Vector Theory (24 pages; 15/3/2014)
Contents
(1) Equation of a line
(i) parametric form
(ii) relation to Cartesian form
(iii) vector product form
(2) Equation of a plane
(i) scalar product form
(ii) parametric form
(iii) converting between scalar product and parametric forms
(3) Angle between two direction vectors
(4) Perpendicular vectors
(i) vector perpendicular to given (2D) vector
(ii) vector perpendicular to two given (3D) vectors
(5) Intersections (lines are 3D)
(i) point of intersection of two lines
(ii) point of intersection of a line and a plane
(iii) line of intersection of two planes
(6) Shortest distances
(i) from a point to a plane
(ii) between two parallel planes
(iii) between parallel lines / from a point to a line
(iv) between two skew lines
1
(1) Equation of a line
(i) parametric form (2D example; but can be extended to 3D)
The vector equation of the line 𝑙 through the points A & B can be
written in various forms:
(a) r = a + 𝜆d
(b) r = a + 𝜆(𝑏 − 𝑎)
(c) r = (1 − 𝜆)a + 𝜆𝑏
(a weighted average of 𝑎 & 𝑏; when 𝜆 = 0, 𝑟 = 𝑎; when 𝜆 = 1, 𝑟 =
1
𝑏; when 𝜆 = , 𝑟 is the average of 𝑎 & 𝑏; the diagram shows 𝜆 =
2
1
)
3
(d) (𝑦𝑥 ) = (𝑎𝑎1 ) + 𝜆 (𝑑𝑑1 )
2
2
1
or (𝑎𝑎1 +𝜆𝑑
)
+𝜆𝑑
2
2
where a = (𝑎𝑎1 ) and d = (𝑑𝑑1 ) is any vector in the direction from A
2
2
to B
(normally d1 & d2 are chosen to be integers with no common
factor)
2
Note the difference between (a) the vector equation of the line
through the points A & B and (b) the vector ⃗⃗⃗⃗⃗
𝐴𝐵 : The vector ⃗⃗⃗⃗⃗
𝐴𝐵
has magnitude |AB| (the distance between A & B) and is in the
direction from A to B.
Whereas the vector equation of the line through A & B is the
position vector 𝑟 of a general point P on the line, with completely
different magnitude and direction to that of the vector ⃗⃗⃗⃗⃗
𝐴𝐵 .
(ii) relation to Cartesian form
(𝑦𝑥 ) = (𝑎𝑎1 ) + 𝜆 (𝑑𝑑1 ) ⇒ 𝜆 =
2
⇒ 𝑦 = 𝑎2 +
2
𝑑2
𝑑1
𝑥−𝑎1
𝑑1
=
𝑦−𝑎2
𝑑2
. (𝑥 − 𝑎1 )
the straight line through (𝑎1 , 𝑎2 ) with gradient
(iii) vector product form (3D lines only)
r = a + 𝜆d can be written as (𝑟 − 𝑎) × 𝑑 = 0
(since 𝑟 − 𝑎 and 𝑑 are parallel)
or 𝑟 × 𝑑 = 𝑎 × 𝑑
eg line through (1, 0, 1) and (0, 1, 0):
0
1
−1
𝑑 = (1 ) − ( 0 ) = ( 1 )
0
1
−1
𝑖
𝑎 × 𝑑 = |𝑗
𝑘
1
0
1
−1
−1
1 | = −𝑖 + 𝑘 = ( 0 )
1
−1
3
𝑑2
𝑑1
−1
−1
Thus equation is 𝑟 × ( 1 ) = ( 0 )
−1
1
Note: Textbooks tend to write the determinant with the elements
transposed (it gives the same result though).
(2) Equation of a plane
(i) scalar product form
Let 𝑎 be the position vector of a point in the plane,
𝑥
and 𝑟 = (𝑦) be a general point in the plane.
𝑧
Let 𝑛 be a vector perpendicular to the plane.
