Vectors Cheat Sheet Edexcel Core Pure 1

Equation of a line in three dimensions Finding shortest distances and reflections

You need to know how to express the equation of a straight line in three dimensions in both vector and cartesian form. You need to be able to find the perpendicular distance between two lines, a point and a line as well as a point and a
Finding angles between lines and planes
The equation of a straight line that passes through the position vector 𝒂 and is parallel to the vector 𝒃 is written as: plane. This is also the shortest distance between them. Through the use of examples, we will explain how to find the
𝒂∙𝒃 shortest distance in each of these scenarios. We will also look at how to find the reflection of a point or a line in a plane.
𝑟 is a general point and 𝜆 is a scalar parameter. § The acute angle 𝜃 between two intersecting lines is given by cos 𝜃 = C C.
§ 𝒓 = 𝒂 + 𝜆𝒃 in vector form. |𝒂||𝒃|
𝒃∙𝒏 For any two non-intersecting lines, there The perpendicular from the point P to a The perpendicular from a point P to a
𝑥!"#
−! 𝑎% 𝑦 − 𝑎& 𝑧 − 𝑎' 𝑎" 𝑏" § The acute angle 𝜃 between the line 𝒓 = 𝒂 + 𝜆𝒃 and the plane 𝒓 ∙ 𝒏 = 𝑘 is given by sin 𝜃 = C C. is a unique line segment AB which is line is a line which passes through P and plane is a line which passes through P
§ = = in cartesian form. where 𝒂 = ,𝑎!2 𝑎𝑛𝑑 𝒃 = ,𝑏! 2 |𝒃||𝒏| perpendicular to both lines. meets the line at a right angle. and is parallel to the normal vector, n.
𝑏%
$! 𝑏& 𝑏' 𝑎# 𝑏# 𝒏𝟏 ∙ 𝒏𝟐
§ The acute angle 𝜃 between the plane 𝒓 ∙ 𝒏𝟏 = 𝑘% and the plane 𝒓 ∙ 𝒏𝟐 = 𝑘& is given by cos 𝜃 = C C.
|𝒏𝟏 ||𝒏𝟐 |
Example 1: Find a vector equation of the straight line which Example 2: With respect to the fixed origin 𝑂, the line 𝑙 is given by
passes through the points 𝐴 and 𝐵 , with position vectors 𝑎% 𝑏% These formulas will give you an acute angle. If you instead wish to find the obtuse angle, subtract your answer from 180.
4𝒊 + 5𝒋 − 𝒌 and 6𝒊 + 3𝒋 + 2𝒌 respectively. the equation 𝒓 = $𝑎& & + 𝜆 $𝑏& &. Prove that the cartesian form of
𝑎' 𝑏' Example 4: The lines 𝑙% and 𝑙& have vector equations 𝒓 = (2𝒊 + 𝒋 + 𝒌) + Example 5: Find the acute angle between the line with equation
!"#! *"#( +"#) 𝑡(3𝒊 − 8𝒋 − 𝒌) and 𝒓 = (7𝒊 + 4𝒋 + 𝒌) + 𝑠(2𝒊 + 2𝒋 + 3𝒌) respectively. 𝒓 = (2𝒊 + 𝒋 − 5𝒌) + 𝑡(3𝒊 + 4𝒋 − 12𝒌) and the plane with equation
We know the line 𝑟 = 𝒂 + 𝜆𝒃 𝑙 is given by = = . Given that 𝑙% and 𝑙& intersect, find the size of the acute angle between the 𝒓 ∙ (2𝒊 − 2𝒋 − 𝒌) = 2.
