Exam practice

A scalar is a physical quantity that has only magnitude: it is given completely by a number and a unit. Examples: mass (50 kg), time (2 hours), temperature (30 ∘°C), distance, speed.
A vector is a physical quantity that has both magnitude and direction. Examples: displacement (5 km to the north), velocity (60 km/h towards Pokhara), force (20 N downward), acceleration, weight.

• a.Row vector: the components written in a row, x first: a = (2, −5).
• b.Column vector: the components written one above the other, x on top: $\vec{a} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$.
• c.Position vector: a vector from the origin to a point; for P(4, 1) it is OP = (4, 1).
• d.Zero vector: the vector of magnitude 0, 0 = (0, 0), such as AA.
• e.Equal vectors: the same magnitude and direction: (3, 1) and (3, 1), say AB for A(0, 0), B(3, 1) and CD for C(2, 2), D(5, 3).
• f.Negative vector: the same magnitude, the opposite direction: (3, 1) and (−3, −1).
• g.Like vectors: the same direction, one a positive multiple of the other: (1, 2) and (3, 6).
• h.Unlike vectors: opposite directions, one a negative multiple of the other: (1, 2) and (−2, −4).

In a regular hexagon each side is parallel to the side opposite it, and the long diagonal FC is parallel to AB and twice as long.

• Equal vectors (same length, same direction): AB = ED, BC = FE, CD = AF.
• Negative vectors (same length, opposite directions): AB and DE, BC and EF, CD and FA; and any side and its reverse, such as AB and BA.
• Like vectors (same direction, any lengths): AB, ED and FC, where FC = 2AB.
• Unlike vectors (opposite directions): AB and CF, or AB and DE.

SR: the opposite side of parallelogram PQRS, pointing the same way.

RT, the opposite side of parallelogram QRTS.

PS and ST: both run along the top line, parallel to QR, the same length and the same way.

TS, SP and RQ: the same length as ST, pointing the other way.

SR and PQ (they are equal to each other).

RT and QS.

An arrow may start at any point, as long as its move is right. Here each starts at a different point, so the four don’t cross.

$\overrightarrow{AB} = (2 - 4, 5 - 3) = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$, $\overrightarrow{BA} = (4 - 2, 3 - 5) = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$.
AB ≠ BA: they are negative vectors, AB = −BA.

$\overrightarrow{AB} = (5 - 6, -4 - 3) = \begin{pmatrix} -1 \\ -7 \end{pmatrix}$, $\overrightarrow{BA} = \begin{pmatrix} 1 \\ 7 \end{pmatrix}$. AB ≠ BA.

$\overrightarrow{AB} = (5 + 6, -2 - 3) = \begin{pmatrix} 11 \\ -5 \end{pmatrix}$, $\overrightarrow{BA} = \begin{pmatrix} -11 \\ 5 \end{pmatrix}$. AB ≠ BA.
No pair is equal: AB and BA are always negative vectors of each other, equal only when A and B are the same point.

$\overrightarrow{AB} = (-1 - 2, 0 - 5) = \begin{pmatrix} -3 \\ -5 \end{pmatrix}$: 3 left, 5 down.

$\overrightarrow{AB} = (-5 - 2, -4 - 3) = \begin{pmatrix} -7 \\ -7 \end{pmatrix}$: 7 left, 7 down.

$\overrightarrow{AB} = (0 + 6, -1 - 4) = \begin{pmatrix} 6 \\ -5 \end{pmatrix}$: 6 right, 5 down.

Count each arrow’s move from tail to head:
$\vec{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$, $\vec{b} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$, $\vec{c} = \begin{pmatrix} -2 \\ -2 \end{pmatrix}$, $\vec{d} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$, $\vec{e} = \begin{pmatrix} 0 \\ 3 \end{pmatrix}$.

Any points work if the move is right. For example, take A(1, 1). Then B = (1 + 2, 1 + 3) = (3, 4), and from B, C = (3 − 1, 4 − 4) = (2, 0).

