MATH1131/1141 Algebra Test 1 2014 S1 v1a · Since neither of these vectors are a scalar multiple of...

MATH1131/1141 Algebra Test 1 2014 S1 v1a

March 27, 2017

These solutions were written and typed up by Harry Gibbs and edited by Henderson Koh,

Vishaal Nathan, Dominic Palanca, with further edits and augmentation provided by Aaron

Hassan. Please be ethical with this resource. It is for the use of MathSoc members, so do

not repost it on other forums or groups without asking for permission. If you appreciate this

resource, please consider supporting us by coming to our events and buying our T-shirts! Also,

happy studying :).

We cannot guarantee that our working is correct, or that it would obtain full marks – please

notify us of any errors or typos at [email protected], or on our Facebook page. There

are often multiple methods of solving the same question. Remember that in the real class test,

you will be expected to explain your steps and working out.

1. (i) The line AB passes through the point A and has a gradient vector which is parallel

to−→B −

−→A (where

−→P refers to the coordinate vector of a point P ). Therefore, a

parametric vector equation for the line is

x =−→A + λ

(−→B −

−→A)

=

4

2

3

+ λ

5− 4

−7− 2

−2− 3

i.e. x =

4

2

3

+ λ

1

−9

−5

,

where λ ∈ R.

(ii) If C lies on the line AB, then there must exist λ ∈ R such that−→C =

−→A+λ

(−→B −

−→A)

.

1

That is,

−→C =

−→A + λ

(−→B −

−→A)

⇐⇒

7

−25

−10

=

4

2

3

+ λ

1

−9

−5

⇐⇒

3

−27

−13

= λ

1

−9

−5

.

However, it is clear that no λ ∈ R can satisfy the above equation (equating first

components shows λ = 3, but if λ = 3, then the third components do not match).

So C does not lie on AB.

2. Let λ = x2 and µ = x3. Solving for x1 we obtain

x1 =7 + 5x2 − x3

2=

7 + 5λ− µ2

.

Hence, x1x2x3

=

7+5λ−µ

2

λ

µ

=

72

0

0

+ λ

52

1

0

+ µ

−12

0

1

,

which is a parametric vector equation for the plane. So, the two direction vectors,

52

1

0

and

−12

0

1

are parallel to the plane. Furthermore, it is clear that they are non-zero and

non-parallel as either is not a scalar multiple of the other (in fact, this is guaranteed to

happen when doing this method in general). Hence,52

1

0

and

−12

0

1

,

are two non-parallel non-zero vectors that are parallel to the plane.

Note. Here we isolated x1 at the start and set x2 and x3 to free parameters. You

could instead isolate a different variable (x2 or x3) and set the remaining two variables

to free parameters, and you would then obtain a different, but equally correct, answer

(remember, the parametric vector equation of a plane is not unique). In fact, choosing x3

2

as the variable to isolate at the start would allow you to completely avoid having fractions

pop up, as x3 has a coefficient of 1 in the Cartesian equation.

3. (i) Recall that the cross product of two vectors in R3, say u =

u1u2u3

and v =

v1v2v3

,

is given by

u× v =

u2v3 − u3v2u3v1 − u1v3u1v2 − u2v1

.

Hence,

−−→AB ×

−→AC =

(−→B −

−→A)×(−→C −

−→A)

=

2

2

−2

×2

1

1

=

2× 1− (−2)× 1

(−2)× 2− 2× 1

2× 1− 2× 2

=

4

−6

−2

.