$! $( $)
passes through both 𝐴 4 lines to one decimal place.
and 𝐵 so we can use 𝒂=$ 5 & Use the direction vector
𝑎% 3 2
−1 Write the general 𝑥 𝑏% of the line and the normal Shortest distance between two parallel Shortest distance between two non-parallel
either as our 𝑎. Use the direction 3 2 , 4 2 ∙ ,−22 = 3(2) + 4(−2) − 12(−1)
6 4 2
𝑥 ,𝑦0 = $𝑎& & + 𝜆 $𝑏& & vectors and take 7−89 ∙ 729 = 3(2) − 8(2) − 1(3) = −13 vector of the plane and −12 −1 lines lines
To find the direction 𝑥 𝑎' = 10 Example 9: Find the shortest distance between the lines with equations Example 10: Find the shortest distance between the lines 𝑙% and 𝑙& with
𝒃 = $3& − $ 5 & = $−2& point 𝒓 as ,𝑦 0. 𝑏' the scalar product. −1 3 take the scalar product.
𝒓 = (𝒊 − 2𝒋 − 𝒌) + 𝜆(5𝒊 + 4𝒋 + 3𝒌) and 𝒓 = (2𝒊 + 𝒌) + 𝜇(5𝒊 + 4𝒋 + 3𝒌). equations 𝒓 = (𝒊) + 𝜆(𝒋 + 𝒌) and 𝒓 = (−𝒊 + 3𝒋 − 𝒌) + 𝜇(2𝒊 − 𝒋 − 𝒌)
vector 𝒃, subtract the 𝑥
2 −1 3 |𝑙% | = <(3)& + (4)& + (−12)& = 13 respectively.
position vectors. Find the magnitude Find the magnitude of the
𝑥 = 𝑎% + 𝜆𝑏% |𝑙% | = <(3)& + (−8)& + (−1)& = √74 |𝑙& | = <(2)& + (−2)& + (−1)& = 3 Let 𝐴 be a general point on the 2 + 5𝜇
4 2 Equate each of the direction direction vectors: 1 + 5𝜆
𝑦 = 𝑎& + 𝜆𝑏& vectors. |𝑙& | = <(2)& + (2)& + (3)& = √17 first line and 𝐵 be a general FFFFF⃗ = , 4𝜇 2 − ,−2 + 4𝜆2
𝐴𝐵 Let A be a general point on the
So, the vector ∴ 𝒓 = $ 5 & + 𝜆 $−2& component. −1 + 2𝜇 1
𝑧 = 𝑎' + 𝜆𝑏' Use 10 point on the second. The 1 + 3𝜇 −1 + 3𝜆 first line and B be a general point FFFFF⃗ = , 3 − 2𝜇 2 − ,𝜆 2
𝐴𝐵
equation of the line is: −1 3 Use cos 𝜃 = ?
'%(
? = 0.367 sin 𝜃 = ? ? = 0.256 shortest distance is the length 1 + 5(𝜇 − 𝜆) on the second. The shortest
)√+,-)√%+- 𝒃∙𝒏 (13)(3) −1 − 𝜇 𝜆
Rearrange each !"#! *"#( +"#) cos 𝜃 = >
𝒂∙𝒃
> sin 𝜃 = > > ofaaaaaa⃗
𝐴𝐵 when 𝐴𝐵 aaaaa⃗ is = > 2 + 4(𝜇 − 𝜆)? distance is the length of aaaaaa⃗
𝐴𝐵 when −2 + 2𝜇
𝜆= , 𝜆= , 𝜆= |𝒂||𝒃| ∴𝜃= 𝑐𝑜𝑠 '% (0.367) = 68.5° |𝒃||𝒏| '%
∴ 𝜃 = 𝑠𝑖𝑛 (0.256) = 14.9° 2 + 3(𝜇 − 𝜆) = , 3 − 2𝜇 − 𝜆 2
equation for 𝜆. $! $( $! perpendicular to both 𝑙% and 𝑙& . aaaaaa⃗
𝐴𝐵 is perpendicular to both 𝑙%
−1 − 𝜇 − 𝜆
1 + 5𝑡 and 𝑙& .