AC = AB + BC = (−5 + 2, 7 + 1) = (−3, 8).

AB = OB − OA = (0 − 2, 2 − (−3)) = (−2, 5).

(i and j are the unit vectors along the axes, from 4.7: xi + yj is (x, y).)
AB = OB − OA = (3i − j) − (i + 2j) = 2i − 3j, that is (2, −3).

$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \begin{pmatrix} 4 + 7 \\ 3 + 8 \end{pmatrix} =$ $\begin{pmatrix} 11 \\ 11 \end{pmatrix}$.

By the triangle law, AB + BC = AC.
So AB + BC + CA = AC + CA = AA = 0, since CA = −AC. Proved.

In △PQR, PQ + QR = PR. In △PRS, PR + RS = PS.
So PQ + QR + RS + SP = PS + SP = 0, since SP = −PS. Proved.

AB + BC = AC (triangle law in △ABC).
AC + CD = AD (in △ACD).
AD + DE = AE (in △ADE).
So AB + BC + CD + DE = AE. Proved. (The pentagon needn’t be regular: the polygon law holds for any.)

Write each diagonal as two sides by the triangle law. In a parallelogram, DC = AB.

In △ABC, AC = AB + BC.
In △BCD, BD = BC + CD.
ABCD is a parallelogram, so DC = AB, and CD = −AB.
Adding: AC + BD = AB + BC + BC − AB = 2BC. Proved.

Put them head to tail: BC + CA = BA.

DC

DA

AD + DO = AO

AC = AB + BC = a + b

The long diagonal AD of a regular hexagon is parallel to BC and twice as long.

AD = 2b: a regular hexagon is six equilateral triangles round its center, so AD passes through the center, is parallel to BC and is twice as long.

DE is the side opposite AB, the same length the other way: DE = −a.
AE = AD + DE = 2b − a

EF = −BC = −b, so AF = AE + EF = 2b − a − b = b − a.

3(4, −2) = (12, −6)

$\frac{1}{2}(4, -2) =$ (2, −1)

$\frac{3}{2}(4, -2) =$ (6, −3)

$\vec{b} = \begin{pmatrix} 2 \times (-5) \\ 2 \times 3 \end{pmatrix} = 2\vec{a}$. A number times a, so $\vec{a} \parallel \vec{b}$ (like vectors, as 2 > 0).

$\vec{d} = -\frac{1}{2}\begin{pmatrix} 3 \\ -2 \end{pmatrix} = -\frac{1}{2}\vec{c}$, so $\vec{c} \parallel \vec{d}$ (unlike vectors, as $-\frac{1}{2} < 0$).

Parallel, so b = ma for some number m: (6, k) = m(3, −2).
From the x-components, 6 = 3m, so m = 2. Then k = −2m = −4.

(1 − 3, 2 + 4) = (−2, 6)

(1 + 3, 2 − 4) = (4, −2)

(−3 − 1, 4 − 2) = (−4, 2), the negative of a − b.

(2 − 3, 4 + 4) = (−1, 8)

(3, 6) − (−6, 8) = (3 + 6, 6 − 8) = (9, −2)

$\left(\frac{1}{2}, 1\right) + \left(-\frac{9}{2}, 6\right) = \left(-\frac{8}{2}, 7\right) =$ (−4, 7)

a = (a − b) + b = (12 + 5, 4 + 7) = (17, 11)

b = (a + b) − a = (5 − 0, 1 − 4) = (5, −3)

3b = (6, −9), so 2a = (0 − 6, −7 + 9) = (−6, 2) and a = (−3, 1).
Check: 2(−3, 1) + 3(2, −3) = (−6 + 6, 2 − 9) = (0, −7). ✓

Write each median as a side plus half of the next side, then use Question 5.

