Edexcel A-Level Further Mathematics Exam
1. How to prove by induction an expression is divisible by a number e.g. uₙ=4ⁿ+6n-1 is
divisible by 9 for all n>=1?
- find the value of expression for n=1 and if divisible by the number then true for base case e.g.
u₁=9 so it's divisible by 9
- assume true for n=k so uₖ=4ᵏ+6k-1 is divisible by 9
- for n=k+1, uₖ₊₁=4ᵏ⁺¹+6(k+1)-1 = 4x4ᵏ+6k+5
- rearrange the assumption to get 4ᵏ=uₖ-6k+1 and sub into expression for n=k+1 to get 4(uₖ-
6k+1)+6k+5 = 4uₖ-18k+9 = 4uₖ-9(2k-1)
- we assumed uₖ is divisible by 9 and 9(2k-1) is also divisible by 9 uₖ₊₁ must be divisible by 9
- blah
2. What to do for a more complicated proof by induction for divisibility if normal method isn't
getting you anywhere?
- find expression for difference between uₖ₊₁ and uₖ and try to show that THIS instead is
divisible by the number
- if it is, since you've assumed uₖ is divisible by the number it follows that uₖ₊₁ has to be
divisible by that number
3. Why is scalar product commutative?
Simply because multiplication of numbers is commutative
,4. Explain vector equation of a line
r=a+λb
- r is position vector of general point on line <x,y,z>
- a is the position vector of a point on the line to "step into the line"
- b is direction vector specifying the "gradient"
- λ is a parameter that you change to move along the line
5. Is r = <1,2,3> + λ<2,5,-2> the only equation for the corresponding line? Explain
No
You could pick any point on the line and any multiple of <2,5,-2> would be the direction vector
as they would all have be on the same line
6. In vector equation of line, explain each case when λ=0, λ=1, 0<λ<1, λ>1 and λ<0
- when λ=0 the point is the position vector of point A
- when λ=1 the point is the position vector of point B, one "step" away from A
- when 0<λ<1 the point is between A and B
- when λ>1 the point is beyond B
- when λ<0 the point is beyond A, opposite direction to B
7. Explain cartesian equation of line in 3D
,- take the vector equation and substitute ₖ=<x,y,z>
- add the vectors and rearrange for λ
- you should get λ = (x-a₁)/d₁ = (y-a₂)/d₂ = (z-a₃)/d₃ where <a₁, a₂, a₃> is position vector of a point
on the plane and <d₁, d₂, d₃> is direction vector
8. What to do if when converting from vector to Cartesian equation of line, the direction
vector has exactly one 0 element? E.g. line through <7,2,3> in direction <0,5,2>
- in the example you end up with λ = (x-7)/0 = (y-2)/5 = (z-3)/2 but obviously you can't divide by
0
- you go back and see that x=7+0λ so x=7
- so you write the Cartesian equation as x=7 and λ=(y-2)/5 = (z-3)/2
9. Arrangement of 2 lines in 3D + conditions
Intersect at unique point
- set the position vectors r equal for both lines and set up simultaneous equations involving λ
and μ where each are the parameters of the respective lines
- if the values for λ and μ found don't satisfy 3rd equation then the lines don't intersect at a
point so either parallel or skew
- else sub either parameter into one of the line equations to get the intersection point
Parallel
- no intersection point and direction vectors are multiples of each other
Skew
- no intersection point and direction vectors aren't parallel to each other
, 10. Angle between two lines
Find angle between their direction vectors using scalar product
11. Why can you find angle between two lines in 3D even if they are skew?
- the angle between them is still the angle between their direction vectors
- imagine translating one of the lines so the two lines DO intersect - you are finding the angle
between lines at this point
12. How to find intersection point between plane and line?
- from the vector equation, the each of x y and z will be some function of λ
- substitute these into the Cartesian equation of the plane and solve for λ
- sub λ into vector equation to get the intersection point
13. How to find acute angle between line and a plane?
- the acute angle is the angle between the line and it's orthogonal projection onto the plane
- find angle between the normal vector and the direction vector of the line and take this from
90°
1. How to prove by induction an expression is divisible by a number e.g. uₙ=4ⁿ+6n-1 is
divisible by 9 for all n>=1?