Equating each Let 𝑡 = (𝜇 − 𝜆), then aaaaa⃗
𝐴𝐵
𝑥 − 𝑎1 𝑦 − 𝑎2 𝑧 − 𝑎3 Finding whether two lines intersect becomes:
= ,2 + 4𝑡 2
Since aaaaa⃗
−2 + 2𝜇 0
equation gives us ∴𝜆= = = 2 + 3𝑡 𝐴𝐵 is perpendicular to 𝑙% ,
, 3 − 2𝜇 − 𝜆 2 ∙ ,12 = 0
𝑏1 𝑏2 𝑏3 You need to be able to determine whether two lines meet or not and find their intersection if they do. The procedure is aaaaa⃗
the result. The shortest distance is when 1 + 5𝑡 5
𝐴𝐵 ∙ (𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑙% ) = 0. −1 − 𝜇 − 𝜆 1
summarised well by a flow chart: aaaaa⃗
𝐴𝐵 is perpendicular to both ,2 + 4𝑡 2 ∙ ,42 = 0 Calculate dot product. 2 − 2 𝜇 − 2𝜆 = 0 [1]
Finding the angle between two vectors Lines do intersect. To find the lines, so: 2 + 3𝑡 3
aaaaa⃗ is also perpendicular to 𝑙& , so −2 + 2𝜇 2
𝐴𝐵
Before we discuss the scalar product, which plays a large role in this chapter, it is important that you know how to correctly Write both equations in intersection, substitute either 𝜆/𝜇 Substitute these values of 𝑠 and , 3 − 2𝜇 − 𝜆 2 ∙ ,−12 = 0
5(1 + 5𝑡) + 4(2 + 4𝑡) + 3(2 + 3𝑡) = 0
aaaaa⃗
𝐴𝐵 ∙ (𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑙&) = 0.
column form and equate −1 − 𝜇 − 𝜆 −1
find the angle between two vectors. back into one of the line equations. 𝑡 into equation [3].
them.
Yes Rearrange for 𝑡. 𝑡 = −0.38 Calculate dot product. −6 + 6 𝜇 + 2𝜆 = 0 [2]
1 + 5(−0.38) −0.9
§ The angle between two vectors a and b is the angle between them when both are pointed away from their FFFFF⃗ = >2 + 4(−0.38)? = , 0.48 2
𝐴𝐵 Solve [1] and [2] simultaneously. 𝜇 = 1, 𝜆=0
Substitute t back into aaaaa⃗
𝐴𝐵 . The
point of intersection. Form three equations in 𝜆 Solve the first two Is the third equation required distance will be given
2 + 3(−0.38) 0.86 −2 + 2(1) 0
Substitute λ and μ into the ∴ M𝐴𝐵 FFFFF⃗ M = A(−0.9)! + (0.48)! + (0.86)! Substitute these values of 𝜇 and 𝜆 FFFFF⃗ = > 3 − 2(1) − (0) ? = , 2 2
𝐴𝐵
and 𝜇 by equating the i, j equations simultaneously satisfied by your values
third equation. aaaaa⃗ c.
by c𝐴𝐵 back into aaaaa⃗
𝐴𝐵 . The shortest −1 − (1) − (0) −2
and k components. for 𝜆 and 𝜇. of 𝜆 and 𝜇? √89
= aaaaa⃗ c. ∴ M𝐴𝐵 FFFFF⃗ M = A(0)! + (2)! + (−2)!
5√2 distance is given by c𝐴𝐵
In the following diagram, the angle In the below diagram, it is tempting to deduce We can re-draw the diagram to show = 2√2
between the vectors 𝒂 and 𝒃 is 30° , that the angle between the vectors 𝒂 and 𝒃 is both vectors pointing away from 𝑋: You also need to be able to determine whether two lines are skew. No
since both vectors point away from the 30°. But this is in fact not true because the Shortest distance between a point and a Shortest distance between a point and a plane
point 𝑋. vectors are not both pointing away from the Lines do not intersect line /'% 0'% 31(
§ Two lines are skew if they do not intersect and are not parallel. Example 11: The line 𝑙 has equation = = , and the point 𝐴 has
point 𝑋. & '& '% § The perpendicular (shortest) distance from the point
coordinates (1, 2, −1). Find the shortest distance between 𝐴 and 𝑙.