P is the midpoint of BC, so $\overrightarrow{BP} = \frac{1}{2}\overrightarrow{BC}$; likewise $\overrightarrow{CQ} = \frac{1}{2}\overrightarrow{CA}$ and $\overrightarrow{AR} = \frac{1}{2}\overrightarrow{AB}$.
$\overrightarrow{AP} = \overrightarrow{AB} + \frac{1}{2}\overrightarrow{BC}$, $\overrightarrow{BQ} = \overrightarrow{BC} + \frac{1}{2}\overrightarrow{CA}$, $\overrightarrow{CR} = \overrightarrow{CA} + \frac{1}{2}\overrightarrow{AB}$.
Adding: $\overrightarrow{AP} + \overrightarrow{BQ} + \overrightarrow{CR} = \frac{3}{2}(\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA}) = \frac{3}{2} \times \vec{0} = \vec{0}$. Proved.

Left side: AB − AC = AB + CA = CA + AB = CB.
Right side: CP − BP = CP + PB = CB.
Both sides equal CB. Proved.

In △ADC, AD + DC = AC. In △ABC, AB + BC = AC.
Both equal AC, so AD + DC = AB + BC. Proved: two routes with the same start and end are the same displacement.

AC = (4 − 1, 5 − (−2)) = (3, 7)
BD = (5 − 2, 2 − (−5)) = (3, 7)
The components are equal, so AC = BD. Proved.

PQ = (−5 + 2, 6 − 3) = (−3, 3)
RS = (5 − 2, 4 − 7) = (3, −3), so −RS = (−3, 3).
So PQ = −RS. Proved.

• a.A unit vector is a vector of magnitude 1, such as i = (1, 0), j = (0, 1) or $\left(\frac{3}{5}, \frac{4}{5}\right)$. The unit vector in the direction of a is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
• b.The magnitude of a vector is the length of its arrow, $|\vec{a}| = \sqrt{x^2 + y^2}$. For (6, 8) it is $\sqrt{36 + 64} = 10$.
• c.The direction of a vector is the angle θ it makes with the positive X-axis, measured anticlockwise, with $\tan\theta = \frac{y}{x}$ and the quadrant from the signs. For (1, 1) it is 45∘°; for (−1, 1) it is 135∘°.

$|\vec{a}| = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
First quadrant: $\theta = \tan^{-1}\left(\frac{3}{3}\right) = \tan^{-1}1 =$ 45∘°.

$|\vec{b}| = \sqrt{16 + 9} = 5$ units.
$\alpha = \tan^{-1}\frac{3}{4} = 36.87^\circ$; second quadrant (x negative, y positive): $\theta = 180^\circ - 36.87^\circ =$ 143.13∘°.

$|\vec{c}| = \sqrt{25 + 75} = \sqrt{100} = 10$ units.
$\alpha = \tan^{-1}\left|\frac{5\sqrt{3}}{-5}\right| = \tan^{-1}\sqrt{3} = 60^\circ$. x is negative and y positive, so c is in the second quadrant: $\theta = 180^\circ - 60^\circ =$ 120∘°.
(Some books give 240∘°, which would point into the third quadrant, with both components negative. Check: $10\cos 120^\circ = -5$ and $10\sin 120^\circ = 5\sqrt{3}$. ✓)

$\overrightarrow{PQ} = \begin{pmatrix} 7 - 2 \\ -5 + 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$, $|\overrightarrow{PQ}| = \sqrt{25 + 9} = \sqrt{34}$ units.
$\alpha = \tan^{-1}\frac{3}{5} = 30.96^\circ$; fourth quadrant (x positive, y negative): $\theta = 360^\circ - 30.96^\circ =$ 329.04∘°.

$\overrightarrow{PQ} = \begin{pmatrix} 6 - 4 \\ 1 + 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$, $|\overrightarrow{PQ}| = \sqrt{4 + 9} = \sqrt{13}$ units.
First quadrant: $\theta = \tan^{-1}\frac{3}{2} =$ 56.31∘°.