As 𝑟 − 𝑎 and 𝑛 are perpendicular, (𝑟 − 𝑎). 𝑛 = 0
⇒ 𝑟. 𝑛 = 𝑎. 𝑛 = 𝑝 (a constant)
⇒ 𝑛𝑥 𝑥 + 𝑛𝑦 𝑦 + 𝑛𝑧 𝑧 = p (Cartesian form)
4
Example
−12
1
If 𝑎 = (2) and 𝑛 = ( 11 ), then
4
−9
−12𝑥 + 11𝑦 − 9𝑧 = −12(1) + 11(2) − 9(4) = −26
(Another way of thinking of this is that, since 𝑎 is a point on the
plane, it is a solution of 𝑟. 𝑛 = 𝑝, so that 𝑝 = 𝑎. 𝑛, or 𝑛. 𝑎)
(ii) parametric form
This is an extension of the parametric form of the vector equation
of a line.
Let 𝑏 and 𝑐 be non-zero vectors in the plane (that are not parallel
to each other).
Then 𝑟 = 𝑎 + (𝜆𝑏 + 𝜇𝑐)
5
Note that 𝑏 and 𝑐 are direction vectors, whilst 𝑎 is a position
vector. 𝑏 and 𝑐 can of course be determined from 2 points 𝑝 and 𝑞
in the plane, as 𝑝 − 𝑎 and 𝑞 − 𝑎 (or 𝑝 − 𝑞)
(iii) converting between scalar product and parametric
forms
(a) to convert from scalar product to parametric form
Example
Suppose that the equation of the plane is 2𝑥 + 3𝑦 + 𝑧 = 4
Let 𝑥 = 𝑠 and 𝑦 = 𝑡 , so that 𝑧 = 4 − 2𝑠 − 3𝑡 and a general point
is
𝑥
𝑠
0
0
1
𝑡
(𝑦) = (
) = ( 0) + 𝑠 ( 0 ) + 𝑡 ( 1 )
4 − 2𝑠 − 3𝑡
𝑧
4
−2
−3
(b) to convert from parametric to scalar product form
Method 1
𝑥
−1
1
2
Example: 𝑟 = (𝑦) = (2) + 𝑠 ( 3 ) + 𝑡 (3)
𝑧
4
5
1
→ 𝑥 = 1 − 𝑠 + 2𝑡
𝑦 = 2 + 3𝑠 + 3𝑡
𝑧 = 4 + 5𝑠 + 𝑡
Then eliminate s and t to obtain an equation in 𝑥, 𝑦 & 𝑧.
6
Method 2
−1
2
For the above example, create normal vector: ( 3 ) × (3)
5
1
𝑖
= |𝑗
𝑘
−1
3
5
2
3| = −12𝑖 + 11𝑗 − 9𝑘
1
giving −12𝑥 + 11𝑦 − 9𝑧 = −12(1) + 11(2) − 9(4) = −26
(3) Angle between two direction vectors
Example 1
To find the acute angle between the line with equation
2
−1
5
𝑟 = (3) + 𝜆 (−2) and the plane with equation 𝑟. ( 1 ) = 1
4
1
−1
−1
5
The two direction vectors in this case are (−2) and ( 1 )
1
−1
−1
1
5
5
Then (−2) . ( 1 ) = | 2 | | 1 | 𝑐𝑜𝑠𝜃 (*),
1
−1 −1
−1
so that 𝑐𝑜𝑠𝜃 =
(−1)(5)+(−2)(1)+(1)(−1)
√1+4+1√25+1+1
=
−8
√162
= −0.62854
This gives 𝜃 = 128.9°
Whether 𝜃 is acute or obtuse depends on the relative direction of
the normal vector to the plane (𝑛) and the direction vector of the
line (𝑑) - see the diagram below.
7
In this case, the angle we want (between the plane and the line) is
𝛼 in the diagram.
Thus 𝛼 = 90 − (180 − 128.9) = 38.9°
Example 2
If we need to find the angle between two planes, then the angle in
question will be 𝛼 in the diagram below. This will be acute, so
that we expect 𝜃 to be obtuse (as 𝜃 = 180 − 𝛼). If one of the
normals to the planes has its direction reversed, then we obtain
an acute angle from the scalar product result (*), and this has to
be converted to the required angle by subtracting from 180°.
8
(4) Perpendicular vectors
(i) vector perpendicular to given (2D) vector
Example
3
2
Given direction vector ( ): gradient is ; hence perpendicular
2
3
2
−3
gradient = − and perpendicular direction vector is ( ) or
3
2
3
( )
−2
(ii) vector perpendicular to two given (3D) vectors
Let given vectors be 𝑎 and 𝑏
Method 1
𝑎×𝑏
Method 2
𝑥
Let 𝑟 = (𝑦) be required vector.