(𝛼, 𝛽, 𝛾) to the plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑 is
/'& 01( |𝑎𝛼 + 𝑏𝛽 + 𝑐𝛾 − 𝑑|
This new diagram shows us that the Example 6: The lines 𝑙% and 𝑙& have vector equations 𝒓 = (2𝒊 + 𝒋 + 𝒌) + Example 7: The lines 𝑙% and 𝑙& have equations = = 𝑧 − 1 and First rewrite the line in vector 1 2
, & You are given this result in the
𝑡(3𝒊 − 8𝒋 − 𝒌) and 𝒓 = (7𝒊 + 4𝒋 + 𝒌) + 𝑠(2𝒊 + 2𝒋 + 3𝒌) respectively. /1% 0 3', form:
𝑙: 𝑟 = , 1 2 + 𝜆 ,−22 √𝑎& + 𝑏 & + 𝑐 &
actual angle between 𝒂 and 𝒃 is Show that the two lines do not intersect. 2
=
,
=
'&
respectively. Prove that 𝑙% and 𝑙& are skew. −3 −1 formula booklet
150°. 1 + 2𝑡 Example 11: Find the perpendicular distance from the point with coordinates
FFFFF⃗ = , 1 − 2𝑡 2, so
𝑂𝐵 (3, 2, −1) to the plane with equation 2𝑥 − 3𝑦 + 𝑧 = 5.
Firstly, we need to show that Let 𝐵 be a general point on 𝑙.
2 + 3𝑡 7 + 2𝑠 2 + 4𝜆 −1 + 5 𝜇 −3 − 𝑡
Write both lines in column these lines don’t intersect. 1 + 2𝑡 1 2𝑡
71 − 8𝑡 9 = 74 + 2𝑠9 7−3 + 2𝜆9 = 7 4 𝜇 9 Then the vector aaaaa⃗
𝐴𝐵 is: |2(3) − 3(2) + 1(−1) − 5| 6
form and equate. Write both lines in vector FFFFF⃗
𝐴𝐵 = , 1 − 2𝑡 2 − , 2 2 = ,−1 − 2𝑡 2
Scalar product (dot product) 1−𝑡 1 + 3𝑠
column form and equate.
1+𝜆 4−2𝜇
−3 − 𝑡 −1 −2 − 𝑡
Use the above formula. =
A(2)! + (−3)! + (1)! √14
The scalar product is a function which takes two vectors and outputs a number. The scalar product of two vectors 𝒂 and 2 + 3𝑡 = 7 + 2𝑠 [1] The shortest distance is when 2𝑡 2
2 + 4𝜆 = −1 + 5 𝜇 [1] aaaaa⃗
𝐴𝐵 is perpendicular to 𝑙, so ,−1 − 2𝑡 2 ∙ ,−22 = 0
𝒃 is written as 𝑎 ∙ 𝑏. You will need to use the scalar product to find angles between two vectors. It is defined as: Form three equations. 1 − 8𝑡 = 4 + 2𝑠 [2]
Form three equations. −3 + 2𝜆 = 4 𝜇 [2] −2 − 𝑡 −1 Reflection of a line in a plane
1 − 𝑡 = 1 + 3𝑠 [3] aaaaa⃗
𝐴𝐵 ∙ (𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑙) = 0
1+𝜆 =4−2𝜇 [3] /'& 0', 318
𝒂∙𝒃 Solving the subsequent 2(2𝑡) − 2(−1 − 2𝑡) − 1(−2 − 𝑡) = 0 Example 14: The line 𝑙% has equation = = . The plane Π has
§ 𝒂 ∙ 𝒃 = |𝒂||𝒃| cos 𝜃 To find angles, rearrange the formula into: cos 𝜃 = |𝒂||𝒃| Solve the first two equations 2 49 4 & '& %
𝑡= , 𝑠=− Solve the first two equations 9 equation: 𝑡=− equation 2𝑥 − 3𝑦 + 𝑧 = 8. The line 𝑙& is a reflection of the line 𝑙% in Π. Find
simultaneously. 11 22 9
simultaneously. 𝜆=− , 𝜇 = −3 the equation of 𝑙& .
