UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
Linear Algebra
Assignment 04 — 2025
⋄
Module Code: MAT1503
Module Name: Linear Algebra
Assignment No.: 04
Due Date: Sunday, 31 August 2026
Semester: 2026
Submitted in partial fulfilment of the requirements for MAT1503
at the University of South Africa.
,UNISA | MAT1503 Assignment 04
Question 1
√
Let ⃗u = ⟨1, 1, 0⟩ and ⃗v = ⟨0, 1, 1⟩. A third vector w
⃗ has magnitude 12 and is in the direction of
the vector projection of ⃗a = ⟨2, 2, 1⟩ onto ⃗b = ⟨1, 0, −2⟩.
1.1 The vector w
⃗
The vector projection of ⃗a onto ⃗b is given by
⃗a · ⃗b ⃗
proj⃗b ⃗a = b.
⃗b · ⃗b
The dot product ⃗a · ⃗b is calculated first:
⃗a · ⃗b = (2)(1) + (2)(0) + (1)(−2) = 2 + 0 − 2 = 0.
Since ⃗a · ⃗b = 0, the vectors ⃗a and ⃗b are orthogonal, and the projection of ⃗a onto ⃗b is the zero
vector:
0 ⃗ ⃗
proj⃗b ⃗a = b = 0.
⃗b · ⃗b
√
The zero vector has no defined direction, so a vector w
⃗ of magnitude 12 cannot be assigned
a direction on this basis. As set out, the question data are inconsistent: ⃗a and ⃗b happen to be
perpendicular, which collapses the very projection the direction of w
⃗ was meant to come from.
For the remainder of the question, w
⃗ is therefore treated as unspecified in direction, and the
analysis proceeds as far as the given vectors ⃗u and ⃗v allow.
1.2 Shortest distance between the two lines
The plane Π1 contains the origin and is spanned by ⃗u and ⃗v , so its normal vector is
i j k
⃗n1 = ⃗u × ⃗v = 1 1 0 .
0 1 1
Page 1 of 9
, UNISA | MAT1503 Assignment 04
Expanding along the first row,
⃗u × ⃗v = i(1 · 1 − 0 · 1) − j(1 · 1 − 0 · 0) + k(1 · 1 − 1 · 0) = ⟨1, −1, 1⟩.
The line L passes through Q(0, 5, 0) and is parallel to ⃗u × ⃗v , giving the parametric form
L: x = t, y = 5 − t, z = t.
The plane Π2 is defined with w
⃗ as its normal vector and passing through P (1, 1, 1). Because
⃗ cannot be assigned a direction from the data in 1.1, the equation of Π2 , and consequently
w
the line of intersection of Π1 and Π2 , cannot be pinned down numerically. What can be stated
with certainty is the general method: once a normal vector ⃗n2 for Π2 is fixed, the direction of
the line of intersection of the two planes is ⃗n1 × ⃗n2 , a point on that line is found by solving the
two plane equations simultaneously, and the shortest distance to L is then obtained by taking
the vector between a point on each line and projecting it onto the common perpendicular
direction (⃗n1 × ⃗n2 ) × ⟨1, −1, 1⟩, divided by the magnitude of that perpendicular. With a concrete
magnitude and direction for w,
⃗ this method returns a single numerical distance; without it, the
distance cannot be evaluated.
Page 2 of 9
College of Science, Engineering and Technology
⋄
Linear Algebra
Assignment 04 — 2025
⋄
Module Code: MAT1503
Module Name: Linear Algebra
Assignment No.: 04
Due Date: Sunday, 31 August 2026
Semester: 2026
Submitted in partial fulfilment of the requirements for MAT1503
at the University of South Africa.
,UNISA | MAT1503 Assignment 04
Question 1
√
Let ⃗u = ⟨1, 1, 0⟩ and ⃗v = ⟨0, 1, 1⟩. A third vector w
⃗ has magnitude 12 and is in the direction of
the vector projection of ⃗a = ⟨2, 2, 1⟩ onto ⃗b = ⟨1, 0, −2⟩.
1.1 The vector w
⃗
The vector projection of ⃗a onto ⃗b is given by
⃗a · ⃗b ⃗
proj⃗b ⃗a = b.
⃗b · ⃗b
The dot product ⃗a · ⃗b is calculated first:
⃗a · ⃗b = (2)(1) + (2)(0) + (1)(−2) = 2 + 0 − 2 = 0.
Since ⃗a · ⃗b = 0, the vectors ⃗a and ⃗b are orthogonal, and the projection of ⃗a onto ⃗b is the zero
vector:
0 ⃗ ⃗
proj⃗b ⃗a = b = 0.
⃗b · ⃗b
√
The zero vector has no defined direction, so a vector w
⃗ of magnitude 12 cannot be assigned
a direction on this basis. As set out, the question data are inconsistent: ⃗a and ⃗b happen to be
perpendicular, which collapses the very projection the direction of w
⃗ was meant to come from.
For the remainder of the question, w
⃗ is therefore treated as unspecified in direction, and the
analysis proceeds as far as the given vectors ⃗u and ⃗v allow.
1.2 Shortest distance between the two lines
The plane Π1 contains the origin and is spanned by ⃗u and ⃗v , so its normal vector is
i j k
⃗n1 = ⃗u × ⃗v = 1 1 0 .
0 1 1
Page 1 of 9
, UNISA | MAT1503 Assignment 04
Expanding along the first row,
⃗u × ⃗v = i(1 · 1 − 0 · 1) − j(1 · 1 − 0 · 0) + k(1 · 1 − 1 · 0) = ⟨1, −1, 1⟩.
The line L passes through Q(0, 5, 0) and is parallel to ⃗u × ⃗v , giving the parametric form
L: x = t, y = 5 − t, z = t.
The plane Π2 is defined with w
⃗ as its normal vector and passing through P (1, 1, 1). Because
⃗ cannot be assigned a direction from the data in 1.1, the equation of Π2 , and consequently
w
the line of intersection of Π1 and Π2 , cannot be pinned down numerically. What can be stated
with certainty is the general method: once a normal vector ⃗n2 for Π2 is fixed, the direction of
the line of intersection of the two planes is ⃗n1 × ⃗n2 , a point on that line is found by solving the
two plane equations simultaneously, and the shortest distance to L is then obtained by taking
the vector between a point on each line and projecting it onto the common perpendicular
direction (⃗n1 × ⃗n2 ) × ⟨1, −1, 1⟩, divided by the magnitude of that perpendicular. With a concrete
magnitude and direction for w,
⃗ this method returns a single numerical distance; without it, the
distance cannot be evaluated.
Page 2 of 9
