Product of Vectors

Grade 11 gave you everything about vectors except how to multiply them. This unit supplies the two products, and they are different animals. The scalar product returns a number measuring how much two vectors point the same way — the tool for angles, perpendicularity and work. The vector product returns a third vector, at right angles to both, whose length is an area — the tool for areas, perpendicular directions and moments. With both in hand, most of the trigonometry of Units 6 and 7 can be proved again in two lines.

9.1 Scalar (dot) product; angle between two vectors

Unit 14 of Grade 11 stated the scalar product and used it. We now start again from the definition and prove the results, as a Grade 12 paper requires.

Two points are worth saying out loud. First, \theta is the angle between the vectors when they start from the same point: given \vec{AB} and \vec{BC}, the angle between them is not the interior angle B, so move one vector until both tails meet. Second, since \cos\theta changes sign at 90^{\circ}, the sign of \vec{a}\cdot\vec{b} alone tells you whether the angle is acute, right or obtuse.

Applying the definition to \hat{i}, \hat{j}, \hat{k} gives a table worth memorising: a base vector makes 0^{\circ} with itself, 90^{\circ} with the other two.

\cdot	\hat{i}	\hat{j}	\hat{k}
\hat{i}	1	0	0
\hat{j}	0	1	0
\hat{k}	0	0	1

So \hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1 (each is a unit vector, \cos 0^{\circ} = 1) and \hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0 (perpendicular, \cos 90^{\circ} = 0).

The component form, and why it agrees with the definition

Almost every calculation uses components rather than |\vec{a}||\vec{b}|\cos\theta, so the two forms must be shown to agree. The bridge is the cosine law of Unit 6.

The angle between two vectors

Reading the two forms as one equation and solving for \cos\theta gives the single most used formula of the unit.

9.2 Geometric interpretation and properties of scalar product

The scalar product is a length times a shadow

Group |\vec{a}||\vec{b}|\cos\theta as |\vec{a}| \times \left(|\vec{b}|\cos\theta\right). The bracket is the shadow \vec{b} casts on the line of \vec{a} when light falls at right angles to it. Drop a perpendicular from the head of \vec{b} onto that line, meeting it at M: then OM is the projection of \vec{b} on \vec{a}.

Algebraic properties, with proofs

The proofs are short, and NEB asks for them.

1. Commutative. The angle between \vec{a} and \vec{b} is the same angle as between \vec{b} and \vec{a}, and multiplication of numbers is commutative:

2. Square of a vector. A vector makes an angle 0^{\circ} with itself, and \cos 0^{\circ} = 1:

This is the identity that turns every length question into a scalar product question. In components it reads a_1^{2}+a_2^{2}+a_3^{2}, the distance formula.

3. Scalars. Take m > 0 first. Multiplying \vec{a} by m multiplies its length by m and leaves its direction alone, so \theta does not change:

If m < 0, then |m\vec{a}| = -m|\vec{a}| and the direction of m\vec{a} is reversed, so the angle becomes 180^{\circ}-\theta and the cosine changes sign:

Both signs give the same rule, and m = 0 makes both sides zero.

5. Perpendicularity. If \vec{a} and \vec{b} are non-zero then |\vec{a}||\vec{b}| \ne 0, so \vec{a}\cdot\vec{b} = 0 forces \cos\theta = 0, i.e. \theta = 90^{\circ}. Running the argument backwards, \theta = 90^{\circ} gives \cos\theta = 0 and the product vanishes.

6. Expansion. Use property 2 on \vec{a}\pm\vec{b}, then expand by distributivity and collect the two middle terms with commutativity:

Replacing \vec{b} by -\vec{b} gives the minus version. These two lines do the work in most Group B problems about |\vec{a}+\vec{b}|.

7. Bound. Since -1 \leq \cos\theta \leq 1,

and |\cos\theta| = 1 only for \theta = 0^{\circ} or 180^{\circ}, that is, only for parallel vectors.

