Which of the following symbols represent vectors?
(Zero or more options can be correct)

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be True.

For example, choice (e) should be False.

For example, choice (f) should be False.

*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 $\overrightarrow{AB}$.

*There is at least one mistake.*

For example, choice (c) should be False.

This is the length, or
magnitude, of the vector $\mathbf{v}$.

*There is at least one mistake.*

For example, choice (d) should be True.

The
difference of two vectors is a vector.

*There is at least one mistake.*

For example, choice (e) should be False.

As
$\mathbf{u}$ is a vector and
$\left|\right|\mathbf{v}\left|\right|$ is a scalar (namely,
the length of $\mathbf{v}$),
this expression does not make sense.

*There is at least one mistake.*

For example, choice (f) should be False.

As
$\overrightarrow{AB}$ is a vector, this is
minus the length of $\overrightarrow{AB}$
and so this is a scalar, not a vector.

*Correct!*

*True**False*This is the length, or magnitude, of the vector $\overrightarrow{AB}$.*False*This is the length, or magnitude, of the vector $\mathbf{v}$.*True*The difference of two vectors is a vector.*False*As $\mathbf{u}$ is a vector and $\left|\right|\mathbf{v}\left|\right|$ is a scalar (namely, the length of $\mathbf{v}$), this expression does not make sense.*False*As $\overrightarrow{AB}$ is a vector, this is minus the length of $\overrightarrow{AB}$ and so this is a scalar, not a vector.

In which of the following cases is the length of
$\mathbf{a}+\mathbf{b}$ strictly smaller
than the length of $\mathbf{a}-\mathbf{b}$
? (Hint: do this question by drawing diagrams!)
(Zero or more options can be correct)

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be True.

*There is at least one mistake.*

For example, choice (a) should be False.

The parallelogram rule for vector addition shows that when
$\mathbf{a}$ and
$\mathbf{b}$
are placed tail to tail, the diagonals of the parallelogram are
$\mathbf{a}+\mathbf{b}$ and
$\mathbf{a}-\mathbf{b}$. With the vectors
$\mathbf{a},\phantom{\rule{0.3em}{0ex}}\mathbf{b}$ as shown here,
we see that $\left|\right|\mathbf{a}+\mathbf{b}\left|\right|>\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.

*There is at least one mistake.*

For example, choice (b) should be False.

Since $\mathbf{a}+\mathbf{b}$
and $\mathbf{a}-\mathbf{b}$
are vectors which form the diagonals of a rectangle in this particular case, we see that
$\left|\right|\mathbf{a}+\mathbf{b}\left|\right|=\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.

*There is at least one mistake.*

For example, choice (c) should be False.

Since $\mathbf{a}+\mathbf{b}$
and $\mathbf{a}-\mathbf{b}$
are vectors which form the diagonals of a rectangle in this particular case, we see that
$\left|\right|\mathbf{a}+\mathbf{b}\left|\right|=\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.

*There is at least one mistake.*

For example, choice (d) should be True.

The parallelogram rule for vector addition shows that when
$\mathbf{a}$ and
$\mathbf{b}$
are placed tail to tail, the diagonals of the parallelogram are
$\mathbf{a}+\mathbf{b}$ and
$\mathbf{a}-\mathbf{b}$. In this case, after
re-drawing $\mathbf{b}$ so that its
tail is at the tail of $\mathbf{a}$,
we see that $\left|\right|\mathbf{a}+\mathbf{b}\left|\right|<\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.

*Correct!*

*False*The parallelogram rule for vector addition shows that when $\mathbf{a}$ and $\mathbf{b}$ are placed tail to tail, the diagonals of the parallelogram are $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$. With the vectors $\mathbf{a},\phantom{\rule{0.3em}{0ex}}\mathbf{b}$ as shown here, we see that $\left|\right|\mathbf{a}+\mathbf{b}\left|\right|>\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.*False*Since $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are vectors which form the diagonals of a rectangle in this particular case, we see that $\left|\right|\mathbf{a}+\mathbf{b}\left|\right|=\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.*False*Since $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are vectors which form the diagonals of a rectangle in this particular case, we see that $\left|\right|\mathbf{a}+\mathbf{b}\left|\right|=\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.*True*The parallelogram rule for vector addition shows that when $\mathbf{a}$ and $\mathbf{b}$ are placed tail to tail, the diagonals of the parallelogram are $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$. In this case, after re-drawing $\mathbf{b}$ so that its tail is at the tail of $\mathbf{a}$, we see that $\left|\right|\mathbf{a}+\mathbf{b}\left|\right|<\left|\right|\mathbf{a}-\mathbf{b}\left|\right|$.

