Core 4 Revision Sheet
Algebra and Functions
Remainder Theorem
−𝑏
To find the remainder when a polynomial f(x) is divided by (ax + b) sub in f( )
𝑎
Eg, Find the remainder when f(x) = xᶟ + 2x² -5x + 3 is divided by (2x + 1)
−1
−1
−1
−1
7
2
2
2
2
8
f( ) = ( ) ᶟ + 2( )² - 5( ) + 3 = 5
𝟕
Therefore the remainder is 5 .
𝟖
Factor Theorem
𝑏
If (ax – b) is a factor of the polynomial f(x) then f( ) = 0
𝑎
Eg, Show (2x -1) is a factor of f(x) = 2xᶟ - x² -2x + 1. Hence, express f(x) as a product of three factors.
1
1
1
1
2
2
2
2
f( ) = 2( )ᶟ - ( ) ² - 2( ) + 1 = 0
therefore (2x -1) is a factor of f(x)
(2x – 1)(ax² + bx + c) = 2xᶟ - x² -2x + 1
(2x – 1)(x² - 1)
-
by inspection, expanding and comparing coefficients, or long division methods
(2x – 1) (x + 1) (x – 1)
Partial Fractions
Express
7𝑥+16
in partial fractions
𝑥²+2𝑥−8
7𝑥+16
𝑥²+2𝑥−8
𝐴
=
𝑥+4
+
𝐵
- multiply both sides by (x + 4)(x - 2)
𝑥−2
7x + 16 = A(x - 2) + B(x + 4)
When x = 2
- substitute values in for x
7(2) + 16 = 0A + 6B
when x = -4
7(-4) + 16 = -6A
30 = 6B
-12 = -6A
B=5
𝟕𝒙+𝟏𝟔
𝒙²+𝟐𝒙−𝟖
A=2
=
𝟐
𝒙+𝟒
+
𝟓
𝒙−𝟐
Repeated factors
Express
2𝑥−5
(𝑥−4)²
in partial fractions
2𝑥−5
(𝑥−4)²
=
𝐴
𝑥−4
+
𝐵
(𝑥−4)²
2x – 5 = A (x – 4) + B
2(4) – 5 = B B = 3
-
multiply both sides by (x – 4)²
-
sub in x = 4
A = 2 (comparing x terms)
𝟐𝒙−𝟓
(𝒙−𝟒)²
=
𝟐
𝒙−𝟒
+
𝟑
(𝒙−𝟒)²
Improper Fractions
Express
𝑥²+7𝑥−14
(𝑥+5)(𝑥−3)
in partial fractions
𝑥²+7𝑥−14
(𝑥+5)(𝑥−3)
=A+
𝐵
(𝑥+5)
+
𝑐
-
(𝑥−3)
multiply both sides by (x + 5)(x – 3)
x² + 7x – 14 = A(x + 5)(x - 3) + B(x – 3) + c(x + 5)
A = 1 (comparing coefficients of x²)
when x = -5
-24 = -8B
𝒙²+𝟕𝒙−𝟏𝟒
(𝒙+𝟓)(𝒙−𝟑)
=1+
B=3
𝟑
(𝒙+𝟓)
+
𝟐
(𝒙−𝟑)
when x = 3
16 = 8C
C=2
Binomial Series
𝑛(𝑛−1)𝑥²
(1 + x)𝑛 = 1 + nx +
2!
+
𝑛(𝑛−1)(𝑛−2)𝑥ᶟ
3!
+ ……. + 𝑥 𝑛
- this is in your formula book on pg 4.
