Vectors 2.pdf

VECTORS
Dr. Gabriel Obed Fosu
Department of Mathematics
Kwame Nkrumah University of Science and Technology
Google Scholar: https://scholar.google.com/citations?user=ZJfCMyQAAAAJ&hl=en&oi=ao
ResearchGate ID: https://www.researchgate.net/profile/Gabriel_Fosu2
Dr. Gabby (KNUST-Maths) Vectors 1 / 37

Lecture Outline
1 Vector Product
Cross product in R3
Triple Product
2 Applications of Vector Products
3 Equation of Lines and Planes
Equation of lines
Equation of Planes

Vector Product
Outline of Presentation
1 Vector Product
Cross product in R3
Triple Product
Equation of lines
Equation of Planes

Vector Product Cross product in R3
Cross Product
Definition
The cross product of two vectors u = [u1,u2,u3] and v = [v1,v2,v3] in R3
is the vector
denoted u×v and defined by
u×v = [u2v3 −u3v2]i+[u3v1 −u1v3]j+[u1v2 −u2v1]k (1)
u×v is also called the vector product or outer product.

Alternative 1 for Cross Product
An easy way to remember the components of u×v is to consider the three vectors i,j,k
and three consecutive elements of the sequence 1,2,3,1,2.
1 We have 123, 231 and 312.
2 The first element points to the position of a basis vector.
The second and third elements indicate the indices of the components of u and v used to
calculate the coefficients.
u×v = [u2v3 −u3v2]i+[u3v1 −u1v3]j+[u1v2 −u2v1]k (2)
For instance:
1 in 123, 1 points to i and 2,3 yield [u2v3 −u3v2];
2 in 231, 2 points to j and 3,1 yield [u3v1 −u1v3]; and
3 in 312, 3 points to k and 1,2 yield [u1v2 −u2v1].

Alternative 2 for Cross Product
Definition
The determinant of order two is defined by
¯
¯
¯
¯
a b
c d
¯
¯
¯
¯ = ad −bc (3)
and the cross product can be written as follows
u×v =
¯
¯
¯
¯
¯
¯
i j k
u1 u2 u3
v1 v2 v3
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
u2 u3
v2 v3
¯
¯
¯
¯i−
¯
¯
¯
¯
u1 u3
v1 v3
¯
¯
¯
¯j+
¯
¯
¯
¯
u1 u2
v1 v2
¯
¯
¯
¯k (4)
= (u2v3 −u3v2)i+(u3v1 −u1v3)j+(u1v2 −u2v1)k (5)

Example
Find u×v where
1 u = [4,3,6] and v = [2,5,−3].
2 u = 2i−j+3k and v = −i+2j+4k,

Example
Find u×v where
1 u = [4,3,6] and v = [2,5,−3].
2 u = 2i−j+3k and v = −i+2j+4k,
1
u × v =
¯
¯
¯
¯
¯
¯
i j k
4 3 6
2 5 −3
¯
¯
¯
¯
¯
¯
(6)
= −39i+24j+14k (7)
= [−39,24,14] (8)

Example
Find u×v where
1 u = [4,3,6] and v = [2,5,−3].
2 u = 2i−j+3k and v = −i+2j+4k,
1
u × v =
¯
¯
¯
¯
¯
¯
i j k
4 3 6
2 5 −3
¯
¯
¯
¯
¯
¯
(6)
= −39i+24j+14k (7)
= [−39,24,14] (8)
2
u × v =
¯
¯
¯
¯
¯
¯
i j k
2 −1 3
−1 2 4
¯
¯
¯
¯
¯
¯
(9)
= −10i−11j+3k. (10)

Properties of Cross Product
1 Cross product is not commutative
u×v ̸= v×u rather u×v = −v×u (11)

2 The associative law for multiplication does not usually hold; that is,
[u×v]×w ̸= u×[v×w] (12)

[u×v]×w ̸= u×[v×w] (12)
3 u×[v+w] = u×v+u×w.

[u×v]×w ̸= u×[v×w] (12)
3 u×[v+w] = u×v+u×w.
4 u×[kv] = ku×v where k ∈ R.

[u×v]×w ̸= u×[v×w] (12)
3 u×[v+w] = u×v+u×w.
5 w·[u×v] = v·[w×u] = u·[v×w].