Given that $\mathbf{u}$ is a
vector of length 2, $\mathbf{v}$
is a vector of length 3 and the angle between them when placed tail to tail is
$4{5}^{\circ}$, which option is closest
to the exact value of $\mathbf{u}\cdot \mathbf{v}$
?
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

Since $\mathbf{u}\cdot \mathbf{v}=\left|\right|\mathbf{u}\left|\right|\phantom{\rule{0.3em}{0ex}}\left|\right|\mathbf{v}\left|\right|cos\theta $,
where $\theta $
is the angle between the vectors when placed tail to tail, we have
$\mathbf{u}\cdot \mathbf{v}=2\times 3\times cos4{5}^{\circ}\approx 4.24.$

*Choice (d) is incorrect*

What is the approximate angle between
$\mathbf{a}$ and
$\mathbf{b}$ if
$\mathbf{a}\cdot \mathbf{b}=3$,
$\left|\right|\mathbf{a}\left|\right|=2$, and
$\left|\right|\mathbf{b}\left|\right|=2.6$?
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

If $\theta $ is the required
angle, then $cos\theta =\frac{\mathbf{a}\cdot \mathbf{b}}{\left|\right|\mathbf{a}\left|\right|\phantom{\rule{0.3em}{0ex}}\left|\right|\mathbf{b}\left|\right|}=\frac{3}{5.2}\approx 0.577$
and hence $\theta \approx 0.955$
radians.

Find non-zero scalars $\alpha $,
$\beta $ such that for
all vectors $\mathbf{a}$
and $\mathbf{b}$,
$$\alpha \left(\mathbf{a}+2\mathbf{b}\right)-\beta \mathbf{a}+\left(4\mathbf{b}-\mathbf{a}\right)=\mathbf{0}.$$
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

If the vector equation is simplified, we get
$$\left(\alpha -\beta -1\right)\mathbf{a}+\left(2\alpha +4\right)\mathbf{b}=\mathbf{0}.$$
Since this holds for all $\mathbf{a}$
and $\mathbf{b}$, it will
hold when $\mathbf{a}$ and
$\mathbf{b}$ are set equal to
$\mathbf{0}$ in turn. This gives
the two conditions $\alpha -\beta -1=0$
and $2\alpha +4=0$, whose
solution is $\alpha =-2,\phantom{\rule{0.3em}{0ex}}\beta =-3.$

*Choice (c) is incorrect*

*Choice (d) is incorrect*

The two vectors $\mathbf{a}$
and $\mathbf{b}$ are
perpendicular. If $\mathbf{a}$
has length $8$
and $\mathbf{b}$ has
length $3$
what is $\left|\right|\mathbf{a}-2\mathbf{b}\left|\right|$?
Enter your answer into the answer box.

*Correct!*

The vectors $\mathbf{a}$,
$-2\mathbf{b}$, and
$\mathbf{a}-2\mathbf{b}$
form the sides of a right-angled triangle, with sides of length
$8$,
$6$ and hypotenuse of
length $\left|\right|\mathbf{a}-2\mathbf{b}\left|\right|$. Therefore by
Pythagoras’ Theorem, $\left|\right|\mathbf{a}-2\mathbf{b}\left|\right|=\sqrt{{8}^{2}+{6}^{2}}=10.$

*Incorrect.*

*Please try again.*

Try drawing a diagram of the vectors $\mathbf{a},\phantom{\rule{0.3em}{0ex}}-2\mathbf{b}$
and $\mathbf{a}-2\mathbf{b}$
and then use Pythagoras’ Theorem.