When n is not a positive integer, the expansion will not terminate and is only valid for -1 < x < 1, or |x| < 1
Therefore, you can only get an approximation of the expansion, usually up to xᶟ
Eg, Obtain the expansion of (1 + 2𝑥)−3 up to and including the term in xᶟ, and state the range of x for which it is valid
(1 + 2𝑥)−3 = 1 + (-3)(2x) +
(−3)(−4)(2𝑥)²
+
2𝑥1
(−3)(−4)(−5)(2𝑥)ᶟ
3𝑥2𝑥1
= 1 – 6x + 24x² - 80xᶟ
|x| < 1
Expansion of (𝐚 + 𝐱)𝒏
Factorise the expression first by taking ‘a’ out as a factor
1
Eg, Expand (4 – 3𝑥)−2 in ascending powers of x up to and including the term x². Hence evaluate
1
, giving your answer
√3.97
to four decimal places
1
3
(4 – 3𝑥)−2 = [4(1 −
4
1
1
x)]−2
3
= 4−2 (1 −
4
1
x)−2
1
2
3
2
3
4
1
1
3
(− )(− )(− )²
2
2
4
2𝑥1
= [1 + (− ) (− 𝑥) +
1
3
2
𝟏
8
𝟑
𝟐
𝟏𝟔
= [1 +
= +
x+
𝑥+
𝟐𝟕
27
128
]
𝑥²]
𝟑
𝟐𝟓𝟔
x² for | 𝒙| < 𝟏
Hence,
𝟒
|x| <
𝟒
1
√3.97
=
1
√4−3𝑥
√3.97 = √4 − 3𝑥
𝟑
3.97 = 4 – 3x
x = 0.01 (sub into expansion)
1
2
+
3
16
(0.01) +
27
256
(0.01)² = 0.50188554…… = 0.5019 (4d.p.)
Applications of partial fractions to series expansions
Express as partial fractions first, and then carry out the binomial expansion.
2−𝑥
Eg, Express (1+𝑥)(1−2𝑥) in partial fractions. Hence obtain the series expansion, giving all terms up to and including xᶟ, and
state the values of x for which the expansion is valid.
2−𝑥
(1+𝑥)(1−2𝑥)
1
(1+𝑥)
=
1
(1+𝑥)
+
1
(1−2𝑥)
= 1(1 + x)−1
= 1(1 + (−1)x +
(−1)(−2)𝑥²
2𝑥1
= 1 – x + x² - xᶟ
1
(1−2𝑥)
= 1(1 − 2
= 1 + 2x + 4x² + 8xᶟ
Therefore,
(1+𝑥)
+
1
(1−2𝑥)
(−1)(−2)(−3)𝑥ᶟ
3𝑥2𝑥1
)
for |x| < 1
x)−1
= 1(1 + (−1)(−2x) +
1
+
(−1)(−2)(−2𝑥)²
2𝑥1
+
(−1)(−2)(−3)(−2𝑥)ᶟ
3𝑥2𝑥1
)
for |x| < 1
= (1 – x + x² - xᶟ) + (1 + 2x + 4x² + 8xᶟ)
= 2 + x + 5x² + 7xᶟ
valid for |x| < 1
Remember, if the range is not |x|< 1 for both, choose the smaller range so that it is valid for both
Trigonometry
Compound Angles
Compound angle formulae are in your formula book on pg 5.
sin (A + B) = sinA cosA + cosA sinB
sin (A – B) = sinA cosB – cosA sinB
cos (A + B) = cosA cosB – sinA sinB
cos (A – B) = cosA cosB + sinA sinB
tan (A+ B) =
𝑡𝑎𝑛𝐴+𝑡𝑎𝑛𝐵
tan (A – B) =
1−𝑡𝑎𝑛𝐴 𝑡𝑎𝑛𝐵
𝑡𝑎𝑛𝐴−𝑡𝑎𝑛𝐵
1+ 𝑡𝑎𝑛𝐴 𝑡𝑎𝑛𝐵
Double Angles
sin 2x = 2sinxcosx
cos 2x = cos²x - sin²x OR
tan 2x =
cos 2x = 2cos²x – 1
OR
cos 2x = 1 – 2sin²x
2 𝑡𝑎𝑛𝑥
1−𝑡𝑎𝑛²𝑥
Triple-angle formulae
sin 3x = 3 sinx – 4 sin³x
cos 3x = 4 cos³x – 3cosx
tan 3x =
3 𝑡𝑎𝑛𝑥−𝑡𝑎𝑛³𝑥
1−3𝑡𝑎𝑛²𝑥
Both the double angle and triple-angle formulae are used in integration with powers of sin x and cos x
𝝅
Eg, work out ∫𝟎𝟒 𝟒 𝒄𝒐𝒔² 𝒙 𝒅𝒙
1
cos²x = (1 + cos2x) -
rearrangement of cos 2x = 2cos²x – 1
2
1
4 cos²x = 4[ (1 + cos2x)]
2
= 2 (1 + cos2x)
= 2 + 2cos2x
𝜋
𝜋
∫04 4 𝑐𝑜𝑠² 𝑥 𝑑𝑥 = ∫04 2 + 2𝑐𝑜𝑠2𝑥 𝑑𝑥
𝜋
= [2𝑥 + 𝑠𝑖𝑛2𝑥]04
𝜋
𝜋
= (2 ( ) + 𝑠𝑖𝑛 2( )) – (2 (0) + 𝑠𝑖𝑛 2(0))
4
𝜋
= +1–0
2
4
=
𝝅+𝟐
𝟐
Harmonic Form
Expressing any function f(𝜃) = a cos𝜃 + b sin𝜃 in the form
R sin (𝜽 ± ∝) or R cos (𝜽 ± ∝)
where R > 0 and ∝ is acute is called expressing in harmonic form.
