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7/29/2019 Vectors Algebra
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Engineering Mathematics 1.2: Vectors Algebra 5
x1 , x2 , x3 axis system
The order and direction of axes are assumed to be 'right handed'. This definition comes from
imagining a right handed screw - if you turn the screw in the direction from axis Ox to Oy the
screw advances along Oz . (and Oy to Oz along Ox, from Oz to Ox along Oy). This is convention
and MUST be adhered to.
The distance from point O to point P, r, can be calculated using the Pythagoras theorem to give
( ) 2/1222 zyx ++=r
2.2 Direction Cosines
We will see later that it is useful to know the angle the line OP makes with each axis.
Fortunately this can be easily calculated. The shaded areas of the figures below each form
right-angled triangles with the axes.
r
Z-axis
Y-axis
X-axis
z
y
x
P(x,y,z)
o
r
-ax s
Y-axis
X-axis
z
y
x
P(x,y,z)
o
Engineering Mathematics 1.2: Vectors Algebra 6
So, knowing the length and the appropriate co-ordinate, each angle can be calculated.
The angle POx is given by
r
xl == cos
The angle POy is given by
r
ym == cos
The angle POz is given by
r
zn == cos
r
Z-axis
Y-axis
X-axis
z
y
x
P(x,y,z)
o
r
Z-axis
Y-axis
X-axis
z
y
x
P(x,y,z)
o
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Engineering Mathematics 1.2: Vectors Algebra 7
The cosines of these angles, ,, are (often written) nml ,, and these are known as the
direction cosines of the line OP.
It can easily be shown that
12
222
2
2
2
2
2
2222 =
++=++=++
r
zyx
r
z
r
y
r
xnml
2.2.1 Examples
1) If P has co-ordinates (5, 4, -9). What is the length OP and its direction cosines.
122
8116252
=
++=
OP
OP
Direction cosines:
122
9
122
4
122
5
=
=
=
n
m
l
2) The position of the top corner of the lecture room was required. To find this a theodolite
was set up as in the figure below. The distance to the side wall, along the x-axis is, 7m
and the following angles were measured: TOx = 70o and TOy = 50o .
Calculate the co-ordinates of the corner relative to the position of he theodolite.
x
yz
T
7m
Engineering Mathematics 1.2: Vectors Algebra 8
Calculate the direction cosines:
6428.050coscos
5.060coscos
====
====
r
ym
r
xl
We need the third direction cosine.
Using 1222 =++ nml
o52.54
58035.0
1cos 22
=
=
===
mlr
zm
The length OT, (r) can be calculated as we know the x co-ordinate, i.e. from l
0.145.0
0.7
5.0
==
==
r
r
xl
From the other direction cosines
125.80.1458035.0
0.90.146428.0
==
==
z
y
Co-ordinates of the corner are: ( ) ( )125.8,0.9,0.7,, =zyx
2.3 A Geometric Representation
We represent vectors geometrically by line segments in space. The length of the line
representing the magnitude and the direction of the line the direction of the vector.
From this definition the starting point is irrelevant.
What is the difference between these two vectors?
a a
O
A A
O
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Engineering Mathematics 1.2: Vectors Algebra 9
Their length is the same.
Their direction is the same.
(The arrows indicate the direction along the line)
These two vectors are equivalent (equal).
By the same argument, the two vectors are also equivalent to a vector of the same length and
direction which starts at the origin. We can therefore use a co-ordinate notation of three
numbers - as used previously for OP - to represent any vector (anywhere in three dimensional
space).
We have shown a vector described in two ways: in terms of co-ordinates (x, y, z) and also
represented geometrically by a line from an origin to this point. We say a dual definition can be
used either a geometrical or a co-ordinate(component) one.
