3-D Cartesian Coordinate System
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VECTOR CALCULUS
3-D Cartesian Coordinate System
YASIR ALI
VECTOR CALCULUS
3-D Cartesian Coordinate System
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VECTOR CALCULUS
VECTORS
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VECTORS
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Vectors
A vector a in 3-dimensional space
may be represented as
a = ax^i+ ay^j + az ^
k = [ax , ay , az ].
We will use a1 , a2 and a3 instead of ax , ay and az , respectively.
Magnitude of a:
|a| =
q
a12 + a2 2 + a3 2 .
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Position Vectors
The four arrows in the plane
(directed line segments) shown
here have the same length and
direction. They therefore
represent the same vector,
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Position Vectors
A vector PQ in standard
position has its initial
point at the origin.
The directed line
segments PQ and ~v are
parallel and have the same
length.
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Vector Addition
Let u = [u1 , u2 , u3 ] and v = [v1 , v2 , v3 ] be vectors
Addition: u + v = [u1 + v1 , u2 + v2 , u3 + v3 ]
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Scalar Multiplication
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In component form, the vector from P1 (x1 , y1 , z1 ) to P2 (x2 , y2 , z2 ) is
P1 P2 = (x2 − x1 )î + (y2 − y1 )ĵ + (z2 − z1 )k̂
P1 P2 = [(x2 − x1 ), (y2 − y1 ), (z2 − z1 )]
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In component form, the vector from P1 (x1 , y1 , z1 ) to P2 (x2 , y2 , z2 ) is
P1 P2 = (x2 − x1 )î + (y2 − y1 )ĵ + (z2 − z1 )k̂
P1 P2 = [(x2 − x1 ), (y2 − y1 ), (z2 − z1 )]
Unit Vector
v̂ =
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v
|v|
VECTOR CALCULUS
In component form, the vector from P1 (x1 , y1 , z1 ) to P2 (x2 , y2 , z2 ) is
P1 P2 = (x2 − x1 )î + (y2 − y1 )ĵ + (z2 − z1 )k̂
P1 P2 = [(x2 − x1 ), (y2 − y1 ), (z2 − z1 )]
Unit Vector
v̂ =
v
|v|
Find a unit vector u in the direction of the vector from P1 (1, 0, 1) to
P2 (3, 2, 0)
1
Find P1 P2
2
Find |P1 P2 |
3
Find û =
u
|u|
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Dot Product
a.b = |a||b| cos θ,
where θ represents the shortest angle between a and b.
If a and b are given then we can
find angle between them by
cos θ =
a.b
.
|a||b|
Let a and b be given as follows
a = a1^i+ a2^j + a3^
k
^
^
b = b1 i + b2 j + b3^
k
then dot product of a and b is
a.b = a1 b1 + a2 b2 + a3 b3
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A small cart is being pulled along a smooth horizontal floor
with a 20-lb force F making a 450 angle to the floor. What
is the effective force moving the cart forward?
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Find the angle θ in the triangle ABC determined by the vertices A = (0, 0),
B = (3, 5), and C = (5, 2)
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CROSS PRODUCT
a × b = |a||b| sin θ.^
n,
where n̂ is perpendicular to plane of a and b.
In component
form the cross product of a and b is given by
a×b=
^i ^j ^
k
a1 a2 a3
b1 b2 b3
or
a×b=
(a2 b3 − b2 a3 )^i− (a3 b1 − b1 a3 )^j− (a1 b2 − b1 a2 )^
k.
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CROSS PRODUCT
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Find a vector perpendicular to the plane of P(1, −1, 0),
Q(2, 1, −1), and R(−1, 1, 2)
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Area of Parallelogram
|u × v| represents the area of a
parallelogram with sides |u| and |v|
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Lines
Parametric equations of straight line passing through point (x0 , y0 , z0 ) and
parallel to a = [a1 , a2 , a3 ] are
x = x 0 + a1 t
y = y 0 + a2 t
z = z 0 + a3 t
(t ∈ R).
In vector form
r(t) = r0 + vt.
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Find the line that passes through the point P1 (x1 , y1 , z1 )
and that is parallel to the position vector a = [a1 , a2 , a3 ]
P1 P are parallel to OA
P1 P = tOA
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VECTOR CALCULUS
Find the line that passes through the point P1 (x1 , y1 , z1 )
and that is parallel to the position vector a = [a1 , a2 , a3 ]
P1 P are parallel to OA
P1 P = tOA
[x − x1 , y − y1 , z − z1 ] = t[a1 , a2 , a3 ]
x = x1 + a1 t
y = y1 + a2 t
z = z1 + a3 t
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Find parametric equations for the
line through P(−2, 0, 4) and
parallel to v = 2î + 4ĵ − 2k̂.
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VECTOR CALCULUS
Find parametric equations for the
line through P(−2, 0, 4) and
parallel to v = 2î + 4ĵ − 2k̂.
Requirements
1
Base Point
2
Direction
given P
given v
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VECTOR CALCULUS
Find parametric equations for the
line through P(−2, 0, 4) and
parallel to v = 2î + 4ĵ − 2k̂.
Requirements
1
Base Point
2
Direction
given P
given v
P(−2, 0, 4) & v = [2, 4, −2]
x = −2 + 2t
y = 0 + 4t
z = 4 − 2t
Find parametric equations for the
line through P(−2, 0, 4) and
parallel to v = 2î + 4ĵ − 2k̂.
Requirements
1
Base Point
2
Direction
given P
given v
P(−2, 0, 4) & v = [2, 4, −2]
x = −2 + 2t
z = 4 − 2t
Find parametric equations for the
line through P(−2, 0, 4) and
parallel to v = 2î + 4ĵ − 2k̂.