For f(𝜃) = a cos𝜃 + b sin𝜃
R = √𝑎² + 𝑏²
𝑏
tan ∝ = , for R cos (𝜃± ∝)
𝑎
or
𝑎
tan ∝ = , for R sin (𝜃± ∝)
𝑏
It allows you to solve equations in the form a cos 𝜃 + b sin 𝜃 = c, as well as finding maximum and minimum points of such functions.
Eg, Express 12 sin𝜽 + 5 cos 𝜽 in the form R sin (𝜽+ ∝) where R > 0 and ∝, measured in degrees to 1 decimal place, is acute. Hence, solve
the equation 12 sin𝜽 + 5 cos 𝜽 =
𝟏𝟑
𝟐
for 0 ≤ 𝜽 ≤ 360°, giving your answer to 1 d.p.
R = √12² + 5²,
Let 12 sin 𝜃 + 5 cos 𝜃 = R sin (𝜃 + 𝛼)
tan ∝ =
5
R = 13
∝ = 22.6°
,
12
Therefore 12 sin 𝜃 + 5 cos 𝜃 = 13 sin (𝜽 + 22.6°)
So, 12 sin 𝜃 + 5 cos 𝜃 =
13
2
becomes
13 sin (𝜃 + 22.6°) =
sin (𝜃 + 22.6°) =
sin (𝜃 + 22.6°) =
13
2
13
2(13)
1
2
1
𝜃 + 22.6° = sin−1 ( )
2
𝜃 + 22.6° = 30° , 150°
𝜃 = 7.4° , 127.4°
Remember, the value for R gives the minimum or maximum value of the function. For example, when R = 5, the minimum value will be at 5 and the maximum value at +5. This is due to the fact the original function, sin x or cos x, has been multiplied by the constant term R.
Exponential Models
This uses functions of the form y = A x 𝑎𝑏𝑡+𝑐 where A, a, b, and c are constants and t is time.
Eg, The value of a car depreciates in such a way that t years after it is new its value, £V, is given by the formula
V = 15000 x (𝟏. 𝟒)−𝒌𝒕
a)
What is the initial value of the car?
After 2 years the value of the car is £8000.
b)
c)
a)
Calculate the value of the constant k to three decimal places.
Calculate the time, to the nearest month, when the value of the car reaches £5000.
Initial value, when t = 0,
V = 15000 x (1.4)0 = 15000
b)
When V = 8000, and t = 2
8000 = 15000 x (1.4)−2𝑘
8000
= (1.4)−2𝑘
15000
8000
log(
15000
8000
)
15000
log(
log (1.4)
c)
-
take logs of both sides
-
take logs of both sides
) = -2k log (1.4)
= -2k
k = 0.934 (3d.p)
When V = 5000,
5000 = 15000 x (1.4)−0.934𝑡
5000
= (1.4)−0.934𝑡
15000
5000
log
15000
5000
log(
)
15000
log (1.4)
= -0.934t log (1.4)
= -0.934t
t = 3.5 years or 3 years, 6 months
The Exponential Function
This involves exponential functions to the base e, i.e. y = A𝑒 𝑘𝑥 , where A is a constant.
Remember ln 𝑒 𝑥 = x
Eg, A pan of water is heated and then allowed to cool. Its temperature, T℃, t minutes after cooling has begun, is given
by the formula
T = 80𝒆−𝟎.𝟐𝒕
a)
b)
c)
What is the temperature of the water when it starts to cool?
What is the temperature after 5 minutes?