2.4 Vector notation found in text books
A vector can be written down in many ways. Some of the more common (and acceptable) ways
you will come across are:
( )321 ,, aaaOAa a
2.5 Magnitude (or Modulus)
The magnitude, modulus or length of the vectora is written as a or OA and given in
component form by
( ) 2/1232221 aaa ++=a
Geometrically this is the length of the line OA
2.6 The Unit Vector
A vector whose modulus is 1 is called a unit vector, sometimes written as a and
a
aa =
Engineering Mathematics 1.2: Vectors Algebra 10
The unit vectors i, j, k
The unit vectors in the co-ordinate directions are denoted i, j,andk.
i = (1,0,0), j = (01,0) and k= (0,0,1)
Any vectorr can be expressed in terms of its componentx, y, zwith respect to the unit vector
(i,j,k) by
zkyjxi ++=r
The notation (x,y,z) is interpreted as (xi, yj, zk).
2.6.1 Example:A vector ( ) kji 434,1,3 +==a
The modulus of ( )( ) 264,1,3 2/1222 === aa
So the unit vector in the direction ofa is
+== kji
26
4
26
1
26
3
a
aa
2.7 Equal Vectors
Two vectors a and b are equal if they have the same magnitude and direction i.e.
a = b
Using the component definition
( )321 ,, aaa=a ( )321 ,, bbb=b
then
332211 bababa ===
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Engineering Mathematics 1.2: Vectors Algebra 11
2.8 Parallel Vectors
If is a scalar and a = b
then
if>0, the vectora is in the same direction as b and has magnitude b
if 0
Anti-parallel if < 0.
Engineering Mathematics 1.2: Vectors Algebra 12
3 Vector Addition
3.1 The boat problem
A boat steams at 4 knots due East for one hour. The tide is running North-North-East at 3 knots.
Where will the boat be after one hour?
A diagram of the vectors involved looks like that below, where a represents the velocity of the
boat and b represents the velocity of the tide.
The net velocity of the boat is represented by the line OC which is the sum ofa and b.
3.2 Geometric addition laws
This leads to the parallelogram rule for vector addition which may be written:
The sum, or resultant, of two vectors a and b is found by forming a parallelogram with a and b
as two adjacent sides.
The sum a + b is the vector represented by the diagonal of the parallelogram.
In component form
( )332211 ,, bababa +++=+ ba
b
a
a+b
O
C
A
North
NN
East
b
B
B C
AO
b
a
a+b
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Engineering Mathematics 1.2: Vectors Algebra 13
Because the vectora is the same as any other vector which is parallel and in the same
direction, a orb could be "moved" in the figure above to form a triangle. This leads to the
triangle law.
If two vectors a and b are represented in magnitude and direction by the two sides of a triangle
taken in order, then their sum is represented in magnitude and direction by the closing third
side.
The triangle rule can be made more general to apply to any geometrical shape - or polygon.
This then becomes the polygon law.
If from a point O, say, as in the figures below, lines are drawn to represent the vectors a, b, c, d,
and e. Then the closing side represents the vectorrand ris the sum of the vectors a, b, c, d,
and e.
edcbar ++++=
b
a
a+b
O A
B
CD
E
a
b
c
d
e
r
O
A
B
C
D
E
a
b
c
d
e
r
Engineering Mathematics 1.2: Vectors Algebra 14
3.3 Vector Laws for addition and subtraction
Commutative law
a + b = b + a
Order is NOT important
In component form
(a1 + b1 , a2 + b2 , a3 + b3 ) = (b1 + a1 , b2 + a2 , b3 + a3 )
Associative law
(a + b) + c = a + (b + c)
This follows from the component definition and geometrically - from the triangle or polygon law.
In component form
b
a
a+b
b
a
a+b
a+(b+c)
(a+b)+c
c
b
a
b+c
a+b
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Engineering Mathematics 1.2: Vectors Algebra 15
(a1 + b1 ) + c1 = a1 + (b1 + c1)
(a2 + b2 ) + c2 = a2 + (b2 + c2)
(a3 + b3 ) + c3 = a3 + (b3 + c3)
Distributive law
( ) baba +=+
This follows from the component definition or geometrically from similar triangles
In component form
( ) ( ) ( )( ) ( )332211332211 ,,,, babababababa +++=+++
Subtraction
a - b = a + (-b)
In component form:
( )332211 ,, bababa =ba
Geometrically:
AO
B
b
a
a+b
A O
B
b
a
( + ) = +a b a b
Engineering Mathematics 1.2: Vectors Algebra 16
Note OBBA = OA
This is an important result as it shows that the vector represented by the line joining two points
is given by taking the vector giving the position of the first point from that for the second
baab =+=+== OAOBBA BOOA
3.4 Vector Addition Examples
(1)
( ) ( ) ( )( )( ) ( )( )2,1,0
3,0,5302,212,1222
3,3,1
1,2,13,2,10,1,2
=+
=+=
=+
===
cba
ba
ba
cba
Modulus ofc
( ) 61,2,1 2/1222 ==c
The unit vector in the direction ofc
==
6
1,
6
2,
6
1
c
cc
Not that the modulus of 1 == cc
O A
B
ba b-
a
-b
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Engineering Mathematics 1.2: Vectors Algebra 17
2) From analysis of the forces in a new bridge it is calculated that the following forces are
imposed at one of the bridge supports (in Newtons).
