Mat1503 Assignment 5 2023

MAT1503
ASSIGNMENT 5
2023

, QUESTION 1




Solution:



1.1).



Let: ⃗⃗⃗⃗
n1 be the normal of plane U and ⃗⃗⃗⃗
n2 be the normal of plane V.



n1 = 〈λ, 5, −2λ〉 and ⃗⃗⃗⃗
⃗⃗⃗⃗ n2 = 〈−λ, 1,2〉



a).



Planes U and V are orthogonal if ⃗⃗⃗⃗
n1 ∙ ⃗⃗⃗⃗
n2 = 0



n1 ∙ ⃗⃗⃗⃗
⃗⃗⃗⃗ n2 = 0
〈λ, 5, −2λ〉 ∙ 〈−λ, 1,2〉 = 0

−λ2 + 5 − 4λ = 0

λ2 − 5 + 4λ = 0
(λ − 1)(λ + 5) = 0

λ = 1 and λ = −5



b).



Planes U and V are orthogonal if ⃗⃗⃗⃗
n1 × ⃗⃗⃗⃗
n2 = 0

,n1 × ⃗⃗⃗⃗
⃗⃗⃗⃗ n2 = 0
i j k
|λ 5 −2λ| = 0
−λ 1 2
5 −2λ λ −2λ λ 5
i| |− j| | + k| | = 〈0,0,0〉
1 2 −λ 2 −λ 1
(10 + 2λ)i − (2λ − 2λ2 )j + (λ + 5λ)k = 〈0,0,0〉



10 + 2λ = 0 1 ⇒ λ = −5

2λ − 2λ2 = 0 2 ⇒ λ(2 − 2λ) = 0 ⇒ λ = 0 or λ = 1

λ + 5λ = 0 3 ⇒ 6λ = 0 ⇒ λ=0



We are getting different values of λ from different equation , so no such values of
λ exist.



1.2).



The normal of plane − x + 3y − 2z = 6 is 〈−1,3, −2〉
Since the plane that passes through the origin is parallel to the plane − x + 3y − 2z = 6
then they have the same normal vector.



Let: n
⃗ be the normal of the plane that passes through the origin.



Equation of plane: 〈x − x0 , y − y0 , z − z0 〉 ∙ n
⃗ =0


〈x − x0 , y − y0 , z − z0 〉 ∙ n
⃗ =0 ∴ 〈x0 , y0 , z0 〉 = 〈0,0,0〉 and n
⃗ = 〈−1,3, −2〉
〈x − 0, y − 0, z − 0〉 ∙ 〈−1,3, −2〉 = 0
〈x, y, z〉 ∙ 〈−1,3, −2〉 = 0

−x + 3y − 2z = 0 (equation of the plane passing through the origin)

, 1.3).


|ax + by + cz + d|
Distance =
√(a)2 + (b)2 + (c)2



∴ where x + by + cz + d = 0 is the equation of the plane



x + by + cz + d = 3x − y + 4z + 2 = 0


|ax + by + cz + d|
Distance = ∴ (x, y, z) = (−1, −2,0) , a = 3, b = −1, c = 4 and d = 2
√(a)2 + (b)2 + (c)2


|(3)(−1) + (−1)(−2) + (4)(0) + 2|
Distance =
√(3)2 + (−1)2 + (4)2
|1|
Distance =
√26
1
Distance = units
√26




QUESTION 2