2 9 2 2(−4/9) (−8/9)
where 𝒂 ∙ 𝒃 represents the scalar product of the vectors 𝑎 and 𝑏. In order to compute the scalar product of two vectors, 𝐿𝐻𝑆 = 1 − =
11 11 9 7
FFFFF⃗ = >−1 − 2(−4/9)? = > (−1/9) ?
𝐴𝐵 We begin by drawing a
Substitute these values of 𝑠 Substitute these values of 𝑠 𝐿𝐻𝑆 = 1 − = − Substitute t back into aaaaa⃗
𝐴𝐵 . The −2 − (−4/9) (−14/9) diagram. To find 𝑙& , we need
you can use the following fact: and 𝑡 into equation [3]. and 𝑡 into equation [3]. 2 2 required distance will be given
49 125 𝑅𝐻𝑆 = 4 − 2(−3) = 10 to find two points that lie on
Example 3: Find the angle 𝜃 between vectors a and b. 𝑅𝐻𝑆 = 1 + 3 T− U = − aaaaa⃗ c. 8 ! 1 ! 14 !
by c𝐴𝐵 FFFFF⃗ M = ef− g + f− g + f− g
∴ M𝐴𝐵 𝑙& . One can be found by
22 22 Since the third equation is 9 9 9
𝑎1 𝑏% Since the third equation is 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆, so the three finding the intersection of 𝑙%
𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆, so, the three not satisfied by our solutions, = 1.80
𝒂 ∙ 𝒃 = 1(−1) + 2(3) + 4(2) equations are not consistent.
§ If 𝒂 = /𝑎20 and 𝒃 = $𝑏& & then 𝒂 ∙ 𝒃 = 𝑎% 𝑏% + 𝑎& 𝑏& + 𝑎' 𝑏' . = 13
not satisfied by our solutions,
equations are not consistent. Hence we can conclude the lines and Π and the other by
we can conclude the lines Hence the lines do not intersect. reflecting the point (2,4,-6).
𝑎3 𝑏' |𝒂| = A1! + 2! + 4! = √21 the lines do not intersect. don’t intersect.
don’t intersect. 4 2 2
Now all we need to do to 5
|𝒃| = A1! + 3! + 2! = √14 729 ≠ 𝑘 7 4 9 for any choice of 𝑘. Reflection of a point in a plane 𝑙": 𝒓 = i 4 j + 𝜆 i−2j. Substituting into Π:
prove these lines are skew is Find the intersection of 𝑙%
The following fact is also very important: ∴ cos 𝜃 =
13
= 0.758 1 −2 2 + 2𝜆
−6
2
1

(√21)(√14) show that the direction Example 13: The plane Π has equation −2𝑥 + 𝑦 + 𝑧 = 5. The point 𝑃 has and Π.
Hence, they are not parallel. So, the , 4 − 2𝜆 2 ∙ i−3j = 8
⇒ 𝜃 = 𝑐𝑜𝑠 $" (0.758)
= 40.7° vectors are not parallel. lines are skew. coordinates (1, 0, 3). Find the coordinates of the reflection of the 𝑃 in Π.
−6 + 𝜆 1
§ The non-zero vectors 𝑎 and 𝑏 are perpendicular if and only if 𝒂 ∙ 𝒃 = 0.
Finding the intersection between a line and a plane We begin by drawing a Solve the subsequent 2(2 + 2𝜆) − 3(4 − 2𝜆) + 1(−6 + 𝜆) = 0
diagram. The reflected point equation. ⇒𝜆=2
is P’. We need to first find the Substitute 𝜆 = 2 into 𝑙% to
Equation of a plane in three dimensions § To find the intersection between a line and a plane, first express the plane in the form 𝒓 ∙ 𝒏 = 𝒌 and replace equation of the red line, find the intersection 𝑋.