[u×v]×w ̸= u×[v×w] (12)
3 u×[v+w] = u×v+u×w.
5 w·[u×v] = v·[w×u] = u·[v×w].
6 ∥u×v∥ = ∥u∥∥v∥sinθ where θ is the internal angle between the directions of u and v.

Exercise
1 Show that
i×i = 0, j×i = −k, k×i = j,
i×j = k, j×j = 0, k×j = −i
i×k = −j, j×k = i, k×k = 0.
2 Find u×v where
1 u = [1,−1,1] and v = [2,1,0],
2 u = i+jcosθ +ksinθ and v = i−jsinθ +kcosθ.

Vector Product Triple Product
Triple Product
Definition (Triple Product)
The product u·[v×w] is called the scalar triple product of the vectors u,v and w. We can
write the scalar triple product as a determinant:
u·[v×w] =
¯
¯
¯
¯
¯
¯
u1 u2 u3
v1 v2 v3
w1 w2 w3
¯
¯
¯
¯
¯
¯
(13)

Triple Product
Definition (Triple Product)
The product u·[v×w] is called the scalar triple product of the vectors u,v and w. We can
write the scalar triple product as a determinant:
u·[v×w] =
¯
¯
¯
¯
¯
¯
u1 u2 u3
v1 v2 v3
w1 w2 w3
¯
¯
¯
¯
¯
¯
(13)
Definition (Right-Handed System)
Three vectors u,v and w [in the given order] are said to constitute a right-handed system if
the scalar triple product
u·[v×w] > 0 (14)

Definition (Left-Handed System)
Three vectors u,v and w [in the given order] are said to constitute a left-handed system if
u·[v×w] < 0 (15)

Definition (Left-Handed System)
Three vectors u,v and w [in the given order] are said to constitute a left-handed system if
u·[v×w] < 0 (15)
Linearly Dependent Vectors
The vectors u,v and w are linearly dependent or coplanar (lie in the same plane) if and
only if
u·[v×w] = 0 (16)
These vectors are linearly independent or non-coplanar if and only if
u·[v×w] ̸= 0 (17)

Applications of Vector Products
1 Vector Product
Cross product in R3
Triple Product
Equation of lines
Equation of Planes

Area and Volume
Area of the parallelogram
The area of the parallelogram formed by u and v
is given by
∥u×v∥ (18)
u and v are collinear (lie on the same line) or
linearly dependent if u×v = 0.
u
v
Two vectors or collinear if one is a scalar multiple of the other. Example [2, 1] and [6, 3]
are collinear

Area and Volumes
Paralellepiped is a three-dimensional shape whose faces are all parallelograms.

Area and Volumes
Paralellepiped is a three-dimensional shape whose faces are all parallelograms.
Volume of the paralellepiped
The volume of the paralellepiped formed by u,v
and w is
Vp = ∥u·[v×w]∥ (19)

Area and Volumes
Tetrahedron is a solid having four plane triangular faces; that is a triangular pyramid.

Area and Volumes
Tetrahedron is a solid having four plane triangular faces; that is a triangular pyramid.
Volume of a tetrahedron
The volume of the tetrahedron formed by u,v and
w is
Vt =
1
6
∥u·[v×w]∥ (20)

Example
Use the scalar triple product to show that the vectors u = [1,4,−7], v = [2,−1,4],w = [0,−9,18]
are coplanar.

Example
Use the scalar triple product to show that the vectors u = [1,4,−7], v = [2,−1,4],w = [0,−9,18]
are coplanar.
u·[v×w] =
¯
¯
¯
¯
¯
¯
1 4 −7
2 −1 4
0 −9 18
¯
¯
¯
¯
¯
¯
(21)
= 1
¯
¯
¯
¯
−1 4
−9 18
¯
¯
¯
¯−4
¯
¯
¯
¯
2 4
0 18
¯
¯
¯
¯−7
¯
¯
¯
¯
2 −1
0 −9
¯
¯
¯
¯ (22)
= 1(18)−4(36)−7(−18) (23)
= 0 (24)
This means that u,v and w are coplanar.