Find, to the nearest minute, the time taken for the water to reach a temperature of 12℃.
a)
When t = 0,
T = 80𝑒 0
T = 80℃
b)
When t = 5,
T = 80𝑒 −0.2(5)
T = 29.4℃
c)
When T = 12,
12 = 80𝑒 −0.2𝑡
12
80
= 𝑒 −0.2𝑡
-
take logs of both sides
12
ln( ) = -0.2t
80
t = 9.486,
therefore 9 minutes (nearest min)
Further Calculus
Implicit Functions
These are functions that are not expressed in the form y = f(x), for example x² + 2xy + yᶟ = 5.
Differentiate using the chain rule and product rule.
Eg, Differentiate xy² + 3x - 2y = 6
2xy
𝑑𝑦
𝑑𝑦
+ y² + 3 - 2
𝑑𝑥
𝑑𝑦
for xy², let u = x and v = y²
du
=0
𝑑𝑥
dx
dv
=1
dx
= 2y
dy
dy
dx
dx
= 2xy
dy
dx
+ y²
(2xy - 2) + 3 + y² = 0
𝑑𝑥
𝑑𝑦
𝑑𝑥
(2xy - 2) = -3 – y²
𝑑𝑦
𝑑𝑥
=
−3 – y²
(2xy−2)
=−
𝟑+ 𝐲²
(𝟐𝐱𝐲−𝟐)
Hence, find an equation to the normal to the curve at the point (2, 1)
Gradient of tangent at x = 1
𝑑𝑦
𝑑𝑥
=−
𝟑+(𝟏)²
𝟒
(𝟐(𝟐)(𝟏)−𝟐)
1
Therefore, gradient of normal is
= - = -2
𝟐
2
1
Equation of normal is
y – 1 = (x – 2)
2
2y – 2 = x – 2
2y = x
Parametric Equations
This is where x and y are defined in terms of a third variable t. You may be asked to find a Cartesian equation for
parametric forms.
Eg, 1.
2.
x=t–4
y = 2t²
t=x+4
sub in for t in the equation for y
x = 3sin t,
y = 2cos t
𝑥
sin t =
cos t =
3
y = 2(x + 4)²
𝑦
2
using the identity sin²t + cos²t = 1
𝑥 2
𝑦 2
3
2
( ) + ( ) = 1 or 4x² + 9y² = 1
Parametric Differentiation
In order to find
𝑑𝑦
𝑑𝑥
you must find
Eg,1. find an expression for
𝒅𝒚
𝒅𝒙
𝑑𝑦
𝑑𝑡
and
y = 2t² - 3
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 12tᶟ
=
𝑑𝑡
1
, then apply the chain rule.
in terms of t for x = 3𝒕𝟒 , y = 2t² - 3
x = 3𝑡 4
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑥
= 4t
therefore,
12𝑡 3
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
x
= 4t x
𝑑𝑡
𝑑𝑥
1
12𝑡 3
=
𝟏
𝟑𝒕²
2. Find the equation of the tangent to the curve x = 3t², y = 7 + 12t, at the point where t = 2.
x = 3t²
𝑑𝑥
𝑑𝑡
= 6t
y = 7 + 12t
𝑑𝑡
𝑑𝑥
=
1
𝑑𝑦
6𝑡
𝑑𝑡
When t = 2,
𝑑𝑦
𝑑𝑥
y – 31 = x – 12
2
= =1
2
𝑑𝑦
= 12
𝑑𝑥
= 12 x
1
6𝑡
=
𝟐
𝐭
When t = 2, x = 3(2)² = 12, and y = 7 + 12(2) = 31
y = x + 19
Using Partial Fractions
Expressing functions as partial fractions can enable you to integrate them.
Eg, Evaluate, giving your answer in the form ln
𝟕
𝟏
∫𝟓 (𝒙−𝟑)(𝒙−𝟒)
where a and b are rational numbers.
𝒃
𝒅𝒙
1
(𝑥−3)(𝑥−4)
7 1
∫5
𝒂
−
𝑥−4
1
=
𝑥−4
1
𝑥−3
-
1
1
Express (𝑥−3)(𝑥−4) in partial fractions
𝑥−3
𝑑𝑥
Now integrate
[ln(𝑥 − 4) − ln(𝑥 −
3)]75
(ln 3 – ln 4) – (ln 1 – ln 2)
3
1
4
2
3
𝟑
ln ( ) – ln ( ) = ln ( 41 ) = ln ( )
2
𝟐
Differential Equations
These are equations that include x and y as well as one or more derivatives of y with respect to x.
