MATH 2030: MIDTERM
Q.1. Solve for the vector w in terms of u and v:
• w − u = 2(w − 2u + 2v) + 3(u − v)
• − 32 (w − u) + v = 2(w + u) − 2(2u − 2v)
A.1.
• Distributivity yields,
w−u =
=
2w − 4u + 4v + 3u − 3v
2w − u + v
Moving all w terms to the right hand side we find
−v = w
• Multiply by 2 on both sides and use distributivity
−3w + 3u) + 2v = 4w + 4u − 8u − 8v
Bringing all w terms to the right hand side, and all others to the left hand
side
−u − 10v = 7w
dividing by seven gives w.
Q.2. Find the projection of v onto u where
• ut = [ 32 , − 23 , − 13 ] and vt = [2, −2, 2].
• ut = [ 45 , 1, − 34 ] and v = [ 21 , −1, − 12 ]
A.2. The projection of v onto u is
proju (v) =
u · v
u
u·u
we use this definition to calculation the projections
• Here u · u = 49 + 94 +
of v onto u is then
1
9
= 1 and u · v =
4
3
+
4
3
−
2
3
= 2. Thus the projection
 2   1 
1  32   31 
−3 = −3
proju (v) =
2
− 13
− 16
9
50
• Here u · u = 25
16 + 1 + 16 = 16 =
projection of v onto u is then
25
8
and u · v =
proju (v) = 0
1
5
8
−1+
3
8
= 0. Thus the
2
MATH 2030: MIDTERM
Q.3. Find the distance between the parallel lines:
   
     
 
x
−3
1
x
−2
1
y  =  2  + s 2 , y  =  2  + t 2 .
z
1
2
z
0
2
0.1. A.3. To determine the distance between the two parallel lines we need a vector
between them, in this case we take the difference of the position vectors for the
initial point for each line and denote this as v,
     
−3
−2
−1
v =  2  −  2  =  0 .
1
0
1
Taking this vector we consider the projection of v onto d the direction vector for
the lines
 
1
1
v·d
d = 2 .
projd (v) =
d·d
5
2
Q.4. Use elementary tow operations to reduce the given matrix to a) row echelon
form and b) reduced row echelon form.
•


1 −3
2
 3 −9 10 
−2 6 −10
•


1 2 3 4
 4 3 2 1 .
5 5 5 5
A.4.
• Applying row operations, R2 −3R1 , R3 +2R1 produces the following matrix,
applying R3 + 32 R2 and 41 R2 we find the row echelon form,




1 −3 2
1 −3 2
0
0
4 → 0 0 1 .
0 0−6
0 0 0
To produce the reduced row echelon form we apply the operation R1 − 2R2


1 −3 0
0 0 1 .
0 0 0
• The row operations R3 − R2 followed by R3 − R1 eliminates the third row
entirely, while the row operations R2 −4R1 then − 51 R2 gives the row echelon
form of the matrix






1 2 3 4
1 2
3
4
1 2 3 4
4 3 2 1 → 0 −5 −10 −15 → 0 1 2 3 .
0 0 0 0
0 0
0
0
0 0 0 0
MATH 2030: MIDTERM
3
One more row operation reduces this to reduced row echelon form, R1 −2R2
:


1 0 −1 −2
0 1 2
3 .
0 0 0
0
 
 
 
1
1
2
Q.5. Let p = 2 , u =  1 , and v = 1; describe all points Q = (a, b, c)
3
−1
0
such that the line through Q with direction vector v (i.e., x2 = q + tv where we
denote the vector in standard position with head at Q as q.) - intersects the line
with equations x1 = p + su. That is find values of a, b, c such that x1 = x2 or in
vector form
p + su = q + tv
• Simplify this vector equation and expand it to derive the parametric equations, that is, find the linear system in the variables s and t
• Determine the augmented matrix for this new linear system.
• Use row operations to bring this matrix into reduced row echelon form.
Assuming this system is consistent (i.e. the lines do intersect), identify the
condition for a, b, c in the form of one linear equation.
A.5.
• Rewriting this, we have su − tv = p − q, expanding this we find
s − 2t = 1 − a, s − t = 2 − b, −s = 3 − c
• The augmented system in this case will be


1 −2 1 − a
 1 −1 2 − b 
−1 0 3 − c
• The row operations R2 − R1 and R2 + R1 clears the first column below the
first pivot.


1 −2
1−a
 0 1 1 − b + a ;
0 −2 4 − c − a
to clear the second column, below and above the second pivot in the second
row we apply R1 + 2R2 and R3 + 2R2 to get


3 − 2b + a
1 0
 0 1
;
1−b+a
0 0 6 − c − 2b + a
for this system to be consistent, so that the two lines intersect, the values
a, b, c must satisfy the equation of a plane a − 2b − c = −6.
4
MATH 2030: MIDTERM
Q.6. a) State the definition for a set of n vectors {v1 , v2 , ..., vn } to be linearly
dependent.
b) In both cases, state whether the sets of vectors are linearly independent or
dependent, using an appropriate theorem, by inspection or as a last resort - by
direct calculation. If dependent, find the dependence relation between the vectors.
•
       
2
5
9
3
−3 , 1 , −6 , 1
6
4
−5
5
•
       
5
0
0
0
 0  0 1 6
 ,  ,  ,  
 0  4 3 0
2
4
1
12
A.6. a) A set of n vectors are linearly dependent if there is some non-trivial solution
ct = [c1 , ..., cn ] 6= 0 such that
c1 v1 + c2 v2 + ... + cn vn = 0
if c = 0 we say the vectors are linearly independent.
b)
• In this case there are four vectors in R3 , as the standard basis consists of
three vectors, one of these must be a linear combination of the other three,
forming the augmented matrix and row reducing we find




77
1 0 0
5 2
9 3
53
292 
 1 −3 −6 1  →  0 1 0
53
4 6 −5 5
0 0 1 − 379
53
• Here there are four vectors from R4 , due to the upper-triangular form of
the matrix components we may use vector algebra to simplify and produce
the standard basis, hence the vectors are linearly independent.