Property 4 is the one that needs a picture rather than algebra.

Work done by a force

The scalar product was invented for this.

Only the part of the force along the displacement does work — the projection idea of §9.2. A force at right angles to the motion, such as the string tension on a whirling stone, does zero work because \cos 90^{\circ} = 0.

9.3 Applications of scalar product in geometry and trigonometry

The method never changes: name the sides as vectors, write the one relation they satisfy, and dot both sides with something useful.

For a triangle ABC the standard naming is

taken in order round the triangle, so that by the triangle law

and the side lengths are a = |\vec{a}| = BC, b = |\vec{b}| = CA, c = |\vec{c}| = AB, as in Unit 6.

Care is needed with the angles. \vec{b} and \vec{c} do not make the angle A: \vec{c} = \vec{AB} leaves A while \vec{b} = \vec{CA} arrives at A. Reversing \vec{b} gives \vec{AC}, which does. So the angle between \vec{b} and \vec{c} is 180^{\circ}-A, and

The same argument at the other two vertices gives \vec{c}\cdot\vec{a} = -ca\cos B and \vec{a}\cdot\vec{b} = -ab\cos C. Those three lines are the engine of everything in this section.

Geometry theorems in three lines

Compound angles by scalar product

The addition formulas of trigonometry fall straight out of the two expressions for one scalar product. This is a favourite Group C question.

9.4 Vector (cross) product; geometrical interpretation and properties

The scalar product throws away all direction information. Often the direction is what you need — the normal to a plane, the axis of a turning effect. That calls for a product whose answer is a vector.

Since \sin\theta \geq 0 on 0^{\circ} to 180^{\circ}, the factor |\vec{a}||\vec{b}|\sin\theta is never negative: it is the magnitude |\vec{a}\times\vec{b}|, and all the directional information sits in \hat{n}.

Applying the definition to the base vectors gives the results every calculation rests on. Distinct base vectors are perpendicular, so \sin 90^{\circ} = 1 and the product has magnitude 1; the right-hand rule picks the sign.

Properties, with proofs

1. Anti-commutativity. The magnitude is unchanged, because |\vec{b}||\vec{a}|\sin\theta = |\vec{a}||\vec{b}|\sin\theta and \theta is the same angle. Only the direction changes: turning \vec{b} into \vec{a} instead of \vec{a} into \vec{b} reverses the sense of the right-hand rule, so \hat{n} becomes -\hat{n}. Same length, opposite direction:

This is the single biggest difference from the scalar product, and the commonest source of sign errors. Order matters.

2 and 3. Parallel vectors. A vector makes an angle 0^{\circ} with itself, and \sin 0^{\circ} = 0, so \vec{a}\times\vec{a} = \vec{0}. More generally if \vec{a} and \vec{b} are parallel then \theta = 0^{\circ} (same sense) or 180^{\circ} (opposite sense), and \sin\theta = 0 either way. Conversely, if neither vector is \vec{0} and |\vec{a}||\vec{b}|\sin\theta = 0 then \sin\theta = 0, so \theta is 0^{\circ} or 180^{\circ} and the vectors are parallel. This is the cross-product test for parallelism, and it is neater than comparing ratios of components because it never divides by zero.

4. Scalars. For m > 0 neither the angle nor \hat{n} changes and the magnitude is multiplied by m. For m < 0 the direction of m\vec{a} reverses, so the angle becomes 180^{\circ}-\theta — but \sin(180^{\circ}-\theta) = \sin\theta, so the magnitude picks up only the factor |m| = -m, while \hat{n} reverses. The two sign changes give m overall.

5. Distributivity holds, and is what makes the determinant formula below work. Its proof is longer than the syllabus requires — NEB quotes it — but you may use it freely.

6. No associativity. One counterexample settles it. On the left, \hat{i}\times\hat{j} = \hat{k}, so

On the right, \hat{i}\times\hat{i} = \vec{0}, so

and -\hat{j} \ne \vec{0}. Never move the brackets in a cross product.

