Exam practice

The scalar product of two vectors a and b is the product of their magnitudes and the cosine of the angle θ between them: a · b = |a| |b| cos θ. The answer is a number (a scalar), not a vector, which is where the name comes from. In coordinates, for a = (x₁, y₁) and b = (x₂, y₂), a · b = x₁x₂ + y₁y₂.
Example: for $\vec{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, a · b = 3 × 2 + 4 × 1 = 10. In physics, the work done by a force F moving a body through a displacement d is F · d: a force of 10 N at 60∘° to the movement, over 5 m, does $10 \times 5 \times \cos 60° = 25$ J of work.

Perpendicular: when a · b = 0 (neither vector zero). Reason: a · b = |a| |b| cos θ, and with both lengths not zero this is 0 only when cos θ = 0, that is $\theta = 90°$. In coordinates, x₁x₂ + y₁y₂ = 0.

Parallel: when a · b = ±|a| |b|. Reason: parallel vectors make 0∘° ($\cos 0° = 1$) or 180∘° ($\cos 180° = -1$). In coordinates, when one vector is a number times the other, a = kb, that is x₁/x₂ = y₁/y₂, or x₁y₂ = x₂y₁.

i · j = 1 × 0 + 0 × 1 = 0.

a · b = 2 × 3 + 0 × 0 = 6.

a · b = (−4) × 1 + 2 × 2 = −4 + 4 = 0.

p · q = 3 × (−1) + 1 × 3 = −3 + 3 = 0.

i · j = 0, so cos θ = 0 and $\theta = 90°$.

$\cos\theta = \dfrac{6}{2 \times 3} = 1$, so $\theta = 0°$: both lie along the X-axis, pointing the same way.

a · b = 0, so $\cos\theta = \dfrac{0}{\sqrt{20}\,\sqrt{5}} = 0$ and $\theta = 90°$.

p · q = 0, so cos θ = 0 and $\theta = 90°$.

i · j = 1 × 0 + 0 × 1 = 0, so they are perpendicular.

a · b = 2 × 0 + 0 × 3 = 0, so they are perpendicular (one along the X-axis, one along the Y-axis). They are not parallel: x₁y₂ − x₂y₁ = 6 − 0 ≠ 0.

2 × 3 + m × 2 = 0, so 6 + 2m = 0 and m = −3.

3 × 6 + m × 4 = 0, so 18 + 4m = 0 and $m = -\frac{9}{2}$ (−4.5).

3 × 6 + m × (−2) = 0, so 18 − 2m = 0 and m = 9.

3m + (−2) × 3 = 0, so 3m − 6 = 0 and m = 2.

Parallel: x₁y₂ = x₂y₁, so 2 × 2 = 3m and $m = \frac{4}{3}$. Check: $(2, 4/3) = \frac{2}{3}(3, 2)$. ✓

3 × 4 = 6m, so m = 2: (3, 2) is half of (6, 4).

3 × (−2) = 6m, so m = −1: (3, −1) is half of (6, −2).

3 × 3 = m × (−2), so $m = -\frac{9}{2}$. Check: $(-9/2, 3) = -\frac{3}{2}(3, -2)$, pointing the opposite way. ✓

a · b = x₁x₂ + y₁y₂.

If neither is the zero vector, they are perpendicular: |a| |b| cos θ = 0 with both lengths not zero means cos θ = 0, $\theta = 90°$.

2 × 4 + 3 × 5 = 8 + 15 = 23.

When the two vectors point the same way ($\theta = 0°$). Reason: for given lengths, a · b = |a| |b| cos θ changes only with cos θ, which is at most 1, at $\theta = 0°$. The greatest value is then |a| |b|.
For the vectors of part c, $|\vec{a}|\,|\vec{b}| = \sqrt{13}\,\sqrt{41} \approx 23.09$, only a little more than 23: (2, 3) and (4, 5) point in nearly the same direction (about 5∘° apart).

End minus start: $\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$.

$\overrightarrow{AB} = \begin{pmatrix} -1 - (-2) \\ -3 - 1 \end{pmatrix} =$ $\begin{pmatrix} 1 \\ -4 \end{pmatrix}$; $\overrightarrow{BD} = \begin{pmatrix} 2 - (-1) \\ 2 - (-3) \end{pmatrix} =$ $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$.

$\overrightarrow{AC} = \begin{pmatrix} 3 - (-2) \\ -2 - 1 \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$ and $\overrightarrow{BD} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$.
AC · BD = 5 × 3 + (−3) × 5 = 15 − 15 = 0, so cos θ = 0 and $\theta = 90°$: the diagonals are perpendicular.