Requirements
1
Base Point
2
Direction
given P
given v
P(−2, 0, 4) & v = [2, 4, −2]
x = −2 + 2t
y = 0 + 4t
z = 4 − 2t
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Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
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Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
Requirements
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Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
Requirements
1
Base Point
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VECTOR CALCULUS
Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
Requirements
1
Base Point
2
Direction
any of P or Q may be selected as base point
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VECTOR CALCULUS
Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
Requirements
1
Base Point
2
Direction
any of P or Q may be selected as base point
PQ
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VECTOR CALCULUS
Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
Requirements
1
Base Point
2
Direction
any of P or Q may be selected as base point
PQ
Q(1, −1, 4) & PQ = [4, −3, 7]
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VECTOR CALCULUS
Find parametric equations for the line through
P(−3, 2, −3) and Q(1, −1, 4).
Requirements
1
Base Point
2
Direction
any of P or Q may be selected as base point
PQ
Q(1, −1, 4) & PQ = [4, −3, 7]
x = 1 + 4t
y = −1 − 3t
z = 4 + 7t
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VECTOR CALCULUS
Parametric Equations of Line Can you find parametric equations of line
for
through two given points:
P(1, 2, −1) and Q(−1, 0, 1). A
~ = [−2, −2, 2]
vector parallel to required line is PQ
parametric eq. x = 1 − 2t, y = 2 − 2t and z = −1 + 2t
through a point and a given parallel vector:(or parallel line)
P(1, 2, −1) and [0, 2, 1] then parametric eq. x = 1, y = 2 + 2t and
z = −1 + t
through a point and parallel to given line:
Q(−1, 0, 1) and
parallel line x = 1 − 2t, y = 2 − 2t and z = −1 + 2t. A vector parallel
to required line is [−2, −2, 2] parametric eq. x = −1 − 2t, y = −2t
and z = 1 + 2t
through a point and parallel to a given axis:
P(1, 2, −1) and
parallel to x−axis. A vector parallel to required line is [1, 0, 0]
parametric eq. x = 1 + t, y = 2 and z = −1
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PLANES
If n = [a1 , a2 , a3 ] to a plane and
P0 (x0 , y0 , z0 ) is any point on the plane
then equation of plane is
−−→
n.P0 P = 0
a1 (x − x0 ) + a2 (y − y0 ) + a3 (z − z0 ) = 0,
where P(x, y , z) is any point on the
plane. In general
a1 x + a2 y + a3 z + d = 0.
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Find an equation of the plane containing the point (1, 2, 3) with normal
vector [4, 5, 6], and sketch the plane.
1
Point on Plane
2
Normal to the Plane
3
Use a1 (x − x0 ) + a2 (y − y0 ) + a3 (z − z0 ) = 0
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Find an equation of the plane containing the point (1, 2, 3) with normal
vector [4, 5, 6], and sketch the plane.
1
Point on Plane
2
Normal to the Plane
3
Use a1 (x − x0 ) + a2 (y − y0 ) + a3 (z − z0 ) = 0
Find the plane containing the three points A(1, 2, 2), B(2, −1, 4) and
C (3, 5, −2).
1
Point on Plane
2
Normal to the Plane
3
Use a1 (x − x0 ) + a2 (y − y0 ) + a3 (z − z0 ) = 0
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VECTOR CALCULUS
Find an equation of the plane containing the point (1, 2, 3) with normal
vector [4, 5, 6], and sketch the plane.
1
Point on Plane
2
Normal to the Plane
3
Use a1 (x − x0 ) + a2 (y − y0 ) + a3 (z − z0 ) = 0
Find the plane containing the three points A(1, 2, 2), B(2, −1, 4) and
C (3, 5, −2).
1
Point on Plane
2
Normal to the Plane
3
Use a1 (x − x0 ) + a2 (y − y0 ) + a3 (z − z0 ) = 0
Find AB and AC
Find AB × AC
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NON-COLLINEAR POINTS
AB = [1, −3, 2] and AC = [1, 6, −6]
î
ĵ
k̂
AB × AC = 1 −3 2
1 6 −6
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NON-COLLINEAR POINTS
AB = [1, −3, 2] and AC = [1, 6, −6]
î
ĵ
k̂
AB × AC = 1 −3 2
1 6 −6
n̂ = [6, 8, 9]
P(1, 2, 2) and n̂ = [6, 8, 9]
6(x − 1) + 8(y − 2) + 9(z − 2) = 0
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Angle Between Two planes
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INTERSECTING PLANES
1
Parallel planes have same normal vector.
2
Intersection of the planes forms a line.
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VECTOR CALCULUS
Find the line of intersection of the planes x + 2y + z = 3
and x − 4y + 3z = 5.
1
Solving both equations for a single variable (say x)
x = 3 − 2y − z,
2
&
x = 5 + 4y − 3z
Equating both expression gives value of one of the remaining variables
3 − 2y − z = 5 + 4y − 3z =⇒ z = 3y + 1
3
Using this value in the first expression gives x = −5y + 2.
4
Now letting y = t gives
x = −5t + 2,
y = t,
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and z = 3t + 1
VECTOR CALCULUS
Find a vector parallel to the line of intersection of the planes
3x − 6y − 2z = 15 and 2x + y − 2z = 5.
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VECTOR CALCULUS
Distance from a point to a line and a plane
Distance from a point to a line:
The distance from a point S to a line passing through a point P and
parallel to a is
|PS × a|
.
d=
|a|
Distance from a point to a plane:
Distance from a point P0 (x0 , y0 , z0 ) to the plane ax + by + cz + d = 0 is
d=
|ax0 + by0 + cz0 + d|
√
.
a2 + b 2 + c 2
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VECTOR CALCULUS