## MATH1014 Quizzes

Quiz 2: Length, Dot Product and Cross Product
Question 1 Questions
Which of the following symbols represent vectors? (Zero or more options can be correct)
 a) $\mathbf{v}$ b) $AB$ c) $||\mathbf{v}||$ d) $\mathbf{u}-\mathbf{v}$ e) $\mathbf{u}+||\mathbf{v}||$ f) $-||\stackrel{\to }{AB}||$

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 $\stackrel{\to }{AB}$.
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 $||\mathbf{v}||$ 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 $\stackrel{\to }{AB}$ is a vector, this is minus the length of $\stackrel{\to }{AB}$ and so this is a scalar, not a vector.
Correct!
1. True
2. False This is the length, or magnitude, of the vector $\stackrel{\to }{AB}$.
3. False This is the length, or magnitude, of the vector $\mathbf{v}$.
4. True The difference of two vectors is a vector.
5. False As $\mathbf{u}$ is a vector and $||\mathbf{v}||$ is a scalar (namely, the length of $\mathbf{v}$), this expression does not make sense.
6. False As $\stackrel{\to }{AB}$ is a vector, this is minus the length of $\stackrel{\to }{AB}$ 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)
 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 $\mathbf{a}$ 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 $||\mathbf{a}+\mathbf{b}||>||\mathbf{a}-\mathbf{b}||$.
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 $||\mathbf{a}+\mathbf{b}||=||\mathbf{a}-\mathbf{b}||$.
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 $||\mathbf{a}+\mathbf{b}||=||\mathbf{a}-\mathbf{b}||$.
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 $||\mathbf{a}+\mathbf{b}||<||\mathbf{a}-\mathbf{b}||$.
Correct!
1. 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 $||\mathbf{a}+\mathbf{b}||>||\mathbf{a}-\mathbf{b}||$.
2. 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 $||\mathbf{a}+\mathbf{b}||=||\mathbf{a}-\mathbf{b}||$.
3. 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 $||\mathbf{a}+\mathbf{b}||=||\mathbf{a}-\mathbf{b}||$.
4. 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 $||\mathbf{a}+\mathbf{b}||<||\mathbf{a}-\mathbf{b}||$.
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)
 a) 4.5 b) 6.2 c) 4.2 d) 5.1

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Since $\mathbf{u}\cdot \mathbf{v}=||\mathbf{u}||\phantom{\rule{0.3em}{0ex}}||\mathbf{v}||cos\theta$, where $\theta$ is the angle between the vectors when placed tail to tail, we have $\mathbf{u}\cdot \mathbf{v}=2×3×cos4{5}^{\circ }\approx 4.24.$
Choice (d) is incorrect
What is the approximate angle between $\mathbf{a}$ and $\mathbf{b}$ if $\mathbf{a}\cdot \mathbf{b}=3$, $||\mathbf{a}||=2$, and $||\mathbf{b}||=2.6$? Exactly one option must be correct)
 a) 1.25 radians b) 0.87 radians c) 1.32 radians d) 0.96 radians

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
If $\theta$ is the required angle, then $cos\theta =\frac{\mathbf{a}\cdot \mathbf{b}}{||\mathbf{a}||\phantom{\rule{0.3em}{0ex}}||\mathbf{b}||}=\frac{3}{5.2}\approx 0.577$ 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)
 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) $\alpha =-2,\phantom{\rule{0.3em}{0ex}}\beta =3$

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 $||\mathbf{a}-2\mathbf{b}||$? 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 $||\mathbf{a}-2\mathbf{b}||$. Therefore by Pythagoras’ Theorem, $||\mathbf{a}-2\mathbf{b}||=\sqrt{{8}^{2}+{6}^{2}}=10.$
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)
 a) 2.7 km b) 4.2 km c) 7.4 km d) 3.6 km

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 $||\stackrel{\to }{OQ}||$. Now $||\stackrel{\to }{OP}||=5,$ 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 $||\stackrel{\to }{PQ}||=3$ 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 $\sqrt{{\left(5∕\sqrt{2}-3\right)}^{2}+{\left(-5∕\sqrt{2}\right)}^{2}}\approx 3.6$ 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 $\mathbf{u}×\mathbf{v}$ represent cross product.) (Zero or more options can be correct)
 a) $\left(\mathbf{u}+\mathbf{v}\right)×\left(\mathbf{u}\cdot \mathbf{w}\right)$ b) $||\mathbf{u}||×\left(\mathbf{w}\cdot \mathbf{v}\right)$ c) $\left(\mathbf{u}×\mathbf{v}\right)\cdot \mathbf{w}$ d) $||\mathbf{u}||\left(\mathbf{v}×\mathbf{w}\right)$ e) $\mathbf{u}+\left(\mathbf{v}\cdot \mathbf{w}\right)$

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 $\mathbf{v}×\mathbf{w}.$
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!
1. 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.
2. False This is meaningless. The cross product is defined between two vectors, not two scalars.
3. True This is a dot product of two vectors and the end quantity is a scalar.
4. True This is a vector since it is a scalar multiple of the vector $\mathbf{v}×\mathbf{w}.$
5. False This is meaningless. We can’t add a vector to a scalar.
Find the vector $\mathbf{u}×\mathbf{v}$ when $\mathbf{u}=\left[3,-1,1\right]$ and $\mathbf{v}=\left[2,5,1\right].$ Exactly one option must be correct)
 a) $\left[-6,-1,17\right]$ b) $\left[-6,1,17\right]$ c) $\left[4,5,13\right]$ d) $\left[-4,-1,15\right]$

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 $\mathbf{u}×\mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
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 $\mathbf{u}×\mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
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 $\mathbf{u}×\mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
Find the vector $\mathbf{u}×\mathbf{v}$ when $\mathbf{u}=\left[3,4,6\right]$ and $\mathbf{v}=\left[0,1,1\right].$ Exactly one option must be correct)
 a) $\left[6,2,-1\right]$ b) $\left[-3,1,1\right]$ c) $\left[-2,-3,3\right]$ d) $\left[0,4,6\right]$

Choice (a) 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 $\mathbf{u}×\mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
Choice (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 $\mathbf{u}×\mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$
Choice (c) is correct!
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 $\mathbf{u}×\mathbf{v}=\left[{u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}\right].$