𝑧
Then eliminate two of 𝑥, 𝑦 & 𝑧 from 𝑟. 𝑎 = 0 and 𝑟. 𝑏 = 0 (*)
to give a direction vector in terms of parameter 𝑥, 𝑦 or 𝑧.
𝑥
eg (2𝑥 )
3𝑥
(note: form of eq'ns (*) ensures that 𝑦 and 𝑧 will be multiples of 𝑥)
1
which is equivalent to the direction vector (2)
3
9
(5) Intersections (lines are 3D)
(i) point of intersection of two lines
Note: Lines may not have a point of intersection, if the equations
are not consistent; in which case they are termed 'skew'.
Example: intersection of 𝑙1 & 𝑙2 ,
𝑥
1
1
where 𝑙1 has equation 𝑟 = (𝑦) = (−6) + 𝜆 (2)
𝑧
−1
3
𝑥
9
2
and 𝑙2 has equation 𝑟 = (𝑦) = (7) + 𝜇 ( 3 )
𝑧
2
−1
𝑥
3
Eliminate 𝜆 & 𝜇, to give (𝑦) =(−2)
𝑧
5
(ii) point of intersection of a line and a plane
2
1
Example: 𝑙1 has equation 𝑟 = (3) + 𝜆 ( 2 ) ;
4
−1
5
plane has equation 𝑟. ( 1 ) = 1
−1
2
1
5
Then ((3) + 𝜆 ( 2 )) . ( 1 ) = 1 creates a linear equation in 𝜆.
4
−1
−1
(It is possible that the line is either parallel to the plane or lies in
the plane; in which case the term corresponding to
10
1
5
𝜆 ( 2 ) . ( 1 ) above will vanish, since the scalar product will be
−1
−1
zero; then the remaining numbers will only be consistent if the
2
vector corresponding to (3) lies in the plane; ie if the scalar
4
2
5
product corresponding to (3) . ( 1 ) equals the right-hand side.)
4
−1
(iii) line of intersection of two planes
Method 1
Starting with the equations of the planes in the form 𝑟. 𝑛 = 𝑑, let
𝑥
𝜆
(eg) 𝑥 = 𝜆, to obtain (𝑦) = (𝑎𝜆 + 𝑏); ie 𝑦 & 𝑧 will be expressible
𝑧
𝑐𝜆 + 𝑑
as linear functions of 𝜆.
Note that we are effectively choosing a point on the line which has
𝑥 coordinate 0 (and 𝑦 & 𝑧 coordinates 𝑏 & 𝑑).
Example 2 covers an unusual case.
Example 1
Planes 2𝑥 + 𝑧 = 3 & 𝑥 + 𝑦 − 𝑧 = 2
Let 𝑥 = 𝜆, so that 𝑧 = 3 − 2𝜆 and 𝑦 = 2 + (3 − 2𝜆) − 𝜆 = 5 − 3𝜆
So that the equation of the line of intersection of the planes is:
11
0
1
𝑟 = (5) + 𝜆 (−3)
3
−2
Example 2
Planes 2𝑥 + 𝑧 = 4 & 𝑥 − 𝑧 = −1
This implies that 𝑥 = 1, 𝑧 = 2 and 𝑦 = 𝜆,
so that the equation of the line of intersection of the planes is:
1
0
𝑟 = (𝑝) + 𝜆 (1) , where any value can be chosen for p
0
2
Method 2
The required line will be perpendicular to the normal vectors of
both planes. Therefore the vector product of the normal vectors
to the two planes has the direction vector of the required line.
