# Quiz 1: Vectors

Question

## Question 1

Which of the following expressions represent vectors ?
 a) $v$ b) AB c) $|v|$ d) $\stackrel{⃗}{AB}+3|v|$ e) $-\stackrel{⃗}{PQ}$ f) $2u+\stackrel{⃗}{AB}$

There is at least one mistake.
For example, choice (a) should be true.
There is at least one mistake.
For example, choice (b) should be false.
This is the length, or magnitude, of the vector $\stackrel{⃗}{AB}$.
There is at least one mistake.
For example, choice (c) should be false.
This is the length, or magnitude, of the vector $v$.
There is at least one mistake.
For example, choice (d) should be false.
Although $\stackrel{⃗}{AB}$ is a vector, $|mathbfv|$ represents the length, or magnitude, of $v$.
There is at least one mistake.
For example, choice (e) should be true.
This is the vector pointing in the opposite direction to $\stackrel{⃗}{PQ}$.
There is at least one mistake.
For example, choice (f) should be true.
both $u$ and $\stackrel{⃗}{AB}$ are vectors, so their sum is also a vector.
1. True.
2. False. This is the length, or magnitude, of the vector $\stackrel{⃗}{AB}$.
3. False. This is the length, or magnitude, of the vector $v$.
4. False. Although $\stackrel{⃗}{AB}$ is a vector, $|mathbfv|$ represents the length, or magnitude, of $v$.
5. True. This is the vector pointing in the opposite direction to $\stackrel{⃗}{PQ}$.
6. True. both $u$ and $\stackrel{⃗}{AB}$ are vectors, so their sum is also a vector.

## Question 2

In the notation used in Math1002, which of the following expressions represent vectors ?
 a) $-v$ b) $u+v$ c) $u+|v|$ d) $-|\stackrel{⃗}{AB}|$ e) $2u-3v$

There is at least one mistake.
For example, choice (a) should be true.
Minus a vector is a vector.
There is at least one mistake.
For example, choice (b) should be true.
The sum of two vectors is a vector.
There is at least one mistake.
For example, choice (c) should be false.
As $u$ is a vector and $|v|$ is a number (namely,m the length of $v$), this expression does not even make sense.
There is at least one mistake.
For example, choice (d) should be false.
As $\stackrel{⃗}{AB}$ is a vector, this is minus the length of $\stackrel{⃗}{AB}$.
There is at least one mistake.
For example, choice (e) should be true.
Any linear combination of vectors is again a vector.
1. True. Minus a vector is a vector.
2. True. The sum of two vectors is a vector.
3. False. As $u$ is a vector and $|v|$ is a number (namely,m the length of $v$), this expression does not even make sense.
4. False. As $\stackrel{⃗}{AB}$ is a vector, this is minus the length of $\stackrel{⃗}{AB}$.
5. True. Any linear combination of vectors is again a vector.

## Question 3

How many different vectors are drawn here ?

Vectors are equal when they have the same direction and length, their position in space does not matter.
Not correct. You may try again.
Remember that a vector is specified by its direction and magnitude, so that the two arrows of equal length pointing to the west represent the same vector, while the three arrows of equal length pointing to the north-east also represent the same vector.

## Question 4

How many different vectors are drawn here ?

All of the vectors are different. However, some of these vectors are multiples of each other; for example,

Not correct. You may try again.
Vectors are equal when they have the same direction and length, their position in space does not matter.

## Question 5

Express the vector $u$ in terms of $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c$.
 a) $-a+b+c$ b) $a-b+c$ c) $a+b+c$ d) $-a-b-c$

Not correct. Choice (a) is false.
Trace out the vector $u$ starting at the tail and moving along the vectors $a$, $b$ and $c$.
Not correct. Choice (b) is false.
Trace out the vector $u$ starting at the tail and moving along the vectors $a$, $b$ and $c$.
Not correct. Choice (c) is false.
Trace out the vector $u$ starting at the tail and moving along the vectors $a$, $b$ and $c$.
Starting at the tail of $u$, we see that $u=-a-b-c$.