Alternatively, one can find

2

2

−2

×2

1

1

by evaluating the formal determinant

∣∣∣∣∣∣∣ı̂ ̂ k̂

2 2 −2

2 1 1

∣∣∣∣∣∣∣where ı̂ =

1

0

0

, ̂ =

0

1

0

and k̂ =

0

0

1

are the standard basis vectors, and arrive

at the same answer. In fact, in practice most people would do it this way (as the

3

cross product formula is hard to remember), so here is how it would be done: 2

2

−2

×2

1

1

=

∣∣∣∣∣∣∣ı̂ ̂ k̂

2 2 −2

2 1 1

∣∣∣∣∣∣∣= ı̂

∣∣∣∣∣2 −2

1 1

∣∣∣∣∣− ̂∣∣∣∣∣2 −2

2 1

∣∣∣∣∣+ k̂

∣∣∣∣∣2 2

2 1

∣∣∣∣∣= (2 + 2) ı̂− (2 + 4) ̂+ (2− 4) k̂ = 4ı̂− 6̂− 2k̂

=

4

−6

−2

.X

(ii) The 4ABC has half the area of the parallelogram formed by the vectors−−→AB and

−→AC. The area of the parallelogram is given by

∥∥∥−−→AB ×−→AC∥∥∥, hence

Area of 4ABC =1

2

∥∥∥−−→AB ×−→AC∥∥∥ =√

14,

since

∥∥∥−−→AB ×−→AC∥∥∥ =

∥∥∥∥∥∥∥ 4

−6

−2

∥∥∥∥∥∥∥

=√

16 + 36 + 4

=√

56

= 2√

14.

4. (i) The projection of a vector, a, onto another (non-zero) vector, v, is given by

projb (a) =a · b‖b‖2

b.

Substituting given data yields

projv

(−−→PQ)

=

(−→Q −

−→P)· v

‖v‖2v.

Note that−→Q −

−→P =

1− 1

4− 2

4− 3

=

0

2

1

. So(−→Q −

−→P)· v =

0

2

1

·2

3

1

= 0 × 2 +

4

2× 3 + 1× 1 = 7. Moreover, ‖v‖2 = 22 + 32 + 12 = 4 + 9 + 1 = 14. Hence

projv

(−−→PQ)

=

(−→Q −

−→P)· v

‖v‖2v =

7

14v

=1

2

2

3

1

=

13212

.

(ii) Drawing a rough picture is a good idea for this question.

By construction, the vector that minimises the distance from Q to ` is given by

−−→PQ− projv

(−−→PQ)

. After calculating that−−→PQ− projv

(−−→PQ)

=

−11212

, we find that

d =∥∥∥−−→PQ− projv

(−−→PQ)∥∥∥

=

∥∥∥∥∥∥∥−1

1212

∥∥∥∥∥∥∥

=

√1 +

1

4+

1

4

=

√3

2.

An alternative method is to use Pythagoras’ Theorem, and so d2 =

√∥∥∥−−→PQ∥∥∥2 − ∥∥∥projv

(−−→PQ)∥∥∥2.

5

MATH1131/1141 Algebra Test 1 2014 S1

v1bMarch 27, 2017

These solutions were originally written and typed up by Harry Gibbs, and edited provided by

Henderson Koh, Vishaal Nathan, Aaron Hassan and Dominic Palanca. Please be ethical with

this resource. It is for the use of MathSoc members, so do not repost it on other forums or

groups without asking for permission. If you appreciate this resource, please consider supporting

us by coming to our events and buying our T-shirts! Also, happy studying :).





1. The 3 points are collinear if and only if the line segment−−→AB is parallel to the line segment

−→AC. But

−−→AB =

−→B −−→A =

5− 3

−4− 5

3− 7

=

2

−9

−4

and−→AC =

−→C −−→A =

−5− 3

41− 5

22− 7

=

−8

36

15

.

Since neither of these vectors are a scalar multiple of the other, they are not parallel, and

hence the 3 points are not collinear.

2. To find the parametric vector equation of a plane we need two non-parallel direction

vectors and a fixed vector, which lies on the plane. The point A must be on the plane, so

we shall take that to be the fixed vector. As long as−−→AB is not parallel to

−→AC we can take

our two direction vectors to be−−→AB and

−→AC. (Note that if these two vectors were parallel,

then A,B,C would be collinear and there wouldn’t be a unique plane through A,B and

6

C.) First we calculate−−→AB and

−→AC and find that

−−→AB =

−→B −

−→A =

3− 1

4− 2

2− 1

=

2

2

1

and−→AC =

−→C −

−→A =

5− 1

2− 2

1− 1

=

4

0

0

.