Using Example 1 above, with planes 2𝑥 + 𝑧 = 3 & 𝑥 + 𝑦 − 𝑧 = 2,
𝑖
2
1
(0 ) × ( 1 ) = | 𝑗
1
−1
𝑘
2
0
1
1 | = −𝑖 + 3𝑗 + 2𝑘
−1
1
In order to find the equation of the line, we just need a point on it;
ie a point on both planes, so that 2𝑥 + 𝑧 = 3 & 𝑥 + 𝑦 − 𝑧 = 2
eg let 𝑧 = 0; then 𝑥 =
3
2
&𝑦 =
1
2
3
and the equation of the line is 𝑟 =
2
(1)
2
0
12
−1
+𝜇( 3 )
2
0
1
Note: This can be seen to be equivalent to 𝑟 = (5) + 𝜆 (−3) in
3
−2
Example 1, as follows:
3
Let 0 + 2𝜇 = 3 − 2𝜆, so that 𝜇 = − 𝜆
2
3
Then
2
(1)
2
0
−1
𝜆
+ 𝜇 ( 3 ) = (5 − 3𝜆)
2
3 − 2𝜆
(6) Shortest distances
(i) shortest distance from a point to a plane
(See "Equation of a plane" to convert between the scalar product
and parametric forms of the equation of a plane, if necessary.)
Method 1
Example 1
1
Point, P is (2) ; plane has equation 4𝑥 + 3𝑦 − 12𝑧 = 26 (*)
3
The position vector of the point in the plane at the shortest
distance from P is:
1
4
4
(2) + 𝜆 ( 3 ) for some 𝜆 (to be determined), as ( 3 ) is the
3
−12
−12
direction vector normal to the plane.
Since this point lies in the plane, it satisfies (*);
hence 4(1 + 4𝜆) + 3(2 + 3𝜆) − 12(3 − 12𝜆) = 26 (**)
giving 𝜆 =
4
13
13
The shortest distance is the distance travelled from P to the plane,
4
4
4
4
along the direction vector ( 3 ); ie |( 3 )| = (13) = 4
13
13
−12
−12
Example 2
In the special case where P is the origin O (and the plane has
equation 4𝑥 + 3𝑦 − 12𝑧 = 26 as before), (**) becomes
4(4𝜆) + 3(3𝜆) − 12(−12𝜆) = 26
4
ie 𝜆|𝑛| = 26, where 𝑛 is the normal to the plane, ( 3 )
−12
2
As before, the shortest distance from O to the plane is
𝜆|𝑛| =
26
|𝑛|
=
26
13
=2
Alternatively, if the equation of the plane is given in 'normalised'
form (ie the direction vector has unit magnitude; the word
'normal' being used here in a different sense to that of the normal
to a plane);
ie
4
13
𝑥+
3
13
𝑦−
12
13
𝑧 = 2, then the distance required is simply the
right-hand side of the equation.
Method 2
Using the above example, we can find the equation of the plane
1
parallel to 4𝑥 + 3𝑦 − 12𝑧 = 26 and passing through (2).
3
The equation of the parallel plane will be
14
4𝑥 + 3𝑦 − 12𝑧 = 4(1) + 3(2) − 12(3)
ie 4𝑥 + 3𝑦 − 12𝑧 = −26
From the special case of the Origin in Method 1, the distance
between the two planes (and hence between the point and the
plane) is
26−(−26)
√4 2 +32 +(−12)2
=
52
13
=4
Note: This method gives rise to the standard formula:
𝑝1
, as the shortest distance from the point (𝑝2 )
√𝑛1 2 +𝑛2 2 +𝑛3 2
𝑝3
to the plane 𝑛1 𝑥 + 𝑛2 𝑦 + 𝑛3 𝑧 = 𝑑
|𝑛1 𝑝1 +𝑛2 𝑝2 +𝑛3 𝑝3 −𝑑|
Method 3
1
Using the same example, where P is (2) and the plane has
3
equation 4𝑥 + 3𝑦 − 12𝑧 = 26 (*),
we first of all find a point Q in the plane (as in the diagram above)
and create the vector ⃗⃗⃗⃗⃗
𝑃𝑄
The required distance will then be the projection of ⃗⃗⃗⃗⃗
𝑃𝑄 onto 𝑛
(the normal to the plane); namely
15
⃗⃗⃗⃗⃗ .𝑛|
|𝑃𝑄
|𝑛|
In this case, putting 𝑦 = 𝑧 = 0 (say) in (*) gives 𝑥 =
13
13
2
, so that
11
2
2
𝑄 = ( 0 ) , and ⃗⃗⃗⃗⃗
𝑃𝑄 = (−2)
−3
0
⃗⃗⃗⃗⃗ . 𝑛 = (11) (4) + (−2)(3) + (−3)(−12) = 52
Then 𝑃𝑄
2
and the shortest distance =
52
√4 2 +32 +(−12)2
=
52
13
=4
(ii) distance between two parallel planes
Using the method for finding the shortest distance from the origin
to a plane (method 1, example 2 of "shortest distance from a point
to a plane"), the two planes need first of all to be put into
normalised form; the constant term of each equation then gives
the distance of the plane from the origin, so that the distance
between the planes is then the difference between the constant
terms.
