ASSIGNMENT 4 2025
UNIQUE NO.
DUE DATE: 29 AUGUST 2025
, Linear Algebra I
Question 1
(1.1) Find an equation for the plane that passes through the origin (0,0,0) and is
parallel to the plane −𝑥 + 3𝑦 − 2𝑧 = 6 .
A plane parallel to −𝑥 + 3𝑦 − 2𝑧 = 6 must have the same normal vector 𝐧 = ⟨−1,3, −2⟩.
A plane through the origin with that normal has equation
−𝑥 + 3𝑦 − 2𝑧 = 0.
That is the required plane.
(1.2) Distance between point (−1, −2,0) and the plane 3𝑥 − 𝑦 + 4𝑧 = −2 .
Write the plane in standard form 𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 + 𝐷 = 0 . Move RHS to left:
3𝑥 − 𝑦 + 4𝑧 + 2 = 0,
so 𝐴 = 3, 𝐵 = −1, 𝐶 = 4, 𝐷 = 2 .
Distance formula from point 𝑃(𝑥0 , 𝑦0 , 𝑧0 ) to plane:
|𝐴𝑥0 + 𝐵𝑦0 + 𝐶𝑧0 + 𝐷|
𝑑= .
√𝐴2 + 𝐵 2 + 𝐶 2
Plug in 𝑃(−1, −2,0):
numerator = |3(−1) + (−1)(−2) + 4(0) + 2| = | − 3 + 2 + 0 + 2| = |1| = 1.
Denominator:
√32 + (−1)2 + 42 = √9 + 1 + 16 = √26.
Thus
1
𝑑= .
√26