Since−−→AB and

−→AC are clearly not parallel, we can take them as our direction vectors.

Hence, for λ1, λ2 ∈ R

x =

1

2

1

+ λ1

2

2

1

+ λ2

4

0

0

is a parametric vector equation for the plane through A,B and C.

Note. Remember, there are multiple possible parametric vector equations for a given

plane, so you may get a different answer to this one if you used different direction vectors.

3. (i) The distance between A and B is given by∥∥∥−−→AB∥∥∥. Hence

d(A,B) =∥∥∥−−→AB∥∥∥ =

√(5− 1)2 + (6− 2)2 + (4− 3)2 =

√16 + 16 + 1 =

√33

(ii) The projection of a vector, a, onto another (non-zero) vector, b, is given by

projb (a) =a · b‖b‖2

b.

Therefore, if a · b = 0 (i.e. the two vectors are orthogonal), we see that projb (a) =

0b = 0.

Now, observe that

−−→AB =

−→B −

−→A

=

5

6

4

−1

2

3

=

4

4

1

.

7

Also,

−→AC =

−→C −

−→A

=

2

1

3

−1

2

3

=

1

−1

0

Thus

−−→AB ·

−→AC =

4

4

1

· 1

−1

0

= 4− 4

= 0.

Hence by our earlier comment about orthogonal vectors having zero vector as the

projection, we have proj−→AC

(−−→AB)

= 0 (the zero vector in R3, i.e. (0, 0, 0)T ).

4. First we derive the parametric vector equations for the lines AY and BX.

First, the line AY . The point Y is given by 23B, since Y is on the line we can take it to

be our fixed vector. We can take−→A −

−→Y as our direction vector. Hence, for λ ∈ R, the

parametric vector equation is

x =−→Y + λ

(−→A −

−→Y)

=2

3

0

−3

−6

+ λ

4−(23(0)

)−4−

(23(−3)

)8−

(23(−6)

)

=

0

−2

−4

+ λ

4

−2

12

.

Second, the line BX. The method is the same as above, hence, for µ ∈ R, the parametric

8

vector equation is

x =−→X + µ

(−→B −

−→X)

=

3

−3

6

+ µ

−3

0

−12

.

We will show that P lies on both the lines AY and BX. This implies that P is the

intersection of AY and BX as the lines only intersect once. Hence we will solve P =

X +µ(B−X) and P = Y +λ(A−Y ) for µ and λ (a solution for µ and λ will exist if and

only if P lies on the respective line).

−→P =

−→X + µ

(−→B −

−→X)

2

−3

2

=

3

−3

6

+ µ

−3

0

−12

−1

0

−4

= µ

−3

0

−12

=⇒ µ =1

3.

As there is a solution for µ, P lies on BX.

−→P =

−→Y + λ

(−→A −

−→Y)

2

−3

2

=

0

−2

−4

+ λ

4

−2

12

2

−1

6

= λ

4

−2

12

=⇒ λ =1

2

As there is a solution for λ, P lies on AY . Hence P is the intersection of AY and BX.

9

MATH1131/1141 Algebra Test 1 2014 S1

v2bMarch 27, 2017

These hints were originally written and typed up by Brendan Trinh, and edited by Henderson

Koh, Vishaal Nathan and Dominic Palanca, with further edits and the worked sample solutions

written and typed up by Aaron Hassan. Please be ethical with this resource. It is for the

use of MathSoc members, so do not repost it on other forums or groups without asking for

permission. If you appreciate this resource, please consider supporting us by coming to our

events and buying our T-shirts! Also, happy studying :).

We cannot guarantee that our answers are correct – please notify us of any errors or typos

at [email protected], or on our Facebook page. There are often multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

1. Hints

(i) Let λ = x−23 , λ = y+1

4 and λ = z+31 . Make x, y and z the subject for each equation(∗)

and you should be able to arrive at the following parametric equation

x =

xyz

=

2

−1

−3

+ λ

3

4

1

, where λ ∈ R.