Example: Find the distance between the planes
3𝑥 + 4𝑦 + 12𝑧 = 13 and 3𝑥 + 4𝑦 + 12𝑧 = 39
As √32 + 42 + 122 = 13, the normalised equations are
1
13
(3𝑥 + 4𝑦 + 12𝑧) = 1 and
1
13
(3𝑥 + 4𝑦 + 12𝑧) = 3
so that the distance between the planes is 3 − 1 = 2
(iii) distance between parallel lines / shortest distance from
a point to a line
Assuming that A and B are given points on the two lines, and that
𝑑 is the common direction vector:
16
Method 1
Let C be the point on 𝑙1 with parameter 𝑘, so that 𝑐 = 𝑎 + 𝑘𝑑 (*)
Then we require 𝑑. (𝑐 − 𝑏) = 0
Solving this equation for 𝑘 and substituting for 𝑘 in (*) gives 𝑐,
and the distance between the two lines is then |𝑐 − 𝑏|.
Example
1
3
−2
3
Let lines be 𝑟 = (0) + 𝜆 (−1) and 𝑟 = (−1) + 𝜆 (−1)
2
1
−1
1
1
−2
1
3
If 𝑎 = (0) , 𝑏 = (−1) and 𝑐 = (0) + 𝑘 (−1),
2
−1
2
1
1 + 3𝑘 + 2
3
then (−1) . ( −𝑘 + 1 ) = 0 → 9 + 9𝑘 + 𝑘 − 1 + 𝑘 + 3 = 0
1
2+𝑘+1
→ 11𝑘 + 11 = 0 → 𝑘 = −1
−2
Hence 𝑐 = ( 1 ) and the distance between the lines is
1
√(−2 + 2)2 + (1 + 1)2 + (1 + 1)2 = √8
17
Method 2
1 + 3𝑘
Having obtained the general point, 𝐶 = ( −𝑘 ) on 𝑙1 in
2+𝑘
Method 1, we can minimise the distance BC by finding the
stationary point of either 𝐵𝐶 or 𝐵𝐶 2 :
𝐵𝐶 2 = (1 + 3𝑘 + 2)2 + (−𝑘 + 1)2 + (2 + 𝑘 + 1)2
= 11𝑘 2 + 22𝑘 + 19
Then
and
𝑑
𝑑𝑘
𝑑
𝑑𝑘
(𝐵𝐶 2 ) = 22𝑘 + 22
(𝐵𝐶 2 ) = 0 ⇒ 𝑘 = −1, as before
Method 3
As 𝐵𝐶 = 𝐴𝐵𝑠𝑖𝑛𝜃, 𝐵𝐶 =
⃗⃗⃗⃗⃗ ×𝑑|
|𝐴𝐵
|𝑑|
−3
⃗⃗⃗⃗⃗
In the above example, 𝐴𝐵 =(−1) and ⃗⃗⃗⃗⃗
𝐴𝐵 × 𝑑 =
−3
𝑖
|𝑗
−3
−1
𝑘
−3
3
2
−1| = −4𝑖 − 6𝑗 + 6𝑘 = −2 ( 3 )
−3
1
Then 𝐵𝐶 =
2√4+9+9
√9+1+1
=
2√22
√11
= √8
(iv) shortest distance between two skew lines
Method 1
eg 𝑙1 : 𝑟 = 𝑎 + λ𝑏 & 𝑙2 : 𝑟 = 𝑐 + μ𝑑 ; A has position vector 𝑎
etc
18
XY is shortest distance, as it is perpendicular to both 𝑙1 and 𝑙2
unit vector in direction of XY is
𝑏×𝑑
|𝑏×𝑑|
AE = XY ; ∠𝐶𝐸𝐴 = 90° (as CE is in the plane of the 'back wall' of
the cuboid - because C lies on 𝑙2 )
So AE is the projection of 𝑐 − 𝑎 onto the direction of XY; ie onto
𝑏×𝑑
|𝑏×𝑑|
So XY = AE = |(𝑐 − 𝑎).
