MAT1503 Assignment 4
Memo (COMPLETE
ANSWERS) Due 29 August
2025
For assistance contact
Email:
,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 will have the same normal vector. The normal vector for the
given plane is
n =⟨−1,3,−2⟩.
The equation of a plane can be written as
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.
Since the new plane passes through the origin (0,0,0), we can use this as our point:
−1(x−0)+3(y−0)−2(z−0)=0
−x+3y−2z=0
(1.2) Find the distance between the point (−1,−2, 0) and the plane 3x−y+4z=−2.
The distance (
d) 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∣
The plane equation 3x−y+4z=−2 can be rewritten as 3x−y+4z+2=0. Here, A=3, B=−1, C=4,
and D=2. The point is
(−1,−2,0), so x0=−1, y0=−2, and z0=0.
Substituting these values into the 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
Question 2: 15 Marks
(2.1) Find the angle between the two vectors v 1=⟨−1,1,0,−1⟩ and v 2
=⟨1,−1,3,−2⟩. Determine whether both vectors are perpendicular, parallel or neither.
The angle
θ between two vectors can be found using the dot product formula:
, $$\cos \theta = \frac{\vec{v}_1 \cdot \vec{v}_2}{||\vec{v}_1|| \cdot ||\vec{v}_2||}$$First,
calculate the dot product of the two vectors:$$\vec{v}_1 \cdot \vec{v}_2 = (-1)(1) + (1)(-1) +
(0)(3) + (-1)(-2)$$
v 1⋅v 2=−1−1+0+2=0
Since the dot product is 0, the vectors are perpendicular.
(2.2) Find the direction cosines and the direction angles for the vector r
=⟨0,−1,−2,3,4⟩.
First, find the magnitude of the vector
∣∣r ∣∣:
∣∣r ∣∣=02+(−1)2+(−2)2+32+42 =0+1+4+9+16 =30
The direction cosines are the components of the vector divided by its magnitude:
• cosα=30 0=0
• cosβ=30 −1
• cosγ=30 −2
• cosδ=30 3
• cosϵ=30 4
The direction angles are the inverse cosines of the direction cosines:
• α=arccos(0)=90∘ or 2π radians
• β=arccos(30 −1)
• γ=arccos(30 −2)
• δ=arccos(30 3)
• ϵ=arccos(30 4)
Memo (COMPLETE
ANSWERS) Due 29 August
2025
For assistance contact
Email:
,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 will have the same normal vector. The normal vector for the
given plane is
n =⟨−1,3,−2⟩.
The equation of a plane can be written as
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.
Since the new plane passes through the origin (0,0,0), we can use this as our point:
−1(x−0)+3(y−0)−2(z−0)=0
−x+3y−2z=0
(1.2) Find the distance between the point (−1,−2, 0) and the plane 3x−y+4z=−2.
The distance (
d) 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∣
The plane equation 3x−y+4z=−2 can be rewritten as 3x−y+4z+2=0. Here, A=3, B=−1, C=4,
and D=2. The point is
(−1,−2,0), so x0=−1, y0=−2, and z0=0.
Substituting these values into the 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
Question 2: 15 Marks
(2.1) Find the angle between the two vectors v 1=⟨−1,1,0,−1⟩ and v 2
=⟨1,−1,3,−2⟩. Determine whether both vectors are perpendicular, parallel or neither.
The angle
θ between two vectors can be found using the dot product formula:
, $$\cos \theta = \frac{\vec{v}_1 \cdot \vec{v}_2}{||\vec{v}_1|| \cdot ||\vec{v}_2||}$$First,
calculate the dot product of the two vectors:$$\vec{v}_1 \cdot \vec{v}_2 = (-1)(1) + (1)(-1) +
(0)(3) + (-1)(-2)$$
v 1⋅v 2=−1−1+0+2=0
Since the dot product is 0, the vectors are perpendicular.
(2.2) Find the direction cosines and the direction angles for the vector r
=⟨0,−1,−2,3,4⟩.
First, find the magnitude of the vector
∣∣r ∣∣:
∣∣r ∣∣=02+(−1)2+(−2)2+32+42 =0+1+4+9+16 =30
The direction cosines are the components of the vector divided by its magnitude:
• cosα=30 0=0
• cosβ=30 −1
• cosγ=30 −2
• cosδ=30 3
• cosϵ=30 4
The direction angles are the inverse cosines of the direction cosines:
• α=arccos(0)=90∘ or 2π radians
• β=arccos(30 −1)
• γ=arccos(30 −2)
• δ=arccos(30 3)
• ϵ=arccos(30 4)