A boat sails 5 km south-east then 3 km due west. Approximately how far is it from
its starting position?
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

The boat’s journey can be represented by the following diagram, where the origin
$O$ is taken to be the
starting position and $Q$
is the finishing position.
The required distance from the starting position is then
$\left|\right|\overrightarrow{OQ}\left|\right|$.
Now $\left|\right|\overrightarrow{OP}\left|\right|=5,$
and since initially the boat sails in a south-easterly direction, the coordinates of
$P$ must be
$\left(\frac{5}{\sqrt{2}},-\frac{5}{\sqrt{2}}\right)$. Since
$\left|\right|\overrightarrow{PQ}\left|\right|=3$ and
$Q$ is due west
of $P$, the
coordinates of $Q$
must be $\left(\frac{5}{\sqrt{2}}-3,-\frac{5}{\sqrt{2}}\right)$.
Hence the required distance from the origin (by Pythagoras’ Theorem) is
$\sqrt{{\left(5\u2215\sqrt{2}-3\right)}^{2}+{\left(-5\u2215\sqrt{2}\right)}^{2}}\approx 3.6$
km.

Which of the following expressions make sense? (There may be more than one. Note that
$\mathbf{u}\cdot \mathbf{v}$ represent dot
product while $\mathbf{u}\times \mathbf{v}$
represent cross product.) (Zero or more options can be correct)

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be True.

For example, choice (d) should be True.

For example, choice (e) should be False.

*There is at least one mistake.*

For example, choice (a) should be False.

This is
meaningless. The second bracket is a scalar quantity and we can’t take a cross product of a vector
with a scalar.

*There is at least one mistake.*

For example, choice (b) should be False.

This is meaningless. The cross product is defined between two vectors, not two
scalars.

*There is at least one mistake.*

For example, choice (c) should be True.

This is a dot product of two vectors and the end quantity is a scalar.

*There is at least one mistake.*

For example, choice (d) should be True.

This is a vector since it is a scalar multiple of the vector
$\mathbf{v}\times \mathbf{w}.$

*There is at least one mistake.*

For example, choice (e) should be False.

This
is meaningless. We can’t add a vector to a scalar.

*Correct!*

*False*This is meaningless. The second bracket is a scalar quantity and we can’t take a cross product of a vector with a scalar.*False*This is meaningless. The cross product is defined between two vectors, not two scalars.*True*This is a dot product of two vectors and the end quantity is a scalar.*True*This is a vector since it is a scalar multiple of the vector $\mathbf{v}\times \mathbf{w}.$*False*This is meaningless. We can’t add a vector to a scalar.

Find the vector $\mathbf{u}\times \mathbf{v}$
when $\mathbf{u}=\left[3,-1,1\right]$ and
$\mathbf{v}=\left[2,5,1\right].$ Exactly one option
must be correct)

*Choice (a) is correct!*

*Choice (b) is incorrect*

Recall
that if $\mathbf{u}=\left[{u}_{1},{u}_{2},{u}_{3}\right]$
and $\mathbf{v}=\left[{v}_{1},{v}_{2},{v}_{3}\right]$
then $\mathbf{u}\times \mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$

*Choice (c) is incorrect*

Recall
that if $\mathbf{u}=\left[{u}_{1},{u}_{2},{u}_{3}\right]$
and $\mathbf{v}=\left[{v}_{1},{v}_{2},{v}_{3}\right]$
then $\mathbf{u}\times \mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$

*Choice (d) is incorrect*

Recall
that if $\mathbf{u}=\left[{u}_{1},{u}_{2},{u}_{3}\right]$
and $\mathbf{v}=\left[{v}_{1},{v}_{2},{v}_{3}\right]$
then $\mathbf{u}\times \mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$

Find the vector $\mathbf{u}\times \mathbf{v}$
when $\mathbf{u}=\left[3,4,6\right]$ and
$\mathbf{v}=\left[0,1,1\right].$ Exactly one option
must be correct)
Recall that if $\mathbf{u}=\left[{u}_{1},{u}_{2},{u}_{3}\right]$
and $\mathbf{v}=\left[{v}_{1},{v}_{2},{v}_{3}\right]$
then $\mathbf{u}\times \mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
Recall that if $\mathbf{u}=\left[{u}_{1},{u}_{2},{u}_{3}\right]$
and $\mathbf{v}=\left[{v}_{1},{v}_{2},{v}_{3}\right]$
then $\mathbf{u}\times \mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
Recall
that if $\mathbf{u}=\left[{u}_{1},{u}_{2},{u}_{3}\right]$
and $\mathbf{v}=\left[{v}_{1},{v}_{2},{v}_{3}\right]$
then $\mathbf{u}\times \mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

*Choice (d) is incorrect*