(𝑏×𝑑)
|𝑏×𝑑|
| (the modulus sign ensuring that the
distance is +ve)
Note: This method can't be used to find the distance between two
parallel lines, as |𝑏 × 𝑑| = 0, since 𝑠𝑖𝑛𝜃 = 0
19
Example 1a: To find the shortest distance between the lines
0
−1
7
2
𝑙1 : 𝑟 = (−2) + 𝜆 ( 2 ) and 𝑙2 : 𝑟 = ( 12 ) + 𝜆 (−14)
0
−1
5
4
The direction normal to the two lines is
𝑖
𝑏 × 𝑑 = |𝑗
𝑘
2
2
−1
7
−6
2
−14| = (−15) ; and we can take ( 5 ) instead
−42
14
4
As √4 + 25 + 196 = 15, the unit vector in this direction is
2
(5)
15
14
1
0
−1
1
2
2
1
1
We then require ((−2) − ( 12 )) . ( 5 ) = (−14) . ( 5 )
15
15
0
5
14
−5
14
=
1
15
(2 − 70 − 70) = −
138
15
=−
so that the required distance is
46
5
46
5
or 9.2
Method 2
Find the vector perpendicular to both 𝑏 and 𝑑 , as in Method 1:
𝑛 =𝑏×𝑑
Then the equation of the plane with normal 𝑛 , containing line 𝑙1
(ie the front face of the cuboid in Method 1) will be 𝑟. 𝑛 = 𝑎. 𝑛
Similarly the equation of the plane with normal 𝑛 , containing line
𝑙2 (ie the back face of the cuboid) will be 𝑟. 𝑛 = 𝑐. 𝑛
20
The distance between these two planes (ie XY) is obtained by first
adjusting the equations of the planes, so that they are based on a
normal vector of unit magnitude.
𝑟.𝑛
Thus
|𝑛|
=
𝑎.𝑛
|𝑛|
Then 𝑋𝑌 = |
and
𝑎.𝑛
|𝑛|
−
𝑟.𝑛
|𝑛|
𝑐.𝑛
|𝑛|
=
𝑐.𝑛
|𝑛|
| [See "Distance between two parallel
planes"]
[Note that this method is algebraically equivalent to method 1.]
Example 1b (Lines as in 1a)
2
From Example 1a, |𝑛| = ( 5 )
15
14
𝑛
1
Then the equations of the planes in which the front and back faces
of the cuboid in method 1 lie are
1
15
(2𝑥 + 5𝑦 + 14𝑧) =
and
1
15
1
15
[2(0) + 5(−2) + 14(0)] = −
(2𝑥 + 5𝑦 + 14𝑧) =
1
15
10
15
[2(−1) + 5(12) + 14(5)] =
128
15
So the distance between the two planes, and hence between the
two lines is =
128−(−10)
15
=
138
15
=
46
5
= 9.2
Method 3
Referring to the earlier diagram, suppose that X and Y have
position vectors 𝑟 = 𝑎 + 𝜆𝑋 𝑏 & 𝑟 = 𝑐 + 𝜇𝑌 𝑑 respectively.
Then, if 𝑛 is the vector normal to both 𝑏 and 𝑑,
𝑐 + 𝜇𝑌 𝑑 = 𝑎 + 𝜆𝑋 𝑏 + 𝑘𝑛 (*)
21
(ie Y is reached by travelling first to X and then along XY) and XY
will then = 𝑘|𝑛|
(*) gives 3 simultaneous equations in 𝜆, 𝜇 & 𝑘:
𝑐1 + 𝜇𝑌 𝑑1
𝑎1 + 𝜆𝑋 𝑏1 + 𝑘𝑛1
(𝑐2 + 𝜇𝑌 𝑑2 ) = (𝑎2 + 𝜆𝑋 𝑏2 + 𝑘𝑛2 ) , from which 𝑘 can be found
𝑐3 + 𝜇𝑌 𝑑3
𝑎3 + 𝜆𝑋 𝑏3 + 𝑘𝑛3
Example 1c: (Lines as in 1a)
2
From Example 1a, 𝑛 = ( 5 )
14
We need to find 𝑘 such that
−1
7
0
2
2
( 12 ) + 𝜇𝑌 (−14) = (−2) + 𝜆𝑋 ( 2 ) + 𝑘 ( 5 )
5
4
0
−1
14
So
7𝜇𝑌 − 2𝜆𝑋 − 2𝑘 = 1
−14𝜇𝑌 − 2𝜆𝑋 − 5𝑘 = −14
4𝜇𝑌 + 𝜆𝑋 − 14𝑘 = −5
𝜇𝑌
1
7
−2 −2
or (−14 −2 −5 ) (𝜆𝑋 ) = (−14)
𝑘
4
1 −14
−5
.