F1 = (1,1,1)
F2 is 6N and acts in the direction (1,2,-2)
F3 is 10N and acts in the direction (1,2,-2)
In order to design the bridge support the resultant force is required. What is the force the
support must impose on the bridge to reduce the resultant force to zero.
The first force is given in the usual vector form. The second two are in valid forms but need to
be converted so that we can perform the usual vector addition.
First calculate the unit vectors in the direction ofF2 and F3
( )( )
( )( )0,4,3
5
1
0169
0,4,3
2,2,1
3
1
441
2,2,1
2
33
2
22
=++
==
=
++
==
F
FF
F
FF
So the in vector for the two forces are
( )
( )0,8,60,5
4,
5
310
4,4,23
2,
3
2,
3
16
2
2
=
=
=
=
F
F
The resultant force F is found by vector addition
( ) ( ) ( ) ( )3,3,90,8,64,4,21,1,1111 =++=++= FFFF
The force the support must impose on the bridge is equal and in the opposite direction as the
resultant i.e.
( )3,3,9forceReaction =
Engineering Mathematics 1.2: Vectors Algebra 18
4 Vector Multiplication
We have seen how vectors can be added, another natural operation in maths is to multiply
quantities together. The idea of multiplication of vectors becomes very useful in engineering. In
particular we will show how to use them to calculate moments and work-done with great ease.
Two types of multiplication exist. They are called the vectorand scalarproducts.
4.1 The Scalar Product
(also known as the dot product or the inner product)
The component of the vectora in the direction of OP is easily calculated as equivalent to the
length ON by
cosa=ON
Considering the constant force F in the figure below, which acts through the point O. If this force
is moved along the line OA, along the vectora, then we can calculate the work done by the
force. (Remember: work done is the component of the force in the direction of movement
multiplied by distance moved by the point of application.)
The component ofF along OA is cosF . This force is moved the distance a , giving,
cosdonework aF=
a
O
A
P
N
O
F
Aa
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Engineering Mathematics 1.2: Vectors Algebra 19
This is the scalar product of the two vectors F and a and can be seen as a geometric
definition.
An equivalent component definition can be written.
Scalar product definition:
The scalar product of two vectors ( )321 ,, aaa=a and ( )321 ,, bbb=b is written with a dot ( )
between the to vectors.
It is defined in component form as
332211 bababa ++=ba
And in geometrical form
cosabba =
Where is the angle between the two vectors and 0 .
The two definitions can be proved to be equivalent by the cosine rule for a triangle,
( )( ) cos2222 OBAOOBOAAB +=
Which can be expanded to show that
cos332211 ab=++ bababa
Three important points about a scalar product:
(1) The scalar product of two vectors gives a number (a scalar)
(2) The scalar product is a product of vectors
(it cannot be of two scalars nor a vector and a scalar)
aO A
b
B
Engineering Mathematics 1.2: Vectors Algebra 20
(3) Use of the "dot" is essentialto indicate that the calculation is a scalar product.
4.2 Rules for the scalar (or dot) product.
Commutative Law
coscos abba
abba
=
=
Associative Law
You cannot have a scalar product of three vectors as 'dotting' the first two gives a scalar.
Distributive Law (for a scalar multiplier)
Brackets are multiplied out in the usual way.