To solve these, you must separate the variables and integrate both sides with respect to x. Remember, you only need to
add one constant value, c.
Eg,
𝒅𝒚
𝒅𝒙
=
𝒙+𝟏
, y = 3 at x = -2. Find the particular solution to the differential equation.
𝒚
𝑑𝑦
𝑑𝑥
=
𝑥+1
𝑦
y
𝑑𝑦
=x+1
𝑑𝑥
separate the variables
∫ 𝑦 𝑑𝑦 = ∫ 𝑥 + 1 𝑑𝑥
𝑦²
2
=
𝑥²
2
+x+c
when y = 3, x = -2
9
2
integrate both sides with respect to x
sub in the values
=2–2+c
c=
9
therefore,
2
𝑦²
2
=
𝑥²
2
+x+
9
2
y² = x² + 2x + 9
y = √𝐱² + 𝟐𝐱 + 𝟗
Modelling with differential equations
You may be asked to solve differential equations within a context.
Eg, At an air temperature of -T℃ the thickness, x cm, of the ice on a pond at a time t minutes satisfies the equation
𝒅𝒙
𝒅𝒕
a)
=
𝑻
𝟏𝟒𝟎𝟎𝟎𝒙
, where x = 0 at t = 0
Show that 7000x² = Tt
14000x
𝑑𝑥
𝑑𝑡
=T
∫ 14000x dx = ∫ 𝑇 𝑑𝑡
14000𝑥²
2
= Tt + c
7000x² = 𝑇𝑡 + c
x = 0, when t = 0, therefore c = 0
7000x² = 𝑻𝒕 (as required)
b)
Given T = 10, when the ice starts forming, calculate how long it takes for the thickness of the ice to reach 2cm.
7000x² = 𝑇𝑡
7000 (2)² = 10t
28000 = 10t
T = 2800 seconds or 46 minutes, 40 seconds
Applications to exponential laws of growth and decay
In general, if the rate of growth of x is proportional to x, then
If the rate of decay of x is proportional to x, then
𝒅𝒙
𝒅𝒕
𝒅𝒙
𝒅𝒕
= kx, where k is a positive constant.
= -kx, where k is a positive constant.
Eg, The value of a car depreciates in such a way that when it is t years old, the rate of decrease in its value is
proportional to the value, £V, of the car at that time. The car costs £12,000 when new.
a)
Show that V = 12000𝒆−𝒌𝒕
𝑑𝑉
𝑑𝑡
1 𝑑𝑉
𝑉
= -kV
= -k
𝑉 𝑑𝑡
1
𝑡
∫12000 𝑉 dV = ∫0 −𝑘 dt
[ln
𝑉
V]12000
= [−kt
-
separate the variables
-
integrate both sides
]𝑡0
ln V – ln 12000 = -kt
ln (
𝑉
12000
𝑉
) = -kt
12000
= 𝑒 −𝑘𝑡
V = 12000𝒆−𝒌𝒕 (as required)
When the car is three years old its value has dropped to £4000.
b)
𝟏
Show that k = ln 3
𝟑
t = 3, when V = 4000
4000 = 12000𝑒 −3𝑘
𝑒 −3𝑘 =
𝑒 −3𝑘 =
4000
12000
1
3
−3k = ln
1
3
1
𝑘 = − ( ln 1 – ln 3)
3
1
k = 0 + ln 3
3
𝟏
k = ln 3 (as required)
𝟑
The owner decides to sell the car when its value reaches reaches £2000.
c)
Calculate, to the nearest month, the age of the car at that time.
V = 12000𝑒 −𝑘𝑡
when V = 2000:
2000 = 12000𝑒 − 𝑘𝑡
𝑒 −𝑘𝑡 =
1
6
−𝑘𝑡 = ln
1
- ln (3)t = ln
3
t=
1
6
1
6
1
6
1
− ln (3)
3
ln
t = 4.9 which is 4 years and 11 months (nearest month)
Vectors
⃗⃗⃗⃗⃗ = (5) means to go from O to A is 5 right and 2 up. It is called a column vector.