𝜇𝑌
1
→ (𝜆𝑋 ) =
( .
7(33) + 14(30) + 4(6) −6
𝑘
→𝑘=
414
675
22
.
.
−15
.
1
. ) (−14)
−42
−5
and 𝑋𝑌 =
414
675
× √4 + 25 + 196 = 9.2
Method 4
As in method 3, suppose that X and Y have position vectors 𝑟 =
𝑎 + 𝜆𝑋 𝑏 & 𝑟 = 𝑐 + 𝜇𝑌 𝑑 respectively.
⃗⃗⃗⃗⃗ = 𝑐 + 𝜇𝑌 𝑑 − (𝑎 + 𝜆𝑋 𝑏)
Then 𝑋𝑌
⃗⃗⃗⃗⃗ . 𝑏 = 𝑋𝑌
⃗⃗⃗⃗⃗ . 𝑑 = 0 (*)
and 𝑋𝑌
Solving (*) enables 𝜆𝑋 & 𝜇𝑌 to be determined,
⃗⃗⃗⃗⃗ | can be found
from which |𝑋𝑌
Example 1d: (Lines as in 1a)
−1
7
0
2
⃗⃗⃗⃗⃗
𝑋𝑌 = ( 12 ) + 𝜇𝑌 (−14) − (−2) − 𝜆𝑋 ( 2 )
5
4
0
−1
−1 + 7𝜇𝑌 − 2𝜆𝑋
= (14 − 14𝜇𝑌 − 2𝜆𝑋 )
5 + 4𝜇𝑌 + 𝜆𝑋
−1 + 7𝜇𝑌 − 2𝜆𝑋
2
Then (14 − 14𝜇𝑌 − 2𝜆𝑋 ) . ( 2 ) = 0
5 + 4𝜇𝑌 + 𝜆𝑋
−1
−1 + 7𝜇𝑌 − 2𝜆𝑋
7
and (14 − 14𝜇𝑌 − 2𝜆𝑋 ) . (−14) = 0
5 + 4𝜇𝑌 + 𝜆𝑋
4
→ −2 + 14𝜇𝑌 − 4𝜆𝑋 + 28 − 28𝜇𝑌 − 4𝜆𝑋 − 5 − 4𝜇𝑌 − 𝜆𝑋 = 0
23
and
−7 + 49𝜇𝑌 − 14𝜆𝑋 − 196 + 196𝜇𝑌 + 28𝜆𝑋 + 20 + 16𝜇𝑌 + 4𝜆𝑋
=0
ie → 21 − 18𝜇𝑌 − 9𝜆𝑋 = 0 or 7 − 6𝜇𝑌 − 3𝜆𝑋 = 0
and −183 + 261𝜇𝑌 + 18𝜆𝑋 = 0 or −61 + 87𝜇𝑌 + 6𝜆𝑋 = 0
−6 −3 𝜇𝑌
−7
or (
) (𝜆 ) = ( )
61
87 6
𝑋
𝜇𝑌
1
6
3
−7
→ (𝜆 ) =
(
)( )
−36+261 −87 −6
61
𝑋
=
1
225
(
1 47
141
)= ( )
75 81
243
−1 + 7𝜇𝑌 − 2𝜆𝑋
92
1
Then ⃗⃗⃗⃗⃗
𝑋𝑌 = (14 − 14𝜇𝑌 − 2𝜆𝑋 ) = (230)
75
5 + 4𝜇𝑌 + 𝜆𝑋
644
⃗⃗⃗⃗⃗ | =
and |𝑋𝑌
1
75
√476100 = 9.2
24