (1)2 + (0)2 + (0)2 = 1
‖𝑎

⃗⃗⃗⃗3 ‖ = √(0)2 + (0)2 + (1)2 + (0)2 = 1
‖𝑎

⃗⃗⃗⃗4 ‖ = √(0)2 + (0)2 + (0)2 + (1)2 = 1
‖𝑎



𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑖𝑛𝑒𝑠:


3
𝑎1 ∙ 𝑟 〈1,0,0,0〉 ∙ 〈0, −1, −2, 〉
cos(𝛼) = = 4 = 0 =0
‖𝑟‖‖𝑎⃗⃗⃗⃗1 ‖ √89 √89
( 4 ) (1) ( 4 )