## Question 6

Express the vector $u$ in terms of $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c$.
 a) $-a+b+c$ b) $a-b-c$ c) $a+b+c$ d) $-a-b-c$

Not correct. Choice (a) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 7

Find non-zero scalars $\alpha$, $\beta$ such that for all vectors $a$ and $b$, $\alpha \left(a+2b\right)-\beta a+\left(4b-a\right)=0.$
 a) $\alpha =2,\phantom{\rule{0.3em}{0ex}}\beta =1$ b) $\alpha =-2,\phantom{\rule{0.3em}{0ex}}\beta =-3$ c) $\alpha =1,\phantom{\rule{0.3em}{0ex}}\beta =3$ d) $\alpha =-2,\phantom{\rule{0.3em}{0ex}}\beta =3$

Not correct. Choice (a) is false.
If the vector equation is simplified, we get $\left(\alpha -\beta -1\right)a+\left(2\alpha +4\right)b=0.$ Since this holds for all $a$ and $b$, it will hold when $a$ and $b$ are set equal to $0$ in turn. This then gives two conditions $\alpha -\beta -1=0$ and $2\alpha +4=0$, whose solution is $\alpha =-2,\phantom{\rule{0.3em}{0ex}}\beta =-3.$
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 8

Find non-zero scalars $\alpha$, $\beta$ such that for all vectors $a$ and $b$, $\alpha \left(2a-b\right)-\beta \left(a+2b\right)=4a-b.$
 a) $\alpha =\frac{7}{3},\phantom{\rule{0.3em}{0ex}}\beta =-\frac{2}{3}$ b) $\alpha =-\frac{7}{3},\phantom{\rule{0.3em}{0ex}}\beta =\frac{2}{3}$ c) $\alpha =\frac{9}{5},\phantom{\rule{0.3em}{0ex}}\beta =-\frac{2}{5}$ d) $\alpha =\frac{9}{5},\phantom{\rule{0.3em}{0ex}}\beta =\frac{2}{5}$

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
We have two equations $2\alpha -\beta =4,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}-\alpha -2\beta =-1.$ The substitution $\alpha =1-2\beta$ implies $2\left(1-2\beta \right)-\beta =5$ and so $\beta =-\frac{2}{5}$, $\alpha =1+\frac{4}{5}=\frac{9}{5}$.
Not correct. Choice (d) is false.

## Question 9

The two vectors $a$ and $b$ are perpendicular. If $a$ has magnitude $8$ and $b$ has magnitude $3$ what is $|a-2b|$?

The vectors $a$, $-2b$, and $a-2b$ form the sides of a right-angled triangle, with sides of length $8$, $6$ and hypotenuse of length $|a-2b|$. Therefore by Pythagoras’ Theorem, $|a-2b|=\sqrt{{8}^{2}+{6}^{2}}=10.$
Not correct. You may try again.
Try drawing a diagram!

## Question 10

In which of the following cases is the length of $a+b$ strictly smaller than the length of $a-b$ ?
 a) b) c) d)

There is at least one mistake.
For example, choice (a) should be false.
The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|>|a-b|$.
There is at least one mistake.
For example, choice (b) should be false.
The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|=|a-b|$.
There is at least one mistake.
For example, choice (c) should be false.
The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|=|a-b|$.
There is at least one mistake.
For example, choice (d) should be true.
The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|<|a-b|$.
1. False. The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|>|a-b|$.
2. False. The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|=|a-b|$.
3. False. The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|=|a-b|$.
4. True. The parallelogram rule for vector addition shows that when $a$ and $b$ are placed tail to tail, the diagonals of the parallelogram are $a+b$ and $a-b$. Hence, in this case, $|a+b|<|a-b|$.