∴ 𝑋 = (6, 0, −4)

You also need to be able to express the equation of a plane in both vector and cartesian form. The direction of a plane is the general point r with the vector equation of the line. Then solve the resultant equation. which is normal to the plane
We know that a normal of Π is (2i-3j+k).
and passes through P. To find the second point, we
described by a normal vector, often denoted as 𝒏. This is simply a vector that is perpendicular to the plane. can find the reflection of the So, the equation of the line that is normal
Example 8: Find the coordinates of the point of intersection of the line 𝑙 and the plane Π where 𝑙 has Find an equation for the red 1 −2 point (2, 4, −6) in Π. To do to the plane and passes through (2, 4, -6)
the equation 𝒓 = −𝒊 + 𝒋 − 5𝒌 + 𝜆(𝒊 + 𝒋 + 2𝒌) and Π has equation 𝒓 ∙ (𝒊 + 2𝒋 + 3𝒌) = 4. Equation of red line: 𝒓 = ,02 + 𝜆 , 1 2 2 2
𝒂 is a point that lies on the plane and 𝒃 and 𝒄 are both line. this we use the same
§ 𝒓 = 𝒂 + 𝜆𝒃 + 𝜇𝒄 3 1 is: 𝒓 = , 4 2 + 𝜆 ,−32
vectors that lie on the plane. 𝜆 and 𝜇 are scalar parameters. process as in example 13. −6 1
First write the equation of the line in −1 + 𝜆 Substitute values into ∏
Equation of line: 𝒓 = , 1 + 𝜆 2 −2(1 − 2𝜆) + 𝜆 + (3 + 𝜆) = 5 2 + 2𝜆 2
§ 𝑛% 𝑥 + 𝑛& 𝑦 + 𝑛' 𝑧 = 𝑑 vector column form. equation. Find where this new line , 4 − 3𝜆 2 ∙ ,−32 = 8
−5 + 2𝜆
1 + 6𝜆 = 5 intersects Π, and use the −6 + 𝜆 1
𝑛% Replace 𝒓 in the equation of the plane −1 + 𝜆 1 Solve the subsequent 2 solution of 𝜆 to find the 2 +2K L
22

where 𝒏 = 7𝑛& 9 𝑎𝑛𝑑 𝒅 = 𝒂 ∙ 𝒏 , 1 + 𝜆 2 ∙ ,2 2 = 4 equation. ⇒𝜆= 7 58/7

The cartesian form can be simplified using the scalar product: 𝑛(
with the vector equation of the line. −5 + 2𝜆 3 3 reflection of (2, 4, -6) which ⇒𝜆=
%%
so 𝑂𝑅′
⎛
⎜ 7 ⎟
⎞
FFFFFFF⃗ = 4 − 3 K22L = P−38/7U
+
Use the dot product. −1 + 𝜆 + (1 + 𝜆)(2) + (−5 + 2𝜆)(3) = 4 To find the reflected point, we call 𝑅. 22 −20/7
⎝−6 + K 7 L ⎠
Solve the subsequent equation for 𝜆. 𝜆=2 we simply substitute 𝜆 = 1 − 2(4/3) −5/3
§ 𝒓∙𝒏=𝒂∙𝒏 &
2( ) into our line. This is ∴ 𝑂𝑃 % = > (4/3) ? = , 4/3 2
To find the direction of the
aaaaa⃗
6 58/7 16/7
−1 + (2) 1 𝑅𝑋 = q 0 r − 7−38/79 = 7−38/79
Substitute 𝜆 back into the equation of (
3 + (4/3) 13/3 aaaaa⃗ or 𝑋𝑅
line 𝑙& , find 𝑅𝑋 aaaaa⃗.
∴ point of intersection = > 1 + (2) ? = , 3 2 because the reflected point P’ −4 −20/7 8/7
the line to find the point of intersection.
𝒓 is a general point. This form is often more useful for the problems you will encounter in this chapter. To convert back −5 + 2(2) −1 lies twice as far from P as X. 6 16/7
𝑥 𝑙& passes through 𝑋 (and 𝑅). 𝑙& : 𝒓 = q 0 r + 𝜆 7−38/79
−4 8/7
into the form 𝑛% 𝑥 + 𝑛& 𝑦 + 𝑛' 𝑧 = 𝑑, you need to replace 𝒓 with ,𝑦0 and expand the scalar product.
𝑧