MAT2615 Assignment 1 (COMPLETE ANSWERS) 2025 - DUE 15 May 2025

,MAT2615 Assignment 1 (COMPLETE ANSWERS)
2025 - DUE 15 May 2025; 100% TRUSTED
Complete, trusted solutions and explanations.
MULTIPLE CHOICE,ASSURED EXCELLENCE
1. (Sections 2.11,2.12) The parametric equations of two lines
are given below: ℓ1 : (x, y, z) = (1, 0, 0) + t(1, 0, 1), t ∈ R ℓ2 :
(x, y, z) = (1, 0,−1) + t(0, 1, 1), t ∈ R Calculate the equation of
the plane containing these two lines. [5] 2. (Sections
2.11,2.12) MAT2601 Given the two planes 3x + 2y − z − 4 = 0
and −x − 2y + 2z = 0. Find a parametric equation for the
intersection. [5] 3. (Sections 2.11,2.12) MAT2611 Find the
point of intersection of the line ℓ : (x, y, z) = (5, 4,−1)+t(1, 1,
0), t ∈ R and the plane 2x + y − z = 3. [5] 4. (Sections
2.5,2.6,4.3) MAT2614 Consider the R2 − R function defined
by f (x, y) = 2x + 2y − 3. Prove from first principles that
lim(x,y)→(−1,1) f (x, y) = −3 [5] 5. (Sections 4.3,4.4,4.5)
Determine whether the following limits exist. If you suspect
that a limit does not exist, try to prove so by using limits
along curves. If you suspect that the limit does exist, you
must use the ϵ − δ definition, or the limit laws, or a
combination of the two. (a) lim (x,y)→(0,0) sin(x + y) x + y (5)
(b) lim (x,y)→(1,1) y + 1 x − 1 (5) (c) lim (x,y)→(0,0) x2 + y2 xy
(5) 15 (d) lim (x,y)→(π/2,π/2) cos x sin y + y tan x (5) [20] 6.
(Sections 4.4,4.7) Consider the R2 − R function given by f (x,
y) = ( −2x2+xy+y2 y2+2xy if 2x ≠ −y 3 2 if (x, y) = (1,−2) or (x,
y) = (2,−4). (a) Write down the domain Df of f . (2) (b)

, Determine lim (x,y)→(1,−2) f (x, y) and lim (x,y)→(2,−4) f (x,
y). (3) (c) Calculate f (1,−2) and f (2,−4). (4) (d) Is f continuous
at (x, y) = (1,−2)? (2) (e) Is f continuous at (x, y) = (2,−4)? (2)
(f) Is f a continuous function? (2) Give reasons for your
answers to (d), (e) and (f). [15]
1. Equation of the plane containing two lines
The given parametric equations are:
ℓ1:(x,y,z)=(1,0,0)+t(1,0,1)\ell_1: (x, y, z) = (1, 0, 0) + t(1, 0, 1) ℓ1:
(x,y,z)=(1,0,0)+t(1,0,1) ℓ2:(x,y,z)=(1,0,−1)+t(0,1,1)\ell_2: (x, y, z)
= (1, 0, -1) + t(0, 1, 1)ℓ2:(x,y,z)=(1,0,−1)+t(0,1,1)
These can be rewritten as:
ℓ1:x=1+t,y=0,z=t\ell_1: x = 1 + t, \quad y = 0, \quad z = tℓ1
:x=1+t,y=0,z=t ℓ2:x=1,y=t,z=−1+t\ell_2: x = 1, \quad y = t, \quad
z = -1 + tℓ2:x=1,y=t,z=−1+t
To find the plane containing both lines, we need a normal
vector to the plane, which can be found using the cross product
of two direction vectors from the lines:
v1=(1,0,1),v2=(0,1,1)\mathbf{v_1} = (1, 0, 1), \quad \
mathbf{v_2} = (0, 1, 1)v1=(1,0,1),v2=(0,1,1)
Computing the cross product:
n=v1×v2=∣ijk101011∣\mathbf{n} = \mathbf{v_1} \times \
mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \
mathbf{k} \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix}n=v1×v2=i10
j01k11 =(0⋅1−1⋅1)i−(1⋅1−1⋅0)j+(1⋅1−0⋅1)k= (0\cdot1 - 1\cdot1)\