8. Lagrange's identity. Start from the magnitude and use \sin^{2}\theta = 1-\cos^{2}\theta:

This identity links the two products, and it is the quickest way to get one of |\vec{a}\times\vec{b}| and \vec{a}\cdot\vec{b} from the other.

The determinant form

Geometrical interpretation: |\vec{a}\times\vec{b}| is an area

9.5 Applications of vector product in geometry and trigonometry

Everything here follows from two proved facts: |\vec{a}\times\vec{b}| is an area, and the product vanishes exactly when the vectors are parallel.

Areas

The third needs a one-line proof. With sides \vec{p}, \vec{q} the diagonals are \vec{d_1} = \vec{p}+\vec{q} and \vec{d_2} = \vec{q}-\vec{p}, so by distributivity

The two self-products are \vec{0}, and -\vec{q}\times\vec{p} = +\vec{p}\times\vec{q} by anti-commutativity, so \vec{d_1}\times\vec{d_2} = 2(\vec{p}\times\vec{q}). Taking magnitudes and halving gives the area |\vec{p}\times\vec{q}| as claimed.

Collinear points and parallel lines

Three points A, B, C are collinear exactly when \vec{AB} and \vec{AC} are parallel, so by property 3 the test is a single cross product:

Geometrically, triangle ABC has zero area. The same product tests two lines for parallelism: their directions satisfy \vec{d_1}\times\vec{d_2} = \vec{0}.

The sine law

Moment of a force

|\vec{M}| = |\vec{F}|d is the "force times perpendicular distance" rule of elementary mechanics, with the cross product supplying the axis as well. A door turns easily at the handle because d is large; pushing along the line of the hinge does nothing, since then \vec{r}\times\vec{F} = \vec{0}.

Compound angles by vector product

In §9.3 the scalar product gave \cos(A\pm B). The same two unit vectors, crossed instead of dotted, give the sine formulas.

Chapter summary

• Scalar product: \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1b_1+a_2b_2+a_3b_3, a number. The two forms agree by the cosine law applied to the triangle on \vec{a}, \vec{b} and \vec{b}-\vec{a}.
• Angle and perpendicularity: \cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}; for non-zero vectors \vec{a}\perp\vec{b} \Leftrightarrow \vec{a}\cdot\vec{b} = 0. Also \vec{a}\cdot\vec{a} = |\vec{a}|^{2}, \hat{i}\cdot\hat{i} = 1, \hat{i}\cdot\hat{j} = 0.
• Properties of \cdot: commutative, distributive over addition, (m\vec{a})\cdot\vec{b} = m(\vec{a}\cdot\vec{b}), and |\vec{a}\cdot\vec{b}| \leq |\vec{a}||\vec{b}|. You may not cancel a vector factor.
• Projection: the scalar projection of \vec{b} on \vec{a} is \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|} (signed), and the vector projection is \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^{2}}\vec{a}. Work done is W = \vec{F}\cdot\vec{d}.
• Dot-product geometry: cosine law a^{2} = b^{2}+c^{2}-2bc\cos A; projection law a = b\cos C + c\cos B; \cos(A\pm B) = \cos A\cos B \mp \sin A\sin B; diagonals of a rhombus are perpendicular because (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^{2}-|\vec{b}|^{2}; the angle in a semicircle is 90^{\circ}.
• Vector product: \vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n} with \hat{n} given by the right-hand rule — a vector. In components it is the 3\times3 determinant with first row \hat{i},\hat{j},\hat{k}, i.e. (a_2b_3-a_3b_2)\hat{i}+(a_3b_1-a_1b_3)\hat{j}+(a_1b_2-a_2b_1)\hat{k}.
• Properties of \times: anti-commutative \vec{b}\times\vec{a} = -(\vec{a}\times\vec{b}); \vec{a}\times\vec{a} = \vec{0}; distributive; not associative; \hat{i}\times\hat{j} = \hat{k} round the cycle; and \vec{a}\times\vec{b} = \vec{0} \Leftrightarrow \vec{a}\parallel\vec{b} for non-zero vectors.
• Areas and Lagrange: |\vec{a}\times\vec{b}| is the area of the parallelogram on \vec{a},\vec{b}; a triangle is \tfrac{1}{2}|\vec{AB}\times\vec{AC}|; a parallelogram from its diagonals is \tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|; and |\vec{a}\times\vec{b}|^{2}+(\vec{a}\cdot\vec{b})^{2} = |\vec{a}|^{2}|\vec{b}|^{2}.
• Cross-product geometry: sine law \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} from \vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a}; \sin(A\pm B) = \sin A\cos B \pm \cos A\sin B; collinearity of A,B,C from \vec{AB}\times\vec{AC} = \vec{0}; moment \vec{M} = \vec{r}\times\vec{F} with |\vec{M}| = |\vec{F}|d.