Alternatively, just do what the worked sample solution does.

(ii) One way is to find the parametric equation for Π 1, equate this equation with the

equation found in part (i) and then solve the system of three linear equations.

1Refer to Question 2 in Test 1 2014 S1 v1a

10

The faster way is to substitute the equations found earlier marked (*) into the Carte-

sian equation of Π. You should find that

P = (11, 11, 0) .

Worked Sample Solution

(i) Recall that if a line has Cartesian equation

x− av1

=y − bv2

=z − cv3

,

then it has a parametric vector form

x =

abc

+ λ

v1v2v3

.

So by inspection, a parametric vector equation for ` is

x =

xyz

=

2

−1

−3

+ λ

3

4

1

, where λ ∈ R.

(ii) On `, we have x = 2 + 3λ, y = −1 + 4λ, z = −3 +λ, for λ ∈ R. Substitute these into

the plane equation to obtain the value of λ at the point of intersection (a solution

for λ will exist if and only if the line has an intersection with the plane):

3 (2 + 3λ)− 2 (−1 + 4λ)− 4 (−3 + λ) = 11

=⇒ 6 + 9λ+ 2− 8λ+ 12− 4λ = 11

=⇒ −3λ = −9

=⇒ λ = 3.

Substitute into the line’s parametric equation to find the point of intersection:

x =

2

−1

−3

+ 3

3

4

1

=

11

11

0

.

Hence the coordinates of the point P are (11, 11, 0).

2. Hints

11

(i) It can be found that2

−−→AB ×

−→AC =

4

−2

5

.

(ii) Area of a parallelogram spanned by two vectors is the Euclidean norm of the cross

product of them, i.e.∥∥∥−−→AB ×−→AC∥∥∥. It can be found that2

A = 3√

5 units2.

Worked Sample Solutions

(i) Note that

−−→AB =

−→B −

−→A

=

3

1

−1

−1

2

1

=

2

−1

−2

.

Also,

−→AC =

−→C −

−→A

=

2

4

1

−1

2

1

=

1

2

0

.


12

Thus

−−→AB ×

−→AC =

∣∣∣∣∣∣∣ı̂ ̂ k̂

2 −1 −2

1 2 0

∣∣∣∣∣∣∣=

∣∣∣∣∣−1 −2

2 0

∣∣∣∣∣ ı̂−∣∣∣∣∣2 −2

1 0

∣∣∣∣∣ ̂+

∣∣∣∣∣2 −1

1 2

∣∣∣∣∣ k̂= 4ı̂− 2̂+ 5k̂

=

4

−2

5

.

(ii) The area of the parallelogram is the norm of the relevant cross product, i.e. it is

∥∥∥−−→AB ×−→AC∥∥∥ =

∥∥∥∥∥∥∥ 4

−2

5

∥∥∥∥∥∥∥

=√

16 + 4 + 25

=√

45

= 3√

5.

3. Hints

(i) It can be found that3

projv

(−−→PQ)

=

3212

1

.

(ii) The shortest distance between a point and a line is the perpendicular distance. Form

a right angled triangle to find that3

d =

√10

2units.

(iii) This can most easily be done by drawing a picture and noticing that the vector−−→OM

is found by simply adding the vectors−−→OP and projv

(−−→PQ)

. You should find that

the co-ordinates of the point M are (5

2,5

2, 4

).


13

Worked Sample Solutions

(i) Note that

−−→PQ =

−→Q −

−→P

=

2

4

4

−1

2

3

=

1

2

1

.

Since v =

3

1

2

, we have−−→PQ · v = 3 + 2 + 2 = 7 and ‖v‖2 = 9 + 1 + 4 = 14. Thus

projv

(−−→PQ)

=

−−→PQ · v‖v‖2

v

=7

14

3

1

2

=

3212

1

(ii) By drawing a diagram and using Pythagoras’ theorem, we see that the distance d is

given by

d =

√∥∥∥−−→PQ∥∥∥2 − ∥∥∥projv

(−−→PQ)∥∥∥2

Now, since−−→PQ =

1

2

1

, we have∥∥∥−−→PQ∥∥∥2 = 1 + 4 + 1 = 6. Also, from the answer to

14

the previous question, we have that

∥∥∥projv

(−−→PQ)∥∥∥2 =

(3

2

)2

+

(1

2

)2

+ 12

=9

4+

1

4+

4

4

=14

4

=7

2.