- find the value of expression for n=1 and if divisible by the number then true for base case e.g.
u₁=9 so it's divisible by 9
- assume true for n=k so uₖ=4ᵏ+6k-1 is divisible by 9
- for n=k+1, uₖ₊₁=4ᵏ⁺¹+6(k+1)-1 = 4x4ᵏ+6k+5
- rearrange the assumption to get 4ᵏ=uₖ-6k+1 and sub into expression for n=k+1 to get 4(uₖ-
6k+1)+6k+5 = 4uₖ-18k+9 = 4uₖ-9(2k-1)
- we assumed uₖ is divisible by 9 and 9(2k-1) is also divisible by 9 uₖ₊₁ must be divisible by 9
- blah
2. What to do for a more complicated proof by induction for divisibility if normal method isn't
getting you anywhere?
- find expression for difference between uₖ₊₁ and uₖ and try to show that THIS instead is
divisible by the number
- if it is, since you've assumed uₖ is divisible by the number it follows that uₖ₊₁ has to be
divisible by that number
3. Why is scalar product commutative?
Simply because multiplication of numbers is commutative
,4. Explain vector equation of a line
r=a+λb
- r is position vector of general point on line <x,y,z>
- a is the position vector of a point on the line to "step into the line"
- b is direction vector specifying the "gradient"
- λ is a parameter that you change to move along the line
5. Is r = <1,2,3> + λ<2,5,-2> the only equation for the corresponding line? Explain
No
You could pick any point on the line and any multiple of <2,5,-2> would be the direction vector
as they would all have be on the same line
6. In vector equation of line, explain each case when λ=0, λ=1, 0<λ<1, λ>1 and λ<0
- when λ=0 the point is the position vector of point A
- when λ=1 the point is the position vector of point B, one "step" away from A
- when 0<λ<1 the point is between A and B
- when λ>1 the point is beyond B
- when λ<0 the point is beyond A, opposite direction to B
7. Explain cartesian equation of line in 3D
,- take the vector equation and substitute ₖ=<x,y,z>
- add the vectors and rearrange for λ
- you should get λ = (x-a₁)/d₁ = (y-a₂)/d₂ = (z-a₃)/d₃ where <a₁, a₂, a₃> is position vector of a point
on the plane and <d₁, d₂, d₃> is direction vector
8. What to do if when converting from vector to Cartesian equation of line, the direction
vector has exactly one 0 element? E.g. line through <7,2,3> in direction <0,5,2>
- in the example you end up with λ = (x-7)/0 = (y-2)/5 = (z-3)/2 but obviously you can't divide by
0
- you go back and see that x=7+0λ so x=7
- so you write the Cartesian equation as x=7 and λ=(y-2)/5 = (z-3)/2
9. Arrangement of 2 lines in 3D + conditions
Intersect at unique point
- set the position vectors r equal for both lines and set up simultaneous equations involving λ
and μ where each are the parameters of the respective lines
- if the values for λ and μ found don't satisfy 3rd equation then the lines don't intersect at a
point so either parallel or skew
- else sub either parameter into one of the line equations to get the intersection point
Parallel
- no intersection point and direction vectors are multiples of each other
Skew
- no intersection point and direction vectors aren't parallel to each other
, 10. Angle between two lines
Find angle between their direction vectors using scalar product
11. Why can you find angle between two lines in 3D even if they are skew?
- the angle between them is still the angle between their direction vectors
- imagine translating one of the lines so the two lines DO intersect - you are finding the angle
between lines at this point
12. How to find intersection point between plane and line?
- from the vector equation, the each of x y and z will be some function of λ
- substitute these into the Cartesian equation of the plane and solve for λ
- sub λ into vector equation to get the intersection point
13. How to find acute angle between line and a plane?
- the acute angle is the angle between the line and it's orthogonal projection onto the plane
- find angle between the normal vector and the direction vector of the line and take this from
90°