Practice questions

Group A — Multiple choice (1 mark each)

1. The value of \hat{i}\cdot\hat{j} + \hat{j}\cdot\hat{k} + \hat{k}\cdot\hat{i} is [1]a0b1c3d-1
2. \hat{j}\times\hat{k} equals [1]a\hat{i}b-\hat{i}c\hat{k}d\vec{0}
3. If |\vec{a}| = 3, |\vec{b}| = 4 and the angle between them is 60^{\circ}, then \vec{a}\cdot\vec{b} is [1]a6b12c6\sqrt{3}d\dfrac{1}{2}
4. The angle between 2\hat{i}+3\hat{j}+\hat{k} and \hat{i}-\hat{j}+\hat{k} is [1]a0^{\circ}b45^{\circ}c60^{\circ}d90^{\circ}
5. |\vec{a}\times\vec{b}|^{2} + (\vec{a}\cdot\vec{b})^{2} equals [1]a|\vec{a}|^{2}+|\vec{b}|^{2}b|\vec{a}|^{2}|\vec{b}|^{2}c2|\vec{a}||\vec{b}|d0
6. For non-zero vectors, \vec{a}\times\vec{b} = \vec{0} means that \vec{a} and \vec{b} are [1]aperpendicularbparallelcequaldunit vectors
7. If the diagonals of a parallelogram are \vec{d_1} and \vec{d_2}, its area is [1]a|\vec{d_1}\times\vec{d_2}|b\dfrac{1}{2}|\vec{d_1}\times\vec{d_2}|c\vec{d_1}\cdot\vec{d_2}d2|\vec{d_1}\times\vec{d_2}|
8. If \hat{a} and \hat{b} are unit vectors inclined at 60^{\circ}, then |\hat{a}-\hat{b}| is [1]a1b\sqrt{2}c\sqrt{3}d2
9. The projection of 2\hat{i}-\hat{j}+\hat{k} on \hat{i}+2\hat{j}-2\hat{k} is [1]a-\dfrac{2}{3}b\dfrac{2}{3}c-2d\dfrac{2}{\sqrt{6}}
10. A force 2\hat{i}+3\hat{j}-\hat{k} newtons moves a body through a displacement \hat{i}+\hat{j}+\hat{k} metres. The work done is [1]a2 Jb4 Jc6 Jd-4 J

Answers to Group A

1. (a) — every term is the dot product of two different base vectors, and each of those is 0.

2. (a) — reading \hat{j}\to\hat{k} with the arrows of the cycle gives the remaining letter with a plus sign.

3. (a) — \vec{a}\cdot\vec{b} = (3)(4)\cos 60^{\circ} = 12\left(\tfrac12\right) = 6.

4. (d) — the dot product is 2-3+1 = 0, and neither vector is \vec{0}.

