MAT2615 Assignment 1
(COMPLETE
ANSWERS) 2025 - DUE
15 May 2025
For assistance contact
Email:
, 1. Equation of the plane containing two lines
The two lines are given by: ℓ1 : (x, y, z) = (1, 0, 0) + t(1, 0, 1) ℓ2 : (x, y, z) = (1, 0,−1) + s(0, 1,
1) (using 's' as the parameter for the second line to avoid confusion)
A point on ℓ1 is P1 = (1, 0, 0) and its direction vector is d1 = (1, 0, 1). A point on ℓ2 is P2 = (1,
0, -1) and its direction vector is d2 = (0, 1, 1).
Since both lines lie on the plane, their direction vectors are parallel to the plane. The normal
vector to the plane (n) will be perpendicular to both d1 and d2. We can find n by taking the cross
product of d1 and d2:
n = d1 × d2 = (1, 0, 1) × (0, 1, 1) = | i j k | | 1 0 1 | | 0 1 1 | = i(01 - 11) - j(11 - 10) + k(11 - 00) =
i(-1) - j(1) + k(1) = (-1, -1, 1)
The equation of the plane can be written in the form n · (r - a) = 0, where n is the normal vector, r
= (x, y, z) is a general point on the plane, and a is a known point on the plane (we can use P1 or
P2). Let's use P1 = (1, 0, 0):
(-1, -1, 1) · ((x, y, z) - (1, 0, 0)) = 0 (-1, -1, 1) · (x - 1, y, z) = 0 -1(x - 1) - 1(y) + 1(z) = 0 -x + 1 -
y + z = 0 -x - y + z = -1 x + y - z = 1
The equation of the plane is x + y - z = 1.
2. Parametric equation for the intersection of two planes
The two planes are: 3x + 2y − z − 4 = 0 (Plane 1) −x − 2y + 2z = 0 (Plane 2)
To find the intersection, we need to solve this system of linear equations. We can eliminate one
variable. Let's add the two equations:
(3x + 2y − z − 4) + (−x − 2y + 2z) = 0 + 0 2x + z - 4 = 0 z = 4 - 2x
Now substitute this expression for z into the second equation: −x − 2y + 2(4 - 2x) = 0 −x − 2y +
8 - 4x = 0 −5x − 2y + 8 = 0 2y = -5x + 8 y = -5/2 x + 4
Now we can write the solution in parametric form by letting x = t, where t ∈ R: x = t y = -5/2 t +
4 z = 4 - 2t
The parametric equation for the intersection is (x, y, z) = (0, 4, 4) + t(1, -5/2, -2), t ∈ R. (We
can also write it as (x, y, z) = (0, 4, 4) + s(2, -5, -4), s ∈ R by letting t = 2s to avoid fractions).
3. Point of intersection of a line and a plane
The line is given by: ℓ : (x, y, z) = (5, 4,−1)+t(1, 1, 0) So, x = 5 + t, y = 4 + t, z = -1 + 0t = -1.
The plane is given by: 2x + y − z = 3
Substitute the parametric equations of the line into the equation of the plane: 2(5 + t) + (4 + t) - (-
1) = 3 10 + 2t + 4 + t + 1 = 3 15 + 3t = 3 3t = 3 - 15 3t = -12 t = -4
(COMPLETE
ANSWERS) 2025 - DUE
15 May 2025
For assistance contact
Email:
, 1. Equation of the plane containing two lines
The two lines are given by: ℓ1 : (x, y, z) = (1, 0, 0) + t(1, 0, 1) ℓ2 : (x, y, z) = (1, 0,−1) + s(0, 1,
1) (using 's' as the parameter for the second line to avoid confusion)
A point on ℓ1 is P1 = (1, 0, 0) and its direction vector is d1 = (1, 0, 1). A point on ℓ2 is P2 = (1,
0, -1) and its direction vector is d2 = (0, 1, 1).
Since both lines lie on the plane, their direction vectors are parallel to the plane. The normal
vector to the plane (n) will be perpendicular to both d1 and d2. We can find n by taking the cross
product of d1 and d2:
n = d1 × d2 = (1, 0, 1) × (0, 1, 1) = | i j k | | 1 0 1 | | 0 1 1 | = i(01 - 11) - j(11 - 10) + k(11 - 00) =
i(-1) - j(1) + k(1) = (-1, -1, 1)
The equation of the plane can be written in the form n · (r - a) = 0, where n is the normal vector, r
= (x, y, z) is a general point on the plane, and a is a known point on the plane (we can use P1 or
P2). Let's use P1 = (1, 0, 0):
(-1, -1, 1) · ((x, y, z) - (1, 0, 0)) = 0 (-1, -1, 1) · (x - 1, y, z) = 0 -1(x - 1) - 1(y) + 1(z) = 0 -x + 1 -
y + z = 0 -x - y + z = -1 x + y - z = 1
The equation of the plane is x + y - z = 1.
2. Parametric equation for the intersection of two planes
The two planes are: 3x + 2y − z − 4 = 0 (Plane 1) −x − 2y + 2z = 0 (Plane 2)
To find the intersection, we need to solve this system of linear equations. We can eliminate one
variable. Let's add the two equations:
(3x + 2y − z − 4) + (−x − 2y + 2z) = 0 + 0 2x + z - 4 = 0 z = 4 - 2x
Now substitute this expression for z into the second equation: −x − 2y + 2(4 - 2x) = 0 −x − 2y +
8 - 4x = 0 −5x − 2y + 8 = 0 2y = -5x + 8 y = -5/2 x + 4
Now we can write the solution in parametric form by letting x = t, where t ∈ R: x = t y = -5/2 t +
4 z = 4 - 2t
The parametric equation for the intersection is (x, y, z) = (0, 4, 4) + t(1, -5/2, -2), t ∈ R. (We
can also write it as (x, y, z) = (0, 4, 4) + s(2, -5, -4), s ∈ R by letting t = 2s to avoid fractions).
3. Point of intersection of a line and a plane
The line is given by: ℓ : (x, y, z) = (5, 4,−1)+t(1, 1, 0) So, x = 5 + t, y = 4 + t, z = -1 + 0t = -1.
The plane is given by: 2x + y − z = 3
Substitute the parametric equations of the line into the equation of the plane: 2(5 + t) + (4 + t) - (-
1) = 3 10 + 2t + 4 + t + 1 = 3 15 + 3t = 3 3t = 3 - 15 3t = -12 t = -4