Thus

d =

√∥∥∥−−→PQ∥∥∥2 − ∥∥∥projv

(−−→PQ)∥∥∥2 =

√6− 7

2

=

√5

2=

√10

4

=

√10

2.

15


March 27, 2017

These solutions were written and typed up by Harry Gibbs and edited by Henderson Koh,

Vishaal Nathan, Aaron Hassan and Dominic Palanca. Please be ethical with this resource. It is

for the use of MathSoc members, so do not repost it on other forums or groups without asking

for permission. If you appreciate this resource, please consider supporting us by coming to our






1. (i) The line AB passes through the point A and has a gradient vector which is parallel

to−→A −

−→B . Therefore

x =−→A + λ

(−→A −

−→B)

=

3

2

1

+ λ

−3

−1

3

where λ ∈ R.

(ii) Suppose that (x1, x2, x3) is a point on the line AB. Then, for some real number λ

we have x1x2x3

=

3

2

1

+ λ

−3

−1

3

.

16

Rewriting this equation coordinate-wise yields

x1 = 3− 3λ,

x2 = 2− λ,

x3 = 1 + 3λ.

By solving for λ we obtain

λ =3− x1

3,

λ = 2− x2,

λ =x3 − 1

3.

Thus, by eliminating λ from the above equations we obtain

x1 − 3

−3=x2 − 2

−1=x3 − 1

3,

which is the Cartesian form of the line AB.

2. It can be found that1x1x2x3

=

1−2λ+µ

7

λ

µ

=

17

0

0

+ λ

−27

1

0

+ µ

17

0

1

,

is a parametric vector equation for the plane. Also, two non-parallel, non-zero vectors

which are parallel to the plane are−27

1

0

and

17

0

1

.

3. (i) The cosine of the angle θ between two vectors, say u and v, is given by

cos θ =u · v‖u‖ ‖v‖

.

In this question, ∠BAC is the angle between the vectors−−→AB and

−→AC. Substituting

u =−→B −

−→A and v =

−→C −

−→A yields

cos (∠BAC) =

(−→B −

−→A)·(−→C −

−→A)

∥∥∥−→B −−→A∥∥∥∥∥∥−→C −−→A∥∥∥ =1√91.


17

(ii) The projection of a vector, u, onto another vector, v, is given by

projv(u) =u · vv · v

v.

Substituting u =−→B −

−→A and v =

−→C −

−→A yields

proj−→AC

(−−→AB)

=

(−→B −

−→A)·(−→C −

−→A)

(−→C −

−→A)·(−→C −

−→A) (−→C −−→A) =

1

13

4

−3

1

.

4. (i) Drawing a picture makes this question much easier.

The line OC goes through the origin and has direction parallel to the line segment−−→OC. By drawing a relevant parallelogram, it is easy to see that

−−→OC is simply the

vector addition of vectors−→OA and

−→AC. But since opposite sides of a parallelogram

are equal, then it is clear that−→AC =

−−→OB. And so, we can conclude that

−−→OC =

−→OA+

−−→OB = a + b. Hence OC has parametric vector form

x = λ(a + b),

where λ ∈ R.

The line AB goes through the point A and has direction parallel to the line segment−−→AB. The line segment

−−→AB is parallel to the vector b−a. Hence AB has parametric

vector form

x = a + µ(b− a),

where µ ∈ R. Note that there are a number of similar parameterisations for this

question, the solution listed is not the only solution.

(ii) There are a number of ways to do this question, so here are two:

1. This solution is more geometrically inclined.1 Note that the diagonals of any

parallelogram, which in this case are the vectors−−→OC and

−−→AB, bisect each other.