5. (b) — this is Lagrange's identity, |\vec{a}|^{2}|\vec{b}|^{2}\left(\sin^{2}\theta+\cos^{2}\theta\right).

6. (b) — \sin\theta = 0, so \theta = 0^{\circ} or 180^{\circ}.

7. (b) — \vec{d_1}\times\vec{d_2} = 2(\vec{p}\times\vec{q}) for sides \vec{p},\vec{q}, so the area is half the diagonal product.

8. (a) — |\hat{a}-\hat{b}|^{2} = 1+1-2\cos 60^{\circ} = 1.

9. (a) — the dot product is 2-2-2 = -2 and the second vector has magnitude 3, so the projection is -\tfrac{2}{3}. It is negative because the angle is obtuse.

10. (b) — W = \vec{F}\cdot\vec{d} = 2+3-1 = 4 J.

Group B — Short answer (5 marks each)

1. Find the angle between the vectors \hat{i}-2\hat{j}+3\hat{k} and 3\hat{i}-2\hat{j}+\hat{k}. [5]
2. Prove by vector method that the diagonals of a rhombus bisect each other at right angles. [5]
3. Find the value of \lambda for which \lambda\hat{i}+2\hat{j}+\hat{k} is perpendicular to 4\hat{i}-9\hat{j}+2\hat{k}, and state the magnitude of the first vector for that \lambda. [5]
4. If |\vec{a}| = 5, |\vec{b}| = 4 and \vec{a}\cdot\vec{b} = 10, find the angle between \vec{a} and \vec{b}, and the values of |\vec{a}+\vec{b}| and |\vec{a}-\vec{b}|. [5]
5. Find a unit vector perpendicular to both 2\hat{i}-6\hat{j}-3\hat{k} and 4\hat{i}+3\hat{j}-\hat{k}. [5]
6. Find the area of the triangle whose vertices are A(1,1,2), B(2,3,5) and C(1,5,5). [5]
7. A force \vec{F} = 2\hat{i}-\hat{j}+2\hat{k} newtons acts at the point A(1,3,4). Find its moment about B(1,2,2) and the perpendicular distance from B to the line of action of \vec{F}. [5]
8. Find the scalar and the vector projection of \vec{a} = \hat{i}-2\hat{j}+\hat{k} on \vec{b} = 2\hat{i}-\hat{j}+2\hat{k}, and the angle between them. [5]

Answers to Group B

1. Dot the two vectors and find both magnitudes.

\vec{a}\cdot\vec{b} = (1)(3)+(-2)(-2)+(3)(1) = 3+4+3 = 10

|\vec{a}| = \sqrt{1+4+9} = \sqrt{14}, \qquad |\vec{b}| = \sqrt{9+4+1} = \sqrt{14}

\cos\theta = \frac{10}{\sqrt{14}\sqrt{14}} = \frac{10}{14} = \frac{5}{7} = 0.7143

\theta = \cos^{-1}(0.7143) = 44.42^{\circ}

2. Let the rhombus be OACB with \vec{OA} = \vec{a} and \vec{OB} = \vec{b}. All four sides are equal, so |\vec{a}| = |\vec{b}|.

Bisection. \vec{OC} = \vec{a}+\vec{b}, so the midpoint of OC has position vector \tfrac{1}{2}(\vec{a}+\vec{b}). The midpoint of AB has position vector \tfrac{1}{2}(\vec{a}+\vec{b}) as well. The two midpoints coincide, so the diagonals bisect each other.

Right angles. The diagonals are \vec{OC} = \vec{a}+\vec{b} and \vec{AB} = \vec{b}-\vec{a}. Then

(\vec{a}+\vec{b})\cdot(\vec{b}-\vec{a}) = \vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} - \vec{b}\cdot\vec{a}

= |\vec{b}|^{2} - |\vec{a}|^{2} = 0 \qquad \text{since } |\vec{a}| = |\vec{b}|

Neither diagonal is \vec{0}, so they are perpendicular.

