MAT1503 Assignment 4 Memo (COMPLETE ANSWERS) Due 29 August 2025

MAT1503 Assignment 4
Memo (COMPLETE
ANSWERS) Due 29 August
2025

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, Question 1
1.1 Plane Equation

The normal vector of the plane −x+3y−2z=6 is n =⟨−1,3,−2⟩. Since the new plane is
parallel to this one, it will have the same normal vector. The equation for a plane is given by
the formula A(x−x0)+B(y−y0)+C(z−z0)=0, where ⟨A,B,C⟩ is the normal vector and (x0,y0,z0
) is a point on the plane.
Given that the new plane passes through the origin (0,0,0), we can substitute the values:
−1(x−0)+3(y−0)−2(z−0)=0
−x+3y−2z=0
This is the equation for the plane that passes through the origin and is parallel to the given
plane.


1.2 Distance from a Point to a Plane
The distance between a point (x0,y0,z0) and a plane Ax+By+Cz+D=0 is given by the
formula:

d=A2+B2+C2 ∣Ax0+By0+Cz0+D∣
In this problem, the point is (−1,−2,0) and the plane is 3x−y+4z=−2, which can be rewritten
as 3x−y+4z+2=0.
Here, A=3, B=−1, C=4, D=2, x0=−1, y0=−2, and z0=0. Substituting these values into the
distance formula:
$$d = \frac{|3(-1) - (-2) + 4(0) + 2|}{\sqrt{3^2 + (-1)^2 + 4^2}}$$$$d = \frac{|-3 + 2 + 0 +
2|}{\sqrt{9 + 1 + 16}}$$$$d = \frac{|1|}{\sqrt{26}}$$

d=26 1

The distance between the point (−1,−2,0) and the plane is 26 1.


Question 2
2.1 Angle Between Vectors

The angle θ between two vectors v 1 and v 2 can be found using the dot product
formula:

cosθ=∣∣v 1∣∣⋅∣∣v 2∣∣v 1⋅v 2