Find all four sides as vectors. Equal vectors on opposite sides mean a parallelogram; then test one angle with a scalar product.

$\overrightarrow{AB} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$, $\overrightarrow{DC} = \begin{pmatrix} 3 - 2 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -4 \end{pmatrix}$, $\overrightarrow{AD} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$.
AB = DC and BC = AD: opposite sides are equal and parallel, so ABCD is a parallelogram.
AB · BC = 1 × 4 + (−4) × 1 = 0, so the angle at B is 90∘°: a parallelogram with a right angle is a rectangle.
In fact more: $|\overrightarrow{AB}| = |\overrightarrow{BC}| = \sqrt{17}$, all four sides are equal, so ABCD is a square (which is why its diagonals in part c are perpendicular).

• a.Drawn head to tail, the three vectors bring you back to where you started: together they make no displacement at all. So they are the three sides of a triangle, taken in order round it. (Three forces that add to zero balance each other.)
• b.a + c = −b. Dot each side with itself: $|\vec{a}|^2 + 2\,\vec{a} \cdot \vec{c} + |\vec{c}|^2 = |\vec{b}|^2$, so 9 + 2 a · c + 16 = 25 and a · c = 0: the sides of 3 and 4 are at right angles.

Square both sides: a length squared is the vector dotted with itself.

Squaring, $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$, so (a + b) · (a + b) = (a − b) · (a − b).
$|\vec{a}|^2 + 2\,\vec{a} \cdot \vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\,\vec{a} \cdot \vec{b} + |\vec{b}|^2$
4 a · b = 0, so a · b = 0.
Hence a and b are perpendicular (for vectors that aren’t zero). In pictures: a + b and a − b are the two diagonals of the parallelogram on a and b, and a parallelogram with equal diagonals is a rectangle.

• a.$\overrightarrow{OR} = \frac{1}{2}(4\vec{i} + 7\vec{j} + 4\vec{i} - 3\vec{j}) = \frac{1}{2}(8\vec{i} + 4\vec{j}) =$ 4i + 2j.
• b.$\overrightarrow{OT} = \frac{1}{2}(5\vec{i} - 4\vec{j} + 3\vec{i} + 6\vec{j}) = \frac{1}{2}(8\vec{i} + 2\vec{j}) =$ 4i + j.

a = 2i − 3j, b = 3i + 2j.

• a.$\overrightarrow{OM} = \dfrac{2\vec{b} + 3\vec{a}}{2 + 3} = \dfrac{(6 + 6)\vec{i} + (4 - 9)\vec{j}}{5} =$ $\frac{12}{5}\vec{i} - \vec{j}$, the point (2.4, −1).
• b.$\overrightarrow{ON} = \dfrac{3\vec{b} - \vec{a}}{3 - 1} = \dfrac{(9 - 2)\vec{i} + (6 + 3)\vec{j}}{2} =$ $\frac{7}{2}\vec{i} + \frac{9}{2}\vec{j}$, the point (3.5, 4.5). Check: AB = (1, 5), and N = A + 1.5 × (1, 5), beyond B, with AN:BN = 1.5:0.5 = 3:1. ✓

• a.$\overrightarrow{OP} = \dfrac{3(2\vec{i} + 3\vec{j}) + 1(6\vec{i} - 2\vec{j})}{3 + 1} = \dfrac{12\vec{i} + 7\vec{j}}{4} =$ $3\vec{i} + \frac{7}{4}\vec{j}$.
• b.$\overrightarrow{OQ} = \dfrac{2(2\vec{i} + 3\vec{j}) - 1(6\vec{i} - 2\vec{j})}{2 - 1} =$ −2i + 8j. Check: Q = B + AB, since AB = (−4, 5) and (2, 3) + (−4, 5) = (−2, 8): twice as far from A as from B. ✓

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it: here $\overrightarrow{FE} = \frac{1}{2}\overrightarrow{BC}$.

BC = BA + AC = AC − AB, since BA = −AB.

In △AEF, EF = EA + AF. E is the midpoint of AC, so $\overrightarrow{EA} = \frac{1}{2}\overrightarrow{CA}$; F is the midpoint of AB, so $\overrightarrow{AF} = \frac{1}{2}\overrightarrow{AB}$.
$\overrightarrow{EF} = \frac{1}{2}\overrightarrow{CA} + \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{CA} + \overrightarrow{AB}) = \frac{1}{2}\overrightarrow{CB}$, by the triangle law in △ABC.
So $\overrightarrow{EF} = \frac{1}{2}\overrightarrow{CB}$, and turning both round, $\overrightarrow{FE} = \frac{1}{2}\overrightarrow{BC}$. Hence proved.