3. Perpendicular means the dot product vanishes:

(\lambda)(4) + (2)(-9) + (1)(2) = 0

4\lambda - 18 + 2 = 0 \; \Longrightarrow \; 4\lambda = 16 \; \Longrightarrow \; \lambda = 4

With \lambda = 4 the first vector is 4\hat{i}+2\hat{j}+\hat{k}, so

|\vec{a}| = \sqrt{16+4+1} = \sqrt{21} = 4.58

4. From the definition of the scalar product,

\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{10}{(5)(4)} = \frac{1}{2} \; \Longrightarrow \; \theta = 60^{\circ}

For the sums, square and expand:

|\vec{a}+\vec{b}|^{2} = |\vec{a}|^{2} + 2\vec{a}\cdot\vec{b} + |\vec{b}|^{2} = 25+20+16 = 61

|\vec{a}+\vec{b}| = \sqrt{61} = 7.81

|\vec{a}-\vec{b}|^{2} = 25-20+16 = 21 \; \Longrightarrow \; |\vec{a}-\vec{b}| = \sqrt{21} = 4.58

5. A vector perpendicular to both is \vec{a}\times\vec{b}, with components (2,-6,-3) and (4,3,-1):

\hat{i}\text{-part} = (-6)(-1)-(-3)(3) = 6+9 = 15

\hat{j}\text{-part} = -\left[(2)(-1)-(-3)(4)\right] = -(-2+12) = -10

\hat{k}\text{-part} = (2)(3)-(-6)(4) = 6+24 = 30

\vec{a}\times\vec{b} = 15\hat{i}-10\hat{j}+30\hat{k}, \quad |\vec{a}\times\vec{b}| = \sqrt{225+100+900} = 35

\hat{n} = \pm\frac{15\hat{i}-10\hat{j}+30\hat{k}}{35} = \pm\left(\frac{3}{7}\hat{i}-\frac{2}{7}\hat{j}+\frac{6}{7}\hat{k}\right)

6. Take two sides from A: \vec{AB} = \hat{i}+2\hat{j}+3\hat{k} and \vec{AC} = 4\hat{j}+3\hat{k}, i.e. components (1,2,3) and (0,4,3).

\hat{i}\text{-part} = (2)(3)-(3)(4) = 6-12 = -6

\hat{j}\text{-part} = -\left[(1)(3)-(3)(0)\right] = -3

\hat{k}\text{-part} = (1)(4)-(2)(0) = 4

\left|\vec{AB}\times\vec{AC}\right| = \sqrt{36+9+16} = \sqrt{61}

\text{area} = \tfrac{1}{2}\sqrt{61} = 3.91\ \text{square units}

7. The position vector of A relative to B is

\vec{r} = \vec{BA} = (1-1)\hat{i}+(3-2)\hat{j}+(4-2)\hat{k} = \hat{j}+2\hat{k}

With components (0,1,2) and (2,-1,2):

\hat{i}\text{-part} = (1)(2)-(2)(-1) = 4, \qquad \hat{j}\text{-part} = -\left[(0)(2)-(2)(2)\right] = 4

\hat{k}\text{-part} = (0)(-1)-(1)(2) = -2

\vec{M} = 4\hat{i}+4\hat{j}-2\hat{k}, \qquad |\vec{M}| = \sqrt{16+16+4} = 6\ \text{N m}

Since |\vec{F}| = \sqrt{4+1+4} = 3 N,

d = \frac{|\vec{M}|}{|\vec{F}|} = \frac{6}{3} = 2\ \text{m}

8. First the dot product and the magnitude of \vec{b}:

\vec{a}\cdot\vec{b} = (1)(2)+(-2)(-1)+(1)(2) = 2+2+2 = 6, \qquad |\vec{b}| = \sqrt{4+1+4} = 3

Scalar projection of \vec{a} on \vec{b}:

\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{6}{3} = 2

Vector projection:

\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^{2}}\,\vec{b} = \frac{6}{9}\left(2\hat{i}-\hat{j}+2\hat{k}\right) = \frac{4}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{4}{3}\hat{k}

For the angle, |\vec{a}| = \sqrt{1+4+1} = \sqrt{6}, so

\cos\theta = \frac{6}{3\sqrt{6}} = \frac{2}{\sqrt{6}} = 0.8165 \; \Longrightarrow \; \theta = 35.26^{\circ}

Group C — Long answer (8 marks each)

1. (a) Define the scalar product of two vectors and prove by vector method that a^{2} = b^{2}+c^{2}-2bc\cos A in any triangle ABC. [6](b) Hence find the largest angle of the triangle whose sides are 3, 5 and 7 units. [2]
2. (a) Prove by vector method that \sin(A-B) = \sin A\cos B - \cos A\sin B. [6](b) Hence evaluate \sin 15^{\circ} in surd form. [2]
3. (a) Prove by vector method the sine law \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} for any triangle ABC. [6](b) In a triangle, A = 30^{\circ}, B = 45^{\circ} and a = 10 cm. Find b. [2]
4. For the points A(1,1,1), B(0,1,3) and C(3,5,5), find(a) the area of triangle ABC,(b) a unit vector perpendicular to the plane ABC, and(c) the angle \angle BAC. Verify your area using \tfrac{1}{2}\,bc\sin A. [8]

Answers to Group C

1. (a) The scalar product of \vec{a} and \vec{b} is \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta, where \theta is the angle between them when they are drawn from a common point; the answer is a real number.

In triangle ABC put \vec{AB} = \vec{c} and \vec{AC} = \vec{b}, both drawn from A, so the angle between them is A and

\vec{b}\cdot\vec{c} = bc\cos A

By the triangle law, \vec{BC} = \vec{AC}-\vec{AB} = \vec{b}-\vec{c}. Square it with the dot product:

a^{2} = \left|\vec{b}-\vec{c}\right|^{2} = (\vec{b}-\vec{c})\cdot(\vec{b}-\vec{c})

Expand by the distributive law:

a^{2} = \vec{b}\cdot\vec{b} - \vec{b}\cdot\vec{c} - \vec{c}\cdot\vec{b} + \vec{c}\cdot\vec{c}

The scalar product is commutative, so the middle terms combine:

a^{2} = |\vec{b}|^{2} + |\vec{c}|^{2} - 2\,\vec{b}\cdot\vec{c}

Substituting \vec{b}\cdot\vec{c} = bc\cos A:

a^{2} = b^{2}+c^{2}-2bc\cos A

(b) The largest angle faces the longest side, 7. Put a = 7, b = 3, c = 5:

\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc} = \frac{9+25-49}{2(3)(5)} = \frac{-15}{30} = -\frac{1}{2}

A = 120^{\circ}

2. (a) Work in the plane z = 0 with A > B. Let \vec{OP} and \vec{OQ} be unit vectors making angles A and B with the positive x-axis:

\vec{OP} = \cos A\,\hat{i}+\sin A\,\hat{j}, \qquad \vec{OQ} = \cos B\,\hat{i}+\sin B\,\hat{j}

Cross them in components, taking \vec{OQ}\times\vec{OP} with (\cos B, \sin B, 0) and (\cos A, \sin A, 0). The \hat{i} and \hat{j} parts both contain a factor 0, so

\vec{OQ}\times\vec{OP} = \left(\cos B\sin A - \sin B\cos A\right)\hat{k}

Now use the definition: both are unit vectors, the angle from \vec{OQ} to \vec{OP} is A-B, and the turn is anticlockwise, so \hat{n} = +\hat{k}:

\vec{OQ}\times\vec{OP} = (1)(1)\sin(A-B)\,\hat{k}

Comparing the coefficients of \hat{k} in the two expressions:

\sin(A-B) = \sin A\cos B - \cos A\sin B

(b) Take A = 45^{\circ} and B = 30^{\circ}:

\sin 15^{\circ} = \sin 45^{\circ}\cos 30^{\circ} - \cos 45^{\circ}\sin 30^{\circ}

= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}

which is 0.2588.