( ) ( ) ( )bababa ==
Distributive Law (for vector addition)
( ) cabacba +=+
Powers of vectors
2
2
3
2
2
2
1
0cos
a
aa
aa
=
=
++= aaa
No other powers of vectors are possible other than 2.
Note that for unit vectors i, j, k
12
== iii 12
== jjj 12
== kkk
Perpendicular Vectors
Ifa and b are perpendicular then2
= and 0
2coscos ==
Hence 0=ba
BUT 0=ba does not imply that a and b are perpendicular, because a could be zero orb could
be zero.
Note: Suppose caba = [ ( ) 0= cba ]
This does not mean that b = c, since a could be zero ora perpendicular to b-c.
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Engineering Mathematics 1.2: Vectors Algebra 21
i.e. vectors cannot be cancelled in the same way as scalars.
For unit vectors which are perpendicular
0=== ikkjji
Perpendicularity is important in engineering, for example pressure acts normal to a surface, so
force per unit area is np , wherep is the pressure and n the unit normal vector.
We often have to find a vector that is normal to another vector.
Component of a vector
The component of a vector, F say, in a given direction a is given by cosF .
And we can write
coscos FaFaF ==
to give the component ofF in the direction on a.
4.3 Scalar Product Examples
(1) Find the angle between the two vectors a = (2, 0, 3) and b = (3, 2, 4)
Using cosabba = and 332211 bababa ++=ba
181206 =++=ba
( ) ( )
( ) ( ) 2916494,2,3
139043,0,2
2/1
2/1
=++==
=++==
b
a
=
==
=
02.22
927.02913
18cos
cos291318
(2) Fora = (1, 2, -2), b = (0, 2, 2) c = (3, 2, 1) find
i) ca ii) ( ) cba + iii) ( )cba
i) ( ) 5243 =+=ca
ii) ( ) ( ) ( ) ( ) 110831,2,30,4,1 =++==+ cba
iii)( ) ( )( ) ( ) ( )
0,0,01,2,301,2,3440 ==+= cba
Note that a and b are perpendicular since neithera orb are zero.
Remember that ba is just a number (not a vector).
Engineering Mathematics 1.2: Vectors Algebra 22
(3) Find the work done by the forceF = (3, 1, 5) in moving a particle from P to Q where position
vectors of P and Q are (1, 2, 3) and (2, 4, 4) respectively.
( )1,2,1=== OPOQPQr
Work done by the force F = rF
( ) ( ) 105231,2,15,1,3 =++= units of work.
(4) Find the component of the vectorF = (2, -2, 3) in
(i) the idirection
(ii) the direction (3, 1, -2)
(i) The i direction is the vector (1,0,0) so the component in the idirection is
( ) ( ) ( ) 20,0,13,2,20,0,1 ==F
(iii) Is the vector (3, 1, -2) a unit vector?
( ) 14419 2/1 =++ Therefore it is not a unit vector.
Unit vector =( )
14
2,1,3
Component ofFin direction (3, 1, -2) is
( )( ) ( ) 14
2
14/2,1,33,2,214
2,1,3
==
F
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Engineering Mathematics 1.2: Vectors Algebra 23
4.4 The Vector (or Cross) Product
The main practical use of the vectorproduct is to calculate the moment of a force in three
dimensions. It is of very limited use in two dimensions.
For two vectors a and b, the vector product is defined as:
( ) nbaba sin=
where is the angle between the vectors a and b, ( 0 ) and n is the unit vector normal to
both a and b.
a , b and n are three vectors which form a right-handed set. (Remember how a right handed
axes system was defined earlier.)
It is very important to realise that the result of a vector product is itself a vector.
Note how from this definition order of multiplication matters:
The vector given by ( )ab is in the opposite direction to ( )ba .
(This follows from the right-hand screw rule), so
( ) ( )abba =
b
a
n
b
a
n
Engineering Mathematics 1.2: Vectors Algebra 24
4.5 The Moment of force, F.
If the force F passes through the pointPand r=OP , then the moment of the force about O is
defined as
FrM =
M is a vector in the direction of the normal n .
So, as with all vectors, moments add by the parallelogram rule.
The use of calculating moments by the vector (cross) product comes into itself when used in
situations (particularly in three dimensions) which are very difficult to visualise.
