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MAT2615 ASSIGNMENT 1 2023
- Institution
- University Of South Africa (Unisa)
This documents contains MAT2615 ASSIGNMENT 1 2023 solutions. Clear step by step calculations are provided.
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By: thembekancamiswa • 1 year ago
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MAT2615ASSIGNMENT 1
2023
,Solution:
l1 : (x, y, z) = (1,0, −1) + t(1,2,1)
l2 : (x, y, z) = (2,2,0) + t(0,1, −1)
Let: v
⃗ be the direction vector of l1
⃗ be the direction vector of l2
u
So
⃗ = (1,2,1)
v ⃗ = (0,1, −1)
and u
Let: n
⃗ be the normal vector of the plane containing l1 and l2
⃗ =v
n ⃗ ×u
⃗
i j k
⃗ = |1 2
n 1|
0 1 −1
2 1 1 1 1 2
⃗ = i|
n |− j| |+k| |
1 −1 0 −1 0 1
⃗ = (−2 − 1)i − (−1 − 0)j + (1 − 0)k
n
⃗ = −3i + j + k
n
⃗ = (−3,1,1)
n
Equation of plane ∶ r ∙ n
⃗ = r0 ∙ n
⃗
, where r = (x, y, z)
r0 = point that lies on the plane = (1,0, −1)
⃗ = (−3,1,1)
n
r∙n
⃗ = r0 ∙ n
⃗
(x, y, z) ∙ (−3,1,1) = (1,0, −1) ∙ (−3,1,1)
−3x + y + z = −3 + 0 − 1
−3x + y + z = −4
3x − y − z = 4
Equation of plane containing l1 and l2 ∶ 3x − y − z = 4
Solution:
2z − x − y = 0 ⇒ −x − y + 2z = 0 1
z+y−x−1=0 ⇒ −x + y + z = 1 2
−x − y + 2z = 0
Solve for the system {
−x + y + z = 1
−1 −1 2 0
Augumented matrix = [ | ]
−1 1 1 1
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