OA
3
⃗⃗⃗⃗⃗ is represented by the length as follows:
The magnitude or modulus of the vector OA
⃗⃗⃗⃗⃗
|OA|² = 5² + 3² = 34
⃗⃗⃗⃗⃗ | = √34
|OA
Addition and Subtraction of Vectors
⃗⃗⃗⃗⃗ and 𝐵𝐶
⃗⃗⃗⃗⃗ , then to find 𝐴𝐶
⃗⃗⃗⃗⃗ :
If given 𝐴𝐵
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
𝐴𝐶 = 𝐴𝐵 + 𝐵𝐶
𝟐
−𝟐
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ = ( 𝟓) and ⃗⃗⃗⃗⃗⃗
Eg, 1. Given that 𝑨𝑩
𝑩𝑪 = ( 𝟑), find 𝑨𝑪
−𝟑
−𝟏
⃗⃗⃗⃗⃗ = AB
⃗⃗⃗⃗⃗ + BC
⃗⃗⃗⃗⃗
AC
2
−5
= ( 5 ) + ( 3)
−1
−3
−𝟑
⃗⃗⃗⃗⃗
𝐀𝐂 = ( 𝟖)
−𝟒
𝟑
−𝟐
⃗⃗⃗⃗⃗ ( 𝟏) , 𝒇𝒊𝒏𝒅𝑩𝑨
⃗⃗⃗⃗⃗⃗
2. Given that ⃗⃗⃗⃗⃗⃗
𝑩𝑪 = ( 𝟒) and 𝑨𝑪
−𝟑
−𝟐
⃗⃗⃗⃗⃗ = 𝐵𝐶
⃗⃗⃗⃗⃗ + 𝐶𝐴
⃗⃗⃗⃗⃗
𝐵𝐴
⃗⃗⃗⃗⃗
𝐵𝐴 = ⃗⃗⃗⃗⃗
𝐵𝐶 - ⃗⃗⃗⃗⃗
𝐴𝐶
3
−2
= ( 4) - ( 1)
−3
−2
𝟓
⃗⃗⃗⃗⃗⃗ = (𝟑)
𝑩𝑨
𝟏
Scalar Multiplication
If you multiply a vector by a scalar (magnitude), you get another vector. Eg, 3 x a = 3a
3a and a are parallel
In general, if u = kv then u is parallel to v
Position vectors and direction vectors
⃗⃗⃗⃗⃗
The position vector of a point P with respect to a fixed origin O is the vector 𝑂𝑃
𝟏
Eg,If P has coordinate (1, 2, 3) then ⃗⃗⃗⃗⃗⃗
𝑶𝑷 = p = (𝟐)
𝟑
𝟑
⃗⃗⃗⃗⃗⃗ = q = (𝟓)
If Q has coordinate (3, 5, 2) then 𝑶𝑸
𝟒
𝟐
⃗⃗⃗⃗⃗⃗ = (𝟑)
⃗⃗⃗⃗⃗⃗ is the direction vector from P to Q.
Therefore, 𝑷𝑸
𝑷𝑸
𝟏
Collinearity (lying on a straight line)
Vectors can be used to show three points lie in a straight line.
𝟕
𝟏𝟏
𝟓
Eg, Points P, Q and R have position vectors (𝟒) , (𝟓) and ( 𝟕 ), respectively.
𝟒
𝟏𝟎
𝟏
⃗⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗⃗
a) Find 𝑷𝑸
𝑸𝑹
b)
4
2
⃗⃗⃗⃗⃗ = (1)
⃗⃗⃗⃗⃗ = (2)
𝑃𝑄
𝑄𝑅
3
6
Deduce that P, Q and R are collinear and find the ratio PQ:QR
4
2
`(2) = 2(1)
6
3
⃗⃗⃗⃗⃗ = 2𝑃𝑄
⃗⃗⃗⃗⃗ which means they are parallel. They share point Q, making them collinear.
and 𝑄𝑅
PQ:QR = 1:2
Scalar product
The scalar product a.b of two vectors a and b is defined by a.b = |𝒂| |𝒃| cos 𝜽 where 𝜽 is the angle between the vectors.