4.6 Area calculation
The area of the parallelogram ABCD is given by
ABADABADABh === sinArea
n
r
OP
F
A B
D
C
h
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Engineering Mathematics 1.2: Vectors Algebra 25
The area of the triangle PQR is given by
PRPQ
PRPQ
PR
=
=
=
2
1
sin21
h2
1Area
4.7 Laws for the Vector (or Cross) products
Anti-Commutative
( ) ( )abba =
This follows from the definition that includes the 'right-handed axes', n changes direction when
the multiplication is reversed.
Non-associative
(The triple vector product)
( ) ( ) cbacba
The vectorbcis perpendicular to both b and c and the plane containing b and c.
Similarly, ( ) cba is in the plane ofa and b, so ( )cba and ( ) cba are different vectors.
Brackets must always be used for more than two vectors in a vector product.
Distributive law with multiplication by a scalar
( ) ( ) baba =
P
R
h
Engineering Mathematics 1.2: Vectors Algebra 26
Distributive law with addition
( ) ( ) ( )cabacba +=+
4.8 Parallel vectors.
From the definition of the vector product, i fa and b are parallel, then
0=
and
0=ba
So we can say that if 0=ba then
a = 0
or
b = 0
or
a and b are parallel.
If caba = it cannot be deduced that b = c. You must first show that a 0 and that a is not
parallel to b - c.
4.9 Cartesian (component) form.
For unit vectors i,j,k
( )0sin,0as0 ===== kkjjii
Because of perpendiculatiry
jikikjkji === ,,
jkiijkkij === ,,
Component form of the vector product:
( )
( ) kbjbibbbb
kajaiaaaa
321321
321321
,,
,,
++==
++==
b
a
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )ibajbaiba
kbajbakba
kkbajkbaikba
kjbajjbaijba
kibajibaiiba
kbjbibkajaia
bbbaaa
+++
++=
+++
+++
++=
++++=
=
231332
123121
332313
322212
312111
321321
321321 ,,,,ba
So, how do you remember this!!?
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Engineering Mathematics 1.2: Vectors Algebra 27
There are several ways, the determinant form is shown here
21
21
31
31
32
32
321
321
bb
aak
bb
aaj
bb
aai
bbb
aaa
kji
+=
=ba
( ) ( ) ( )kbabajbabaibaba 122113312332 +=ba
4.10 Examples
(1) Given the three vectors a = (3, 1, 1), b = (2, -1, 1), c = (0, 1, 2)
( )5,1,2
12
13
12
13
11
11
112
113
=
+
=
=
kji
kji
ba
This vector is perpendicular to the plane containing a and b.
( )
( )2,4,3
10
12
20
52
21
51
210
512
=
+=
=
kji
kji
cba
This lies in the plane containing a and b.
( ) ( )( ) ( )( )2,4,3
1,1,33,3,6
13
=
=
= abacbbca
This is the same as the solution for ( ) cba above.
Engineering Mathematics 1.2: Vectors Algebra 28
It can be shown that in general,
( ) ( ) ( )acbbcacba =
It can also be shown that
( ) ( ) ( )cbabcacba =
(2) Find the area of the triangle having vertices at
P= (3, 2, 2), Q= (1, -1, 2), R = (2, 1, 1)
Area PQR = PRPQ 2
1
( ) ( )1,1,10,3,2 == PRPQ
( )1,2,3
11
32
11
02
11
03
111
032
=
+
=
=
kji
kji
PRPQ
Area = ( )2
14149
2
1
2
1=++=PRPQ
P
R
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Engineering Mathematics 1.2: Vectors Algebra 29
(3) A force F of 6 units acts through the point P=(2,3,1) in the direction of the vector (4,1,2). Find
the moment of the force about the point Q=(1,2,1).
The unit vector in the direction of the force is
=
++
++
21
2,
21
1,
21
4
4116
24 kji
So the force F, has components
=
21
2,
21
1,
21
46F
The position vector of P relative to Q is
( )0,1,1=QP
i.e. r
So the moment of the force about Q is
( )
=
=
+=
==
21
18,
21
12,
21
12
3,2,2
14
11
24
01
21
01
214
011
214
01121
6
M
M
kji
kji
kji
FQP
Q
P
Engineering Mathematics 1.2: Vectors Algebra 30
4.11 Triple products of Vectors
Definitions of the two triple products are
( ) cba is the Triple Vector product.
