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MAT1503 Assignment 4 (COMPLETE SOLUTIONS) 2025 - Due 29 August 2025

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Linear Algebra I - MAT1503 Assignment 4 2025 - Due 29 August 2025; 100 % TRUSTED workings, Expert Solved, Explanations and Solutions. For assistance call or W.h.a.t.s.a.p.p us on ...(.+.2.5.4.7.7.9.5.4.0.1.3.2)...........










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Uploaded on
August 12, 2025
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2025/2026
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MAT1503
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

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