MAT2615 Assignment 1 (COMPLETE ANSWERS) 2025 - DUE 15 May 2025; 100% Correct solutions and explanations.

,MAT2615 Assignment 1 (COMPLETE ANSWERS) 2025 - DUE 15 May 2025;
100% Correct solutions and explanations.
1. (Sections 2.11,2.12)
The parametric equations of two lines are given below:
ℓ1 : (x, y, z) = (1, 0, 0) + t(1, 0, 1), t ∈ R
ℓ2 : (x, y, z) = (1, 0,−1) + t(0, 1, 1), t ∈ R
Calculate the equation of the plane containing these two lines. [5]


𝑺𝒕𝒆𝒑 𝟏: 𝑰𝒅𝒆𝒏𝒕𝒊𝒇𝒚 𝒕𝒉𝒆 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒗𝒆𝒄𝒕𝒐𝒓𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆𝒔

𝑇ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒𝑠 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑜𝑓 𝑡𝑡𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠.

 𝐹𝑜𝑟 ℓ1\𝑒𝑙𝑙_1ℓ1, 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑠 𝑣1 = (1,0,1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑣_1} = (1, 0, 1)𝑣1 =
(1,0,1).
 𝐹𝑜𝑟 ℓ2\𝑒𝑙𝑙_2ℓ2, 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑠 𝑣2 = (0,1,1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑣_2} = (0, 1, 1)𝑣2 =
(0,1,1).

𝑺𝒕𝒆𝒑 𝟐: 𝑭𝒊𝒏𝒅 𝒂 𝒑𝒐𝒊𝒏𝒕 𝒐𝒏 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆

𝐵𝑜𝑡ℎ 𝑙𝑖𝑛𝑒𝑠 𝑙𝑖𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒. 𝐴 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑐𝑎𝑛 𝑏𝑒 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝑎 𝑝𝑜𝑖𝑛𝑡

 𝐴 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 ℓ1\𝑒𝑙𝑙_1ℓ1 𝑖𝑠 𝑃1 = (1,0,0)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑃_1} = (1, 0, 0)𝑃1 = (1,0,0).
 𝐴 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 ℓ2\𝑒𝑙𝑙_2ℓ2 𝑖𝑠 𝑃2 = (1,0, −1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑃_2} = (1, 0, −1)𝑃2 = (1,0, −1).

𝑺𝒕𝒆𝒑 𝟑: 𝑪𝒐𝒎𝒑𝒖𝒕𝒆 𝒕𝒉𝒆 𝒏𝒐𝒓𝒎𝒂𝒍 𝒗𝒆𝒄𝒕𝒐𝒓 𝒕𝒐 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆

𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒, 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑛
\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛}𝑛. 𝑇ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒𝑠. 𝑇𝑜 𝑐𝑜𝑚𝑝𝑢
\𝑚𝑎𝑡ℎ𝑏𝑓{𝑣_1}𝑣1 𝑎𝑛𝑑 𝑣2\𝑚𝑎𝑡ℎ𝑏𝑓{𝑣_2}𝑣2:

𝑛 = 𝑣1 × 𝑣2\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛} = \𝑚𝑎𝑡ℎ𝑏𝑓{𝑣_1} \𝑡𝑖𝑚𝑒𝑠 \𝑚𝑎𝑡ℎ𝑏𝑓{𝑣_2}𝑛 = 𝑣1 × 𝑣2 𝑛 =
∣ 𝑖𝑗𝑘101011 ∣\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛}
= \𝑏𝑒𝑔𝑖𝑛{𝑣𝑚𝑎𝑡𝑟𝑖𝑥} \𝑚𝑎𝑡ℎ𝑏𝑓{𝑖} & \𝑚𝑎𝑡ℎ𝑏𝑓{𝑗} & \𝑚𝑎𝑡ℎ𝑏𝑓{𝑘} \\ 1 & 0 & 1 \
\ 0 & 1 & 1 \𝑒𝑛𝑑{𝑣𝑚𝑎𝑡𝑟𝑖𝑥}𝑛 = 𝑖10𝑗01𝑘11

𝑇ℎ𝑖𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑔𝑖𝑣𝑒𝑠 𝑢𝑠:

, 𝑛 = 𝑖(0 ⋅ 1 − 1 ⋅ 1) − 𝑗(1 ⋅ 1 − 1 ⋅ 0) + 𝑘(1 ⋅ 1 − 0 ⋅ 0)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛}
= \𝑚𝑎𝑡ℎ𝑏𝑓{𝑖}(0 \𝑐𝑑𝑜𝑡 1 − 1 \𝑐𝑑𝑜𝑡 1) − \𝑚𝑎𝑡ℎ𝑏𝑓{𝑗}(1 \𝑐𝑑𝑜𝑡 1
− 1 \𝑐𝑑𝑜𝑡 0) + \𝑚𝑎𝑡ℎ𝑏𝑓{𝑘}(1 \𝑐𝑑𝑜𝑡 1 − 0 \𝑐𝑑𝑜𝑡 0)𝑛
= 𝑖(0 ⋅ 1 − 1 ⋅ 1) − 𝑗(1 ⋅ 1 − 1 ⋅ 0) + 𝑘(1 ⋅ 1 − 0 ⋅ 0) 𝑛
= 𝑖(−1) − 𝑗(1) + 𝑘(1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛}
= \𝑚𝑎𝑡ℎ𝑏𝑓{𝑖}(−1) − \𝑚𝑎𝑡ℎ𝑏𝑓{𝑗}(1) + \𝑚𝑎𝑡ℎ𝑏𝑓{𝑘}(1)𝑛
= 𝑖(−1) − 𝑗(1) + 𝑘(1) 𝑛 = (−1, −1,1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛} = (−1, −1, 1)𝑛
= (−1, −1,1)

𝑇ℎ𝑢𝑠, 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑛 = (−1, −1,1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛} = (−1, −1, 1)𝑛
= (−1, −1,1).

𝑺𝒕𝒆𝒑 𝟒: 𝑾𝒓𝒊𝒕𝒆 𝒕𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆

𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦:

𝑛1(𝑥 − 𝑥0) + 𝑛2(𝑦 − 𝑦0) + 𝑛3(𝑧 − 𝑧0)
= 0𝑛_1(𝑥 − 𝑥_0) + 𝑛_2(𝑦 − 𝑦_0) + 𝑛_3(𝑧 − 𝑧_0)
= 0𝑛1(𝑥 − 𝑥0) + 𝑛2(𝑦 − 𝑦0) + 𝑛3(𝑧 − 𝑧0) = 0

𝑤ℎ𝑒𝑟𝑒 (𝑥0, 𝑦0, 𝑧0)(𝑥_0, 𝑦_0, 𝑧_0)(𝑥0, 𝑦0, 𝑧0
) 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑛𝑑 (𝑛1, 𝑛2, 𝑛3)(𝑛_1, 𝑛_2, 𝑛_3)(𝑛1, 𝑛2, 𝑛3
) 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒.

𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑃1 = (1,0,0)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑃_1} = (1, 0, 0)𝑃1
= (1,0,0) 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑛 = (−1, −1,1)\𝑚𝑎𝑡ℎ𝑏𝑓{𝑛}
= (−1, −1, 1)𝑛 = (−1, −1,1), 𝑤𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛:

−1(𝑥 − 1) − 1(𝑦 − 0) + 1(𝑧 − 0) = 0 − 1(𝑥 − 1) − 1(𝑦 − 0) + 1(𝑧 − 0)
= 0 − 1(𝑥 − 1) − 1(𝑦 − 0) + 1(𝑧 − 0) = 0

𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔:

−(𝑥 − 1) − 𝑦 + 𝑧 = 0 − (𝑥 − 1) − 𝑦 + 𝑧 = 0 − (𝑥 − 1) − 𝑦 + 𝑧 = 0 − 𝑥 + 1 − 𝑦 + 𝑧
= 0−𝑥 + 1 − 𝑦 + 𝑧 = 0−𝑥+1−𝑦+𝑧 =0 −𝑥−𝑦+𝑧
= −1 − 𝑥 − 𝑦 + 𝑧 = −1 − 𝑥 − 𝑦 + 𝑧 = −1

𝑇ℎ𝑢𝑠, 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 𝑖𝑠:

𝑥 + 𝑦 − 𝑧 = 1𝑥 + 𝑦 − 𝑧 = 1𝑥 + 𝑦 − 𝑧 = 1

2. (Sections 2.11,2.12)
Given the two planes 3x + 2y − z − 4 = 0 and −x − 2y + 2z = 0. Find a parametric equation
for the intersection. [5]