( ) cba is the Triple Scalar product.
The triple scalar product has an interesting geometrical meaning:
We know that
( )
( )ba
nbaba
andbydefinedramparallelogtheofarea
sin
=
=
Thus
( ) ( )
( ) cosramparallelogtheofarea
ramparallelogtheofarea
cn
cncba
=
=
But h=cosc =height of the parallelepiped normal to the plane containing a and b . ( is the
angle between c and n ).
So
( ) cba = area of the parallelepiped defined by a, b and c.
It follows then that:
(a) If any two vectors are parallel ( ) 0= cba (zero volume)
(b) If the three vectors a co-planar then ( ) 0= cba (zero volume)
A
BC
O
c
b
a
h
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Engineering Mathematics 1.2: Vectors Algebra 31
(c) If ( ) 0= cba then either
i) a = 0, or
ii) b = 0or
iii) c =0 or
iv) two of the vectors are parallel or
v) the three vectors are co-planar
(d) ( ) ( ) ( )acbcbacba == = the same volume.
Engineering Mathematics 1.2: Vectors Algebra 32
5 Vector Equations of Lines and Planes
5.1 The Vector Equation of a Line
Consider the line which passes through the points A and B with position vectors a and b
respectively and a point P, which has position vectorr, and lies on this line as shown
rba === OPOBOA
From this diagram we can see that
AP
APOAOP
+=
+=
ar
Ift is some multiple of AB such that
ABt+= ar
as ba =+AB , then
( )abar += t
So the equation of the line AB is:
( ) bar tt += 1
for
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Engineering Mathematics 1.2: Vectors Algebra 33
( )
11
1
111
ab
axt
tbatx
=
+=
Similarly fory andz
33
3
22
2
ab
azt
ab
ayt
=
=
Hence the Cartesian equation of a line passing through the points (a1,a2,a3) and (b1,b2,b3).is
tab
az
ab
ay
ab
ax=
=
=
33
3
22
2
11
1
Note: For any equation in Cartesian form we can readily express the equation in vector form.
e.g.
tzyx === 12
19
3
In vector form
( )( ) ( )1,1,32,9,0 +=
+=
t
t
r
abar
5.2 Equation of line examples:( ) ( )ababar +=+= ttt1
(1) Find the equation of the line r1 through the points (0,1,-2) and (3,4,3).
( )( ) ( )( )ttt
ttttt
tt
52,31,3
3,4,322,1,0
1
++=
++=
+= bar
i.e. tztytx 52313 +=+==
In Cartesian form
tzyx
tzyx
=+
=
=
=++
=
=
5
2
2
1
3
0
23
2
13
1
03
0
(2) Find the equation of the line r2 through the points (1,1,0) and (-3,-2,-7)
( )( ) ( )
( )sss
sssss
ss
7,31,41
7,2,30,1,1
1
=
+=
+= bar
Engineering Mathematics 1.2: Vectors Algebra 34
In Cartesian form
szyx
=
=
=
73
1
4
1
(3) Do r1 and r2intersect?
For the lines to intersect then t and s should be such that:
ss
stst
752
3131413
=+
=+=
are all satisfied.
Solving the first two simultaneously s = 1 and t = -1.
Substituting these values into the third gives
-7 = -7
i.e. the three equations are satisfied, so the two lines intersect.
Substituting the values fors ort into the respective Cartesian equations gives the point of
intersection as (-3,-2,-7)
(4) Find the angle between the two lines which are in the same direction as the two vectors c =
(3,3,5) and d = (-4,-3,-7)
5635912
cos
==
=
dc
dcdc
( )
( ) 7449916
432599
2/1
2/1
=++=
=++=
d
c
==
=1.173
9927.07443
56
cos
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Engineering Mathematics 1.2: Vectors Algebra 35
5.3 The Vector Equation of a Plane
We can use the fact that a line joining any two points in the plane is perpendicular to the normal
to the plane.
i.e. n and AP are perpendicular.