Exercise
1 Let u = [u1,u2,u3], v = [v1,v2,v3] and w = [w1,w2,w3] be three nonzero vectors.
a. Express w·[u×v] in terms of their coordinates.
b. Show that u×v is perpendicular to u and v.
2 Find a unit vector perpendicular to the plane that passes through the points P =
(1,4,6),Q = (−2,5,−1) and R = (1,−1,1) and compute the area of the triangle PQR.

Equation of Lines and Planes
1 Vector Product
Cross product in R3
Triple Product
Equation of lines
Equation of Planes

Equation of Lines and Planes Equation of lines
Vector equation of lines
1 A line in the 2D plane can be determined by a point on the line and slope (direction
of the line). Similarly, a line L in the 3D space is determined a point Po(xo, yo,zo) on L
and the direction of L.
2 In the 3D space the direction of a line is conveniently described by a vector, so we let
v = [a,b,c] be a vector parallel to L.
1 Let P(x, y,z) be an arbitrary point on L and let ro =
[xo, yo,zo] and r = [x, y,z] be the position vectors
of Po and P
2 If a is the vector
−
−
→
PoP, then the Triangle Law for
vector addition
r = ro +a =⇒ r = ro + tv (25)
Eqn. (25) is the vector equation of L

Parametric Equation of L
1 The vector equation of L in component form is
r = ro + tv (26)
[x, y,z] = [xo, yo,zo]+[ta,tb,tc] (27)
[x, y,z] = [xo + ta, yo + tb,zo + tc] (28)

Parametric Equation of L
1 The vector equation of L in component form is
r = ro + tv (26)
[x, y,z] = [xo, yo,zo]+[ta,tb,tc] (27)
[x, y,z] = [xo + ta, yo + tb,zo + tc] (28)
2 Two vectors are equal if and only if corresponding components are equal. Thus
x = xo + ta, y = yo + tb, z = zo + tc (29)
3 These equations (29) are called parametric equations of the line L through the point
Po and parallel to the vector v.
4 Each value of the parameter t gives a point (x, y,z) on L.

Example
1 Find a vector equation and parametric equations for the line that passes through the
point (5,1,3) and is parallel to the vector i +4j −2k.
2 Find two other points on the line.

Example
1 ro = [5,1,3] = 5i + j +3k and v = i +4j −2k, so the vector equation is
r = [5i + j +3k]+ t[i +4j −2k]
The parametric equations are
x = 5+ t, y = 1+4t, z = 3−2t

Example
1 ro = [5,1,3] = 5i + j +3k and v = i +4j −2k, so the vector equation is
r = [5i + j +3k]+ t[i +4j −2k]
The parametric equations are
x = 5+ t, y = 1+4t, z = 3−2t
2 Choosing the parameter value t = 1 gives the point (6,5,1) is a point on the line.
Similarly, t = −1 gives the point (4,−3,5).

1 The vector equation and parametric equations of a line are not unique. If we change
the point or the parameter or choose a different parallel vector, then the equations
change.

change.
2 For instance, if, instead of (5,1,3), we choose the point (6,5,1) in above, then the
parametric equations of the line become
x = 6+ t, y = 5+4t, z = 1−2t (30)

change.
x = 6+ t, y = 5+4t, z = 1−2t (30)
3 Or, if we stay with the point (5,1,3) but choose the parallel vector 2i +8j −4k, we arrive
at the equations
x = 5+2t, y = 1+8t, z = 3−4t (31)

change.
x = 6+ t, y = 5+4t, z = 1−2t (30)
at the equations
x = 5+2t, y = 1+8t, z = 3−4t (31)
4 In general, if a vector v = [a,b,c] is used to describe the direction of a line L, then the
numbers a,b, and c are called direction numbers of L.

change.
x = 6+ t, y = 5+4t, z = 1−2t (30)
at the equations
x = 5+2t, y = 1+8t, z = 3−4t (31)
4 In general, if a vector v = [a,b,c] is used to describe the direction of a line L, then the
numbers a,b, and c are called direction numbers of L.
5 Since any vector parallel to v could also be used, we see that any three numbers
proportional to a,b, and c could also be used as a set of direction numbers for L.

Symmetric Equations of L
1 Another way of describing a line L is to eliminate the parameter t from x = xo +ta, y =
yo + tb, z = zo + tc. If none of a,b, or c is 0, we can solve each of these equations for
t:
t =
x − xo
a
, t =
y − yo
b
, t =
z − zo
c
(32)
Equating the results, we obtain
x − xo
a
=
y − yo
b
=
z − zo
c
(33)
These equations (33) are called symmetric equations of L.