When two vectors a and b are perpendicular, 𝜃 = 90° and cos 𝜃 = 0. Therefore, a.b = 0
When the angle between the two vectors a and b is acute, cos 𝜃 > 0 and a.b > 0
𝑎1
𝑏1
If a = (𝑎2 ) and b = (𝑏2 ), then a.b = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3
𝑎3
𝑏3
𝟐
𝟒
Eg, find the angle between the vectors a = ( 𝟏) and b = ( −𝟑 ), giving answers to one decimal place.
−𝟐
𝟏𝟐
a.b = |𝒂| |𝒃| cos 𝜃
2
4
a.b = ( 1 ). (−3) = (2 x 4) + (1 x -3) + (-2 x 12) = -19
−2
12
|𝒂| = √2² + 1² + (−2)² = √9 = 3
|𝒃| = √4² + (−3)2 + 12² = √169 = 13
Therefore, a.b = |𝒂| |𝒃| cos 𝜃 gives -19 = 3 x 13 cos 𝜃
cos 𝜃 =
−19
39
and 𝜽 = 119.2°
Perpendicular vectors


If a.b = 0, then a and b are perpendicular vectors
If a and b are perpendicular vectors, then a.b = 0
𝟐
𝟏
Eg, Given that the vectors ( 𝒕 ) and( −𝟑 )are perpendicular, find the value of the constant t.
𝒕−𝟒
−𝟒
a.b = 0
2
1
( t ) . ( −3 ) = (2 x 1) + (t x -3) + (-4(t – 4)) = 0
t−4
−4
2 – 3t – 4t + 16 = 0
7t = 18
t=
18
7
=2
𝟒
𝟕
Vector equation of a line
r = a + t (b – a)
-
where t is a scalar
Eg, Find a vector equation of the line passing through A( -4, 1, - 3) and B (6, -4, 12). Show that the point Q (-2, 0, 0) lies
on the line AB.
r = a + t (b – a)
−4
−4
6
r = ( 1 ) + t [(−4) − ( 1 ) ]
−3
−3
12
𝒙
−𝟒
−𝟒
−𝟒 + 𝟐𝒕
𝟏𝟎
𝟐
r = ( 𝟏 ) + t (−𝟓) = ( 𝟏) + t (−𝟏)
therefore, (𝒚) = ( 𝟏 − 𝒕 )
𝒛
−𝟑
𝟏𝟓
−𝟑
𝟑
−𝟑 + 𝟑𝒕
−2
compare against Q ( 0 ) . If Q lies on the line then:
0
−4
−2
2
( 0 ) = ( 1 ) + t (−1)
0
−3
3
−4 + 2𝑡
−2
( 0) = ( 1 − 𝑡 )
x -> -2 = -4 + 2t
t=1
0
−3 + 3𝑡
y -> 0 = 1 – t
t=1
z -> 0 = -3 + 3t
t=1
since t = 1 works for x, y and z, Q lies on the line.
Parallel and skews lines
If one vector is a scalar multiple of another the vectors are parallel.
−1
2
The direction vectors ( 3 ) and ( −6 ) are parallel
5
−10
In three dimensions, it is possible that lines are not parallel and also do not intersect. These are called skew. You can find
the angle between such lines by finding the angle between the direction vectors.
Eg, a) show that these two lines are skew:
𝟐
𝟏𝟐
𝟓
𝟓
𝒓𝟏 = (𝟏) + 𝝀 (−𝟏) and 𝒓𝟐 = (𝟒) + µ (−𝟔)
𝟑
𝟒
𝟔
𝟐
You need to show they do not intersect by finding the values of λ and µ using a pair of simultaneous equations and show
they do not satisfy the third component.
x -> 2 + 5𝜆 = 5 + 12µ
solving simultaneously gives λ = 3, µ = 1
y -> 1 - λ = 4 - 6µ
check these values in z -> 4 + 6(3) = 2 +3(1)
22 = 5
b)
no, therefore they do not intersect and are skew.
Calculate the angle between the two lines.
12
5
You need to calculate the angle between the direction vectors a (−1) and b (−6)
3
6
12
5
a.b = (−1) . (−6) = (5 x 12) + (-1 x -6) + (6 x 3) = 84
3
6
|𝒂|= √5² + (−1)2 + 6² = √62
|𝒃| =√12² + (−6)2 + 3² = √189
𝑐𝑜𝑠 𝜃 =
84
√62 √89
𝜃 = 𝑐𝑜𝑠 −1 (
84
√62 √189
) = 39.1°