The vectorn is normal to the plane, a is the position vector of A, r is the position vector of P (r =
(x, y, z)).
The vector
ar == OAOPAP
Now r - ais perpendicular to n if
( ) 0= nar
or
nanr =
or
=nr
The above two are the general form of the vector equation of a plane.
If we can take n = (, , ) the equation in Cartesian form is
pzyx =++
since r = (x, y, z)
n
O
AP
Engineering Mathematics 1.2: Vectors Algebra 36
5.4 Examples involving both lines and planes
(1) Find the equation of the plane through the points
a = (3,2,0) b = (1,3,-1) c = (0,-2,3)
a - b lies in the plane
a-b
= (2, -1, 1)a - c lies in the plane
a - c = (3, 4, -3)
A normal, n, to the plane is
( ) ( ) ( ) ( )3,4,31,1,2 == caban
( ) ( ) ( )( )( )11,9,1
38,36,43
43
12
33
12
34
11
343112
=
+=
+
=
=
n
kji
kji
Equation of the plane is nanr =
( ) ( ) ( )
15
0183
11,9,10,2,311,9,1
=++=
=r
In component form this is
( ) ( )15119
1511,9,1,,
=++
=
zyx
zyx
Check
1533180:
1511271:150183:
=+
=+=++
c
ba
All are consistent.
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Engineering Mathematics 1.2: Vectors Algebra 37
(2) (a) Find the point where the plane ( ) 41,2,1 =r (or 42 =++ zyx ) meets the line
r = (2, 1, -1) + t(1, 1, 2) or ( tzyx
=+
=
=
2
1
1
1
1
2)
(b) Find the angle that the line makes with the plane.
[Remember the equations of a line
( )abar += t
tab
az
ab
ay
ab
ax=
=
=
33
3
22
2
11
1
]
The point of intersection must satisfy both the equation of the plane AND the equations of the
line.
i.e (using the vector form)
( )[ ] ( )( ) ( )
5/1
453
4221122
41,2,1
=
=+
=++++
=+
t
t
t
t aba
Substituting into the equation of the line gives the point of intersection
( )
( ) ( )
=
+=
+=
5
3,
5
6,
5
11
2,1,15
11,1,2
abar t
(b) Find the angle that the line makes with the plane.
The normal to the plane is (1, 2, 1) = n.
A vector in the direction of the line is (1, 1, 2) = b - a
( )
( ) 5221
cos
=++=
=
nab
nabnab
( )
( ) 6141
6411
2/1
2/1
=++=
=++=
n
ab
=
==
56.33
833.06
5cos
Engineering Mathematics 1.2: Vectors Algebra 38
(3) Find the perpendicular distance from the point P (2,-3,4) to the planex+2y+2z=13
The equation of the plane in vector form is
( ) 132,2,1 =r
A vector normal to the plane is
( )2,2,1=n
Hence the equation of a line perpendicular to the plane and passing through P is
( )( ) ( )2,2,14,3,2 t
t
+=
+= abar
This line meets the plane when
( ) ( ) ( ) ( ) ( ) 132,2,12,2,12,2,14,3,22,2,1 =+= tr
i.e.
( ) ( )
1
1394
13441862
==+
=++++
t
t
t
Thus the line meets the plane at N with co-ordinates
( ) ( ) ( )6,1,32,2,114,3,2 =+
Hence the perpendicular distance PN
( ) ( ) ( )[ ] units3463123 2/1222 =+++
(3) Find the equation of the line of intersection of the two planes given by:
x + y + z = 5 and 4x + y + 2z = 15
In vector from these planes can be written:
( )( ) 152,1,4
51,1,1
=
=
r
r
The line of intersection must be perpendicular to both the vectors (1,1,1) and (4,1,2).
Hence a vector in the direction of the line must be in the direction
(1,1,1)(4,1,2) = (1,2,-3)
To complete the equation of the line we need to find any one point on the line. Choosing x=0
then, from the Cartesian equations of the two planes,
y + z = 5
y + 2z = 15
Hence forx=0, y=-5 andz = 10
And the point (0, -5, 10) is a point on the line.
The equation of the line can then be written r = (0, -5, 10) + t (1, 2, -3)