Symmetric Equations of L
1 Another way of describing a line L is to eliminate the parameter t from x = xo +ta, y =
yo + tb, z = zo + tc. If none of a,b, or c is 0, we can solve each of these equations for
t:
t =
x − xo
a
, t =
y − yo
b
, t =
z − zo
c
(32)
Equating the results, we obtain
x − xo
a
=
y − yo
b
=
z − zo
c
(33)
These equations (33) are called symmetric equations of L.
2 If one of a,b, or c is 0, we can still eliminate t. For instance, if a = 0 we could write the
equations of L as
x = xo,
y − yo
b
=
z − zo
c
(34)
This means that L lies in the vertical plane x = xo.

Equation of Lines and Planes Equation of Planes
Equation of Planes
1 Although a line in space is determined by a point and a direction, a plane in space is
more difficult to describe.
2 A single vector parallel to a plane is not enough to convey the ‘direction’ of the plane,
but a vector perpendicular to the plane does completely specify its direction.
1 Thus a plane in space is determined by a point
Po(xo, yo,zo) in the plane and a vector n that is
orthogonal to the plane.
2 This orthogonal vector n is called a normal vector.
3 Let P(x, y,z) be an arbitrary point in the plane, and let
ro and r be the position vectors of Po and P. Then the
vector r−ro is represented by
−
−
→
PoP

1 The normal vector n is orthogonal to every vector in the given plane. In particular, n is
orthogonal to r−ro and so we have
n·[r−ro] = 0 =⇒ n·r = n·ro (35)
2 Expanded as
n·[r−ro] = 0 (36)
[a,b,c]·[x − xo, y − yo,z − zo] = 0 (37)
The scalar equation of the plane through the point Po(xo, yo,zo) with the normal vector
n = [a,b,c] is
a(x − xo)+b(y − yo)+c(z − zo) = 0 (38)

Example
Find an equation of the plane through the point (2,4,−1) with normal vector n = [2,3,4]. Find
the intercepts and sketch the plane.

Example
We first compute the scalar equation of the plane:
a(x − xo)+b(y − yo)+c(z − zo)= 0 (39)
2(x −2)+3(y −4)+4(z +1) = 0 (40)
2x +3y +4z= 12 (41)

Example
We first compute the scalar equation of the plane:
a(x − xo)+b(y − yo)+c(z − zo)= 0 (39)
2(x −2)+3(y −4)+4(z +1) = 0 (40)
2x +3y +4z= 12 (41)
To find the x-intercept we set y = 0,z =
0 in this equation and obtain x = 6.
Similarly, the y-intercept is 4 and the z-
intercept is 3.

Angle Between Two Planes
Two planes are parallel if their normal vectors are parallel. For instance, the planes x+2y −
3z = 4 and 2x +4y −6z = 3 are parallel because their normal vectors are n1 = [1,2,−3] and
n2 = [2,4,−6] and n2 = 2n1.

Angle Between Two Planes
Two planes are parallel if their normal vectors are parallel. For instance, the planes x+2y −
3z = 4 and 2x +4y −6z = 3 are parallel because their normal vectors are n1 = [1,2,−3] and
n2 = [2,4,−6] and n2 = 2n1.
If two planes are not parallel, then they
intersect in a straight line and the angle
between the two planes is defined as the
acute angle between their normal vectors.
Thus
cosθ =
n1 ·n2
||n1||||n2||
(42)

Distance from a Point to a Plane
The distance D from a point P1 = (x1, y1,z1) to the plane ax +by +cz +d = 0 is
D =
|ax1 +by1 +cz1 +d|
p
a2 +b2 +c2
(43)

Example
1 Find parametric equations and symmetric equations of the line that passes through
the points A = (2,4,−3) and B = (3,−1,1).
2 At what point does this line intersect the xy-plane?