Hence the intersection is given by the midpoint of−−→OC, which is simply a+b

2 .

2. This solution is more algebraic and a little more involved, but it is more rigor-

ous. The intersection of the two lines can be obtained by solving the line equations

1This method was probably not intended by the examiner, due to part (iii), but is still good to have anintuitive understanding of this question.

18

simultaneously:

λ(a + b) = a + µ(b− a)

(λ+ µ− 1)a + (λ− µ)b = 0

Since a is not parallel to b, the only possible solution is one where (λ + µ − 1) =

(λ − µ) = 0. Solving for λ and µ yields the solutions λ = µ = 12 . By substituting

this into the line equations, we obtain the coordinates for P , namely a+b2 .

(iii) Calculating−−→OP and

−−→PC yields:

−−→OP =

a + b

2− 0 =

a + b

2,

−−→PC =

−−→OC −

−−→OP = a + b− a + b

2=

a + b

2.

Hence, as−−→OP =

−−→PC,

∥∥∥−−→OP∥∥∥ =∥∥∥−−→PC∥∥∥.

Calculating−→PA and

−−→PB yields:

−→PA =

−→OA−

−−→OP = a− a + b

2=

a− b

2,

−−→PB =

−−→OB −

−−→OP = b− a + b

2=

b− a

2.

Hence, as−→PA = −

−−→PB,

∥∥∥−→PA∥∥∥ =∥∥∥−−→PB∥∥∥.

This question basically showed that the diagonals of any parallelogram bisects each

other.

19


March 27, 2017

These answers were written and typed up by Brendan Trinh and edited by Henderson Koh,

Vishaal Nathan, Aaron Hassan and Dominic Palanca. Please be ethical with this resource. It is

for the use of MathSoc members, so do not repost it on other forums or groups without asking

for permission. If you appreciate this resource, please consider supporting us by coming to our


We cannot guarantee that our answers are correct – please notify us of any errors or typos

at [email protected], or on our Facebook page. There are often multiple methods of

solving the same question. Remember that in the real class test, you will be expected to explain

your steps and working out.

1. (i) The co-ordinates of t can be found by dividing the interval AB in to thirds. By

drawing a picture, it can be seen that T lies ”two-thirds of the way from A to B”.

So,

−→OT =

−→OA+

2

3

−−→AB =

1

2

3

+2

3

4

5

−5

.

And so, the coordinates t of the point T are

t =

(11

3,16

3,−1

3

).

(ii) Draw out a rough diagram to have a general idea of how the vectors look. Use the

fact that the quadrilateral was named in cyclic order, so we know through simple

20

vector addition that−−→OD =

−→OA +

−−→AD. But since, opposite sides of a parallelogram

are equal, we can write this as−−→OD =

−→OA +

−−→BC. And so, the coordinates d of the

point D are

d = (4,−8, 7) .

2. It can be found that1

x =

83

0

0

+ λ

13

1

0

+ µ

−23

0

1

, where λ, µ ∈ R

is a parametric vector equation for the plane. Also, two non-parallel non-zero vectors

parallel to the plane are 13

1

0

and

−23

0

1

.

3. By using the formula or evaluating the relevant pseudo-determinant2 5

8

−7

.

4. (i) It can be found that3 1

3/2

1/2

.

(ii) Draw a diagram to help see which vector you need to determine the shortest distance3

d =

√3

2.

(iii) Drawing a diagram will help us see that−−→OM =

−−→OP +

−−→PM . But

−−→PM is precisely the

projection of−−→PQ onto v, which we have just found in part (i). From this, it can be

found that the co-ordinates of the point M are(2,

7

2,7

2

).

1Refer to Question 2 in Test 1 2014 S1 v1a2Refer to Question 3 in Test 1 2014 S1 v1a3Refer to Question 4 in Test 1 2014 S1 v1a

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MATH1131/1141 Algebra Test 1 2014 S1 v1a · Since neither of these vectors are a scalar multiple of...

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Transcript of MATH1131/1141 Algebra Test 1 2014 S1 v1a · Since neither of these vectors are a scalar multiple of...