VECTORS
Further Pure 1, Chapter 1
Note: it is not possible for these chapter summaries to cover all the possible ways examiners can assess the topic
Using the vector cross product to find areas
Find the area of triangle ππ΄π΅ where π is the origin, π΄ has position
vector (2π’ − π£ + 3π€) and π΅ has position vector (3π’ + 2π£ − 4π€)
ΰ·
π × π = π π sin π π§
π × π = −π × π
π’
a
π×π= 1
b1
π£
a2
b2
a2 b3 − a3 b2
π€
a3 = −(a1 b3 − a3 b1 )
b3
a1 b2 − a 2 b1
1
2
Area of triangle ππ΄π΅ = |π × π|
Find the area of parallelogram π΄π΅πΆπ· where π΄, π΅ and π· have position
vectors (π’ − π£ + ππ€), 2π’ − 2π£ − 4π€ and (ππ’ − 4π€) respectively.
Area of parallelogram = |π × π| where π and π
describe adjacent sides from the same vertex
Using the triple scalar product to find volumes
Find the volume of the parallelepiped shown given that
2
1
−2
π= 0 ,
π= 4
and
π= 1
3
3
7
π1
π. π × π = π1
π1
π2
π2
π2
π3
π3
π3
π
π. π × π = π. π × π = −π. π × π
π
i.e. same result if order maintained, negated if order is ‘switched’
π
Volume of parallelepiped= π. π × π
Volume of tetrahedron =
1
π.
6
π×π
For both formulae, π, π and π must be from the same vertex
Find the volume of a tetrahedron which has vertices at
1, 2, 3 , 2, 0, 4 , −1, 4, 0 , and (2, 5, 5).
Line equations and direction cosines (A2 only)
A line passes through the point 1, −4, 5 , makes an angle of 60° with the
positive π₯-axis, makes an angle of 45° with the negative π¦-axis.
Determine two possible equations of the line in the form π« − π × π = π
Line equations
π«−π ×π =π
π«×π= π×π
π is a point, π is direction
Direction cosines
π₯
If π = π¦ , then…
π§
π₯
π = cos ππ₯ =
π
With respect to a fixed origin π, the line π has vector equation
1
1
r= 0 +π 2
1
1
The point π΄ lies on the line π such that the direction cosine of the position
2
vector ππ΄ with the π₯-axis is 3.
Determine the possible coordinates of π΄.
π = cos ππ¦ =
π¦
π
π = cos ππ§ =
π§
π
π₯ βΆ π¦ βΆ π§ and π βΆ π βΆ π
are in the same ratio
π 2 + π2 + π2 = 1
Planes (A2 only)
Find a Cartesian equation of the plane with parametric equation
π«=
To find the normal vector, use the cross product
on the two vectors parallel to the plane
0
−3
1
2 +π 2 +π 2
−3
4
0
Find the vector equation of the line of intersection of the planes with equations
2π₯ + π¦ − 3π§ = 1 and π₯ − π¦ + 2π§ = 5
The line of intersection is perpendicular to
both normals.
•
Direction of line π = π§π × π§π
•
Find a point, π, that is on both planes
(set π§ = 0 then solve for π₯ and π¦, any
value would work though!)
•
Combine with π« = π + ππ
Shortest distance between skew lines (A2 only)
Find the shortest distance between the skew lines with equations
π«=
−1
0
5 +π 1
−3
3
π«=
3
2
2 +π 2
−4
−3
For π« = π + ππ and π« = π + ππ
shortest distance =
π−π . π×π
π×π
Mixed problem (A2 only)
The line with equation π« − (2π’ + 8π£ − 2π€) × (2π£ + 3π€) = π intersects
the plane π± with equation 2π₯ + π¦ − 2π§ = 4 at point π΄.
Be prepared to use skills from across
Core Pure 1 and Further Pure 1!
The point π΅ has coordinates (−4, 4, 5).
The point πΆ is the reflection of point π΅ in the plane π±.
The cross product vector line isn’t
always that useful – the parametric
(CP1) version is often better!
Find the exact area of triangle π΄π΅πΆ, giving your answer in the form π 26
0