Example
1 Find parametric equations and symmetric equations of the line that passes through
the points A = (2,4,−3) and B = (3,−1,1).
2 At what point does this line intersect the xy-plane?
We first find the vector v. Here, we are not explicitly given
a vector parallel to the line, but observe that the vector v
with representation
−
→
AB is parallel to the line.
v =
−
→
AB =
−
−
→
OB −
−
−
→
OA (44)
= [3,−1,1]−[2,4,−3] (45)
= [1,−5,4] (46)

So v = [a,b,c] = [1,−5,4] and taking the point Po as (2,4,−3), then
parametric equations are
x = 2+ t, y = 4−5t, z = −3+4t (48)

So v = [a,b,c] = [1,−5,4] and taking the point Po as (2,4,−3), then
parametric equations are
x = 2+ t, y = 4−5t, z = −3+4t (48)
symmetric equations
x − xo
a
=
y − yo
b
=
z − zo
c
(49)
x −2
1
=
y −4
−5
=
z +3
4
(50)

2. The line intersects the xy-plane when z = 0 in the symmetric equations and obtain
x −2
1
=
y −4
−5
=
3
4
(51)
Thus
x −2 =
3
4
=⇒ x =
11
4
(52)
y −4
−5
=
3
4
=⇒ y =
1
4
(53)
so the line intersects the xy-plane at the point
µ
11
4
,
1
4
,0
¶

Example
Find an equation of the plane that passes through the points P = (1,3,2), Q = (3,−1,6), R =
(5,2,0)

Example
(5,2,0)
Let the vectors a and b corresponding to
−
−
→
PQ and
−
→
PR that is
a = [2,−4,4], b = [4,−1,−2]
Since both a and b lie in the plane, their cross
product a × b is orthogonal to the plane and can
be taken as the normal vector.

Example
(5,2,0)
Let the vectors a and b corresponding to
−
−
→
PQ and
−
→
PR that is
a = [2,−4,4], b = [4,−1,−2]
Since both a and b lie in the plane, their cross
product a × b is orthogonal to the plane and can
be taken as the normal vector.
Thus
n = a×b =
¯
¯
¯
¯
¯
¯
i j k
2 −4 4
4 −1 −2
¯
¯
¯
¯
¯
¯
= 20i +20j +14k (54)

With the point P = (1,3,2) and the normal vector n, an equation of the plane is
12(x −1)+20(y −3)+14(z −2) = 0 (55)
or
6x +10y +7z = 50 (56)

Example
Find the point at which the line with parametric equations x = 2 + 3t, y = −4t, z = 5 + t
intersect the plane 4x +5y −2z = 18.

Example
Find the point at which the line with parametric equations x = 2 + 3t, y = −4t, z = 5 + t
intersect the plane 4x +5y −2z = 18.
We substitute the expressions for x, y, and z from the parametric equations into the
equation of the plane:
4(2+3t)+5(−4t)−2(5+ t) = 18 =⇒ t = −2 (57)
Therefore the point of intersection occurs when the parameter value is t = −2. Then
x = 2+3(−2) = −4, y = −4(−2) = 8, z = 5−2 = 7 (58)
so the point of intersection is (−4,8,3).

Example
Find the angle between the planes x + y + z = 1 and x −2y +3z = 1

Example
Find the angle between the planes x + y + z = 1 and x −2y +3z = 1
The normal vectors of these planes are n1 = [1,1,1] and n2 = [1,−2,3] so
cosθ =
n1 ·n2
||n1||||n2||
(59)
=
1[1]+1[−2]+1[3]
p
3
p
14
(60)
=
2
p
42
(61)
θ = cos−1
·
2
p
42
¸
(62)
= 72ř (63)

Exercise
1 Find the parametric equations of the line L that passes through the point A = (−1,2,1)
and parallel to the vector i+j−k.
2 Find out if the following points belong to L : B = (0,3,0), C = (1,1,1), D = (−2,1,2).
3 If L ′
is a line with parametric equation x = 1+2s, y = 3,z = −2+ s, show that L and L ′
are not parallel and do not intercept. (They are skew lines.)
4 Show that the planes P : 2x+2y − z −10 = 0 and P ′
: 3
2 x − y + z = 0 are perpendicular.
5 Find the angle of intersection between the two planes P : 2x + 3y + 4z = 5 and P ′
:
2x −6y −3z = 0.
6 Find the component of the force F = 2i−j+2k in the direction of n = i+j+k
p
2.

END OF LECTURE
THANK YOU

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