Write each median as half the sum of the two sides from its vertex.

By the mid-point formula for medians (10.3.1):
$\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$, $\overrightarrow{BE} = \frac{1}{2}(\overrightarrow{BA} + \overrightarrow{BC})$, $\overrightarrow{CF} = \frac{1}{2}(\overrightarrow{CA} + \overrightarrow{CB})$.
Adding, $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \frac{1}{2}\left[(\overrightarrow{AB} + \overrightarrow{BA}) + (\overrightarrow{AC} + \overrightarrow{CA}) + (\overrightarrow{BC} + \overrightarrow{CB})\right]$.
Each bracket is a vector plus its opposite, 0. So AD + BE + CF = 0. Hence proved. (The three medians, moved head to tail, close up into a triangle.)

$\vec{p} = \dfrac{m\vec{b} + n\vec{a}}{m + n}$.

AP:PB = m:n compares lengths only, so n × AP = m × PB. Because P is between A and B, the vectors AP and PB also point the same way, along AB. Two vectors in the same direction with equal lengths are equal, so nAP = mPB as vectors, which turns the ratio into an equation we can solve for p.

$\vec{p} = \dfrac{1(4\vec{i} + 8\vec{j}) + 2(\vec{i} + 2\vec{j})}{1 + 2} = \dfrac{6\vec{i} + 12\vec{j}}{3} =$ 2i + 4j. (Here A, B, P and O all lie on one line, since b = 4a.)

When m = n, a ratio of 1 : 1: then $\vec{p} = \dfrac{m\vec{b} + m\vec{a}}{2m} = \frac{1}{2}(\vec{a} + \vec{b})$, and P is the midpoint of AB.

$\vec{p} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$ (m ≠ n).

A, B and P are on one line with P outside AB, so AP and BP point the same way, and AP:BP = m:n gives nAP = mBP.
AP = p − a and BP = p − b, so n(p − a) = m(p − b).
np − na = mp − mb, so mb − na = (m − n)p, and $\vec{p} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$. Hence proved.

$\vec{p} = \dfrac{1(5\vec{i} - 3\vec{j}) - 2(\vec{i} + 4\vec{j})}{1 - 2} = \dfrac{3\vec{i} - 11\vec{j}}{-1} =$ −3i + 11j. Check: 1 < 2, so P should lie beyond A: AB = (4, −7), and (1, 4) − (4, −7) = (−3, 11), one AB behind A, so AP:BP = 1:2. ✓

No. The formula’s denominator m − n would be 0. In pictures: a point outside AB, on its line, is always nearer one end than the other, so AP = BP can’t happen. The only point on the line equally far from A and B is the midpoint, which is inside AB.

Let ABCD be the quadrilateral, with F, G, H and E the midpoints of AB, BC, CD and DA. Join BD.
In △ABD: $\overrightarrow{EF} = \overrightarrow{EA} + \overrightarrow{AF} = \frac{1}{2}\overrightarrow{DA} + \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}\overrightarrow{DB}$.
In △CBD: $\overrightarrow{HG} = \overrightarrow{HC} + \overrightarrow{CG} = \frac{1}{2}\overrightarrow{DC} + \frac{1}{2}\overrightarrow{CB} = \frac{1}{2}\overrightarrow{DB}$.
So EF = HG: EF and HG are equal and parallel. A quadrilateral with a pair of opposite sides equal and parallel is a parallelogram, so EFGH is a parallelogram. Hence proved.

Let O be the center, AB a diameter and C a point on the circle, radius r. Then |OA| = |OC| = r and OB = −OA.
$\overrightarrow{AC} \cdot \overrightarrow{BC} = (\overrightarrow{OC} - \overrightarrow{OA}) \cdot (\overrightarrow{OC} - \overrightarrow{OB}) = (\overrightarrow{OC} - \overrightarrow{OA}) \cdot (\overrightarrow{OC} + \overrightarrow{OA}) = |\overrightarrow{OC}|^2 - |\overrightarrow{OA}|^2 = r^2 - r^2 = 0$.
So $AC \perp BC$ and $\angle ACB = 90°$. Hence proved.

Part c: b points downward (its j part is −2), so the 45∘° is with the downward part of the Y-axis, the direction of −j.

