Higher Maths

Revision Notes Vectors

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Vectors in three dimensions

use scalar product to find the angle between two directed line segments

know the terms: vector, magnitude (length), direction, scalar multiple, position vector,unit vector, directed line segment, component, scalar product

determine the distance between two points in

three dimensional space

know and apply the equality fact

know and apply the fact that if u and v are vectors that can be

represented by parallel lines then u = kv where k is a constant and

the converse

know and apply the fact that if A, P and B are collinear points such that

determine whether three points with given coordinates

are collinear

know and apply the basis vectors i, j, kknow the scalar product facts:

determine whether or not two vectors, in component form, are

perpendicular

a.b = |a| |b| cos θ

a.b = a1b1 + a2b2 + a3b3

a.(b + c) = a.b + a.c

Test Yourself?

B

A

O

a

b

x

y

z

u

A vector u is represented by the directed line segment .

The vector OA is called the position vector of Aand is usually denoted by a.

The vector OB is called the position vector of Band is usually denoted by b.

From the diagram we see that a+ u = b.giving u = b – a

i.e. = b – a.

Test Yourself?

The converse is also true: If u and v are parallel then u = kv

If AB = u and BC = v AND u = kv then, since they share a common point, B,

AB and BC are line segments of the same line. i.e. A, B, and C are collinear.

If vectors u and v are such that u = kv then u and v are parallel.

Test Yourself?

Section Formula: If A, B and P are collinear with P between A and B,

A

B

P

km

kn

Using position vectors: n(p – a) = m(b – p)

Test Yourself?

Vectors i, j, and k are unit vectors which run in the x, y and z-directions respectively.

Any other vector can be expressed in terms of i, j, and k .

The scalar (dot) product of two vectors a and b is defined as

Since cos 90˚ = 0, when two vectors are perpendicular, their scalar product is zero.

Test Yourself?

A

B

C

θ

• Check BA and BC are tail-to-tail• Find BA using BA = a – b Find BC using BC = c – b • Find

• Find BA.BC using components.• Calculate cos θ• Hence calculate θ.

Test Yourself?

Equality:

When two vectors are perpendicular, their scalar product is zero.

Perpendiculars:

revealreveal

Three points on the ‘Cherry Picker’ have coordinates, A(4, –10, 8), B (1, 2, 4), and C(–3, –1, 16).

Show that B is equidistant from A and C.

Three points on the ‘Cherry Picker’ have coordinates, A(4, –10, 8), B (1, 2, 4), and C(–3, –1, 16).

Show that B is equidistant from A and C.

AB = CB = 13.B is equidistant from A and C.

revealreveal

The photographer was at P(1, 3, 5)The chimney of the castle was at C(2, 9, 20).A hill walker on the skyline was at H(6, 33, k).For what value of k is the photographer, chimney and hill walker in a straight line?

For P, C and H to be collinear, we require PC and PH to have a common point … P in this case,AND PC|| PH … so PH = aPC

Now 5 = a ✕ 1 and 30 = a ✕ 6 … i.e. a = 5

So k – 5 = a ✕ 15 = 5 ✕ 15 = 75

So k = 80 when photographer, chimney and walker are in a straight line.

The photographer was at P(1, 3, 5)The chimney of the castle was at C(2, 9, 20).A hill walker on the skyline was at H(6, 33, k).For what value of k is the photographer, chimney and hill walker in a straight line?

A

P

BA length of rigging starts at A(1, 2, 7) and runs to B(6, 12, 27).

Find the point P that dividesthe length of rope, AB, in the ratio 3:2

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A

P

BA length of rigging starts at A(1, 2, 7) and runs to B(6, 12, 27).

Find the point P that dividesthe length of rope, AB, in the ratio 3:2

P is the point (4, 8, 19)

(a) u = 3j + 4k and v = 2i + 3j – 5k.Find u.v

(b) ABC is an equilateral triangle.Find

(c) ABCD is a square of side 1 unit.BD is a quarter circle centre A.EF cuts the square into two congruent rectangles.G lies on the arc BD.Calculate:

A

B C

DE

F

G

revealreveal

(a) u = 3j + 4k and v = 2i + 3j – 5k.Find u.v

(b) PQRis an equilateral triangle of side 2 units.Find

(c) ABCD is a square of side 1 unit.BD is a quarter circle centre A.EF cuts the square into two congruent rectangles.G lies on the arc BD.Calculate:

A

B C

DE

F

G

(a) u.v = (3j + 4k)(2i + 3j – 5k) = 6ji + 9jj – 15jk + 8ki + 12kj – 20kk = 9 – 20 = –11

(b)

(c)

A(7, 4, 1)

B(9, 6, 15)

C(15, 3, 5)

Calculate the size of angle ABC.

revealreveal

A(7, 4, 1)

B(9, 6, 15)

C(15, 3, 5)

Calculate the size of angle ABC.

Two vectors are defined as follows:

(a) For what value of x is u = v?(b) For what value of x are u and v perpendicular?

revealreveal

Two vectors are defined as follows:

(a) For what value of x is u = v?(b) For what value of x are u and v perpendicular?

(a) u = v when 2x = x + 2 and 4 – x = x This occurs when x = 2

(b) Vectors are perpendicular when u.v = 0