3. (a) Take the sides of triangle ABC as vectors in order: \vec{a} = \vec{BC}, \vec{b} = \vec{CA}, \vec{c} = \vec{AB}. Going once round the triangle returns to the start, so

\vec{a}+\vec{b}+\vec{c} = \vec{0}

Substitute \vec{a} = -\vec{b}-\vec{c} into \vec{a}\times\vec{b}:

\vec{a}\times\vec{b} = (-\vec{b}-\vec{c})\times\vec{b} = -(\vec{b}\times\vec{b}) - (\vec{c}\times\vec{b})

Now \vec{b}\times\vec{b} = \vec{0} and -(\vec{c}\times\vec{b}) = \vec{b}\times\vec{c}, so \vec{a}\times\vec{b} = \vec{b}\times\vec{c}. Repeating with \vec{b} = -\vec{c}-\vec{a} gives \vec{b}\times\vec{c} = \vec{c}\times\vec{a}, so

\vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a}

Take magnitudes. Tail to tail, \vec{a} and \vec{b} are inclined at 180^{\circ}-C and \sin(180^{\circ}-C) = \sin C, so |\vec{a}\times\vec{b}| = ab\sin C, and likewise for the others. Hence

ab\sin C = bc\sin A = ca\sin B

Dividing throughout by abc, which is non-zero:

\frac{\sin C}{c} = \frac{\sin A}{a} = \frac{\sin B}{b} \quad\Longrightarrow\quad \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

(b) Using the sine law with a = 10, A = 30^{\circ}, B = 45^{\circ}:

b = \frac{a\sin B}{\sin A} = \frac{10\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{10(0.7071)}{0.5} = 14.14\ \text{cm}

Exactly, b = 10\sqrt{2} cm.

4. Build both sides from A:

\vec{AB} = (0-1,\; 1-1,\; 3-1) = (-1,\,0,\,2)

\vec{AC} = (3-1,\; 5-1,\; 5-1) = (2,\,4,\,4)

(a) Cross them:

\hat{i}\text{-part} = (0)(4)-(2)(4) = -8, \qquad \hat{j}\text{-part} = -\left[(-1)(4)-(2)(2)\right] = 8

\hat{k}\text{-part} = (-1)(4)-(0)(2) = -4

\vec{AB}\times\vec{AC} = -8\hat{i}+8\hat{j}-4\hat{k}, \qquad \left|\vec{AB}\times\vec{AC}\right| = \sqrt{64+64+16} = 12

\text{area} = \tfrac{1}{2}(12) = 6\ \text{square units}

(b) The cross product is normal to the plane, so a unit normal is

\hat{n} = \pm\frac{-8\hat{i}+8\hat{j}-4\hat{k}}{12} = \pm\left(-\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\right)

(c) For the angle at A use the scalar product of the same two vectors:

\vec{AB}\cdot\vec{AC} = (-1)(2)+(0)(4)+(2)(4) = -2+0+8 = 6

\left|\vec{AB}\right| = \sqrt{1+0+4} = \sqrt{5}, \qquad \left|\vec{AC}\right| = \sqrt{4+16+16} = 6

\cos A = \frac{6}{6\sqrt{5}} = \frac{1}{\sqrt{5}} = 0.4472 \; \Longrightarrow \; \angle BAC = 63.43^{\circ}

Check. With c = |\vec{AB}| = \sqrt{5}, b = |\vec{AC}| = 6 and \sin 63.43^{\circ} = 0.8944,

\tfrac{1}{2}bc\sin A = \tfrac{1}{2}(6)(2.2361)(0.8944) = 6.00

✔ in agreement with part (a).