• a.a · b = (x₁i + y₁j) · (x₂i + y₂j) = x₁x₂ + y₁y₂, since i · i = j · j = 1 and i · j = 0. Here a · b = 3x − 8.
• b.3x − 8 = 0, so $x = \frac{8}{3}$.
• c.The angle between b and the Y-axis’s line is 45∘° when $|\vec{b} \cdot \vec{j}| = |\vec{b}|\cos 45°$: $2 = \sqrt{x^2 + 4} \times \frac{1}{\sqrt{2}}$, so $x^2 + 4 = 8$ and x = ±2. (With −j, the downward Y-axis: b · (−j) = 2, positive, so the angle is 45∘°, not 135∘°.) Check: 2i − 2j and −2i − 2j both slope at 45∘°.
• d.$\frac{1}{2}(2\vec{i} + 4\vec{j} + 4\vec{i} - 2\vec{j}) =$ 3i + j.

• a.$\overrightarrow{OD} = \frac{1}{2}(2\vec{i} + 3\vec{j} + 4\vec{i} + \vec{j}) =$ 3i + 2j.
• b.$\overrightarrow{OE} = \frac{1}{2}(2\vec{i} + 3\vec{j} + 6\vec{i} + 5\vec{j}) =$ 4i + 4j.
• c.In general: $\overrightarrow{DE} = \overrightarrow{DA} + \overrightarrow{AE} = \frac{1}{2}\overrightarrow{BA} + \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}(\overrightarrow{BA} + \overrightarrow{AC}) = \frac{1}{2}\overrightarrow{BC}$ [triangle law].

With the numbers: DE = OE − OD = i + 2j, and BC = OC − OB = 2i + 4j, so $\overrightarrow{DE} = \frac{1}{2}\overrightarrow{BC}$. Hence proved.
d. Yes. DE is a number ($\frac{1}{2}$) times BC, so they are parallel. Check with the parallel test: 1 × 4 = 2 × 2. ✓

• a.$\dfrac{m\overrightarrow{OB} + n\overrightarrow{OA}}{m + n}$, for the line from A to B.
• b.$\vec{r} = \dfrac{3\vec{q} + 2\vec{p}}{3 + 2} = \dfrac{3(\vec{i} - 3\vec{j}) + 2(2\vec{i} + \vec{j})}{5} = \dfrac{7\vec{i} - 7\vec{j}}{5}$ = $\frac{7}{5}(\vec{i} - \vec{j})$.
• c.$\vec{p} \cdot \vec{r} = 2 \times \frac{7}{5} + 1 \times \left(-\frac{7}{5}\right) = \frac{7}{5}$; $|\vec{p}| = \sqrt{5}$, $|\vec{r}| = \frac{7}{5}\sqrt{2}$.

$\cos\angle POR = \dfrac{7/5}{\sqrt{5} \times \frac{7}{5}\sqrt{2}} = \dfrac{1}{\sqrt{10}}$, so $\angle POR = \cos^{-1}\frac{1}{\sqrt{10}} \approx 71.57°$.
d. $\vec{q} \cdot \vec{r} = \frac{7}{5} + \frac{21}{5} = \frac{28}{5}$; $|\vec{q}| = \sqrt{10}$. $\cos\angle QOR = \dfrac{28/5}{\sqrt{10} \times \frac{7}{5}\sqrt{2}} = \dfrac{4}{\sqrt{20}} = \dfrac{2}{\sqrt{5}}$, so $\angle QOR \approx 26.57°$.
∠POR is larger, about 2.7 times ∠QOR. R is nearer Q (3 parts from P, 2 from Q), so it is closer to Q in direction too, though the angles are not in the ratio 3 : 2. Together they make $\angle POQ \approx 98.13°$, and indeed $\cos\angle POQ = \dfrac{\vec{p} \cdot \vec{q}}{|\vec{p}|\,|\vec{q}|} = \dfrac{-1}{\sqrt{50}}$. ✓

• a.$\vec{p} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$.
• b.$\vec{p} = \dfrac{2(\vec{i} - 4\vec{j}) - 3(3\vec{i} + 2\vec{j})}{2 - 3} = \dfrac{-7\vec{i} - 14\vec{j}}{-1}$ = 7i + 14j. Check: 2 < 3, so P is beyond A; BA = (2, 6) and A + 2BA = (7, 14), with AP:BP = 2:3. ✓
• c.Beyond B. AP:BP = m:n with m > n means P is farther from A than from B, so it lies on B’s side, past B. (With m < n, as in part b, it lies beyond A.)
• d.a · b = 3 × 1 + 2 × (−4) = −5; $|\vec{a}| = \sqrt{13}$, $|\vec{b}| = \sqrt{17}$.

$\angle AOB = \cos^{-1}\left(\frac{-5}{\sqrt{221}}\right) \approx 109.65°$, obtuse, as the negative product says.