AB = (0 + 5, 2 − 4) = (5, −2), so $|\overrightarrow{AB}| = \sqrt{25 + 4} = \sqrt{29}$.
CD = (6 − 1, −3 + 1) = (5, −2), so $|\overrightarrow{CD}| = \sqrt{29}$.
|AB| = |CD|. Proved. (Here the vectors themselves are equal too.)

AB = (7 − 4, −3 − 5) = (3, −8), so $|\overrightarrow{AB}| = \sqrt{9 + 64} = \sqrt{73}$.
CD = (2 + 1, −11 + 3) = (3, −8), so $|\overrightarrow{CD}| = \sqrt{73}$.
|AB| = |CD|. Proved.

AB = (5, 2), so $|\overrightarrow{AB}|^2 = 25 + 4 = 29$.
CD = (2 − x, −5), so $|\overrightarrow{CD}|^2 = (2 - x)^2 + 25$.
$(2 - x)^2 + 25 = 29$, so $(2 - x)^2 = 4$ and 2 − x = ±2.
x = 0 or x = 4. Both work: C(0, 3) and C(4, 3) are both $\sqrt{29}$ from D.

$|\vec{a}| = \sqrt{9 + 16} = 5$, so $\hat{a} =$ $\left(-\frac{3}{5}, \frac{4}{5}\right)$.

$|\vec{b}| = \sqrt{4 + 25} = \sqrt{29}$, so $\hat{b} =$ $\left(\frac{2}{\sqrt{29}}, -\frac{5}{\sqrt{29}}\right)$.

AB = (−2 − 5, 0 − 6) = (−7, −6) = −7i − 6j.
$|\overrightarrow{AB}| = \sqrt{49 + 36} = \sqrt{85}$; unit vector $\left(-\frac{7}{\sqrt{85}}, -\frac{6}{\sqrt{85}}\right)$.

AB = (−1 + 2, −2 − 1) = (1, −3) = i − 3j.
$|\overrightarrow{AB}| = \sqrt{1 + 9} = \sqrt{10}$; unit vector $\left(\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right)$.

$\left(\frac{3}{5}\right)^2 + y^2 = 1$, so $y^2 = 1 - \frac{9}{25} = \frac{16}{25}$ and $y = \pm\frac{4}{5}$.
Two answers: $\left(\frac{3}{5}, \frac{4}{5}\right)$ and $\left(\frac{3}{5}, -\frac{4}{5}\right)$ are both 1 unit long.

$\frac{9}{13} + \frac{y^2}{13} = 1$, so $9 + y^2 = 13$, $y^2 = 4$, y = ±2.

The components differ, so they are unequal. In fact b = −3a: they are unlike (parallel, opposite ways), and b is three times as long.

Unequal: the components differ. Here a = 3b, a positive multiple, so they are like vectors (the same direction), but a is three times as long. (Some books call them unlike; they point the same way, so they are like.)

AB = (3 − 2, 3 + 1) = (1, 4)
CD = (−1 + 2, −2 + 6) = (1, 4)
The components are equal, so AB = CD. Proved.

PQ = (4 − 2, −2 + 3) = (2, 1)
SR = (1 + 1, −4 + 5) = (2, 1)
The components are equal, so PQ = SR. Proved. (And RS = (−2, −1) = −PQ, its negative.)

AB = (2 − 0, 5 − 3) = (2, 2).
Let D be (x, y): CD = (x − 2, y − 3) = (2, 2), so x = 4 and y = 5.
D(4, 5). (Some books give (0, 5), the answer for A(0, −3), B(2, 5) and C(−2, −3).)

AB = (0 + 1, 4 − y) = (1, 4 − y) and CD = (x + 1, 6 − 3) = (x + 1, 3).
Equal components: x + 1 = 1, so x = 0; 4 − y = 3, so y = 1.

$\overrightarrow{AB} = \begin{pmatrix} a - 6 \\ b - 9 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, so a = 8 and b = 10.