Elementary Linear Algebra

1
Elementary Linear Algebra

• Vectors in 2-Space and 3-Space

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Contents

• Introduction to Vectors (Geometric)
• Norm of a Vector Vector Arithmetic
• Dot Product Projections
• Cross Product
• Lines and Planes in 3-Space

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Definitions

• If v and w are any two vectors, the sum v w is
  the vector determined as follows
• Position the vector w so that its initial point
  coincides with the terminal point of v. The
  vector v w is represented by the arrow from the
  initial point of v to the terminal point of w.
• If v and w are any two vectors, the difference of
  w from v is defined by v w v (-w).

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Definitions

• If v is a nonzero vector and k is nonzero real
  number (scalar), then the product kv is defined
  to be the vector whose length is k times the
  length of v and whose direction is the same as
  that of v if k gt 0 and opposite to that of v if k
  lt 0. We define kv 0 if k 0 or v 0.
• A vector of the form kv is called a scalar
  multiple.

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Remarks

• Rectangular coordinate systems in 3-space fall
  into two categories, left-handed and right-handed.

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Translation of Axes

• In the figure we have translated the axes of an
  xy-coordinate system to obtain an x?y?-coordinate
  system whose O? is at point (x, y) (k,l).
• A point P in 2-space now has both (x, y)
  coordinates and (x?, y?) coordinates.
• x? x k, y? y l, these formulas are called
  the translation equations.
• In 3-space the translation equations are
• x? x k, y? y l, z? z m,
• where (k, l, m) are the xyz-coordinates of the
  x?y?z?-origin.

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Theorem 3.2.1 (Properties of Vector Arithmetic)

• If u, v and w are vectors in 2- or 3-space and k
  and l are scalars, then the following
  relationships hold.
• u v v u
• (u v) w u (v w)
• u 0 0 u u
• u (-u) 0
• k(lu) (kl)u
• k(u v) ku kv
• (k l)u ku lu
• 1u u

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Norm of a Vector

• The length of a vector u is often called the norm
  of u and is denoted by u.
• It follows from the Theorem of Pythagoras(????)
  that the norm of a vector u (u1,u2,u3) in
  3-space is
• A vector of norm 1 is called a unit vector.
• The distance between two points is the norm of
  the vector.
• The length of the vector ku ku k u.

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Definitions

• Let u and v be two nonzero vectors in 2-space or
  3-space, and assume these vectors have been
  positioned so their initial points coincided. By
  the angle between u and v, we shall mean the
  angle ? determined by u and v that satisfies 0 ?
  ? ? ?.

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Definitions

• If u and v are vectors in 2-space or 3-space and
  ? is the angle between u and v, then the dot
  product or Euclidean inner product u v is
  defined by

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Example

• If the angle between the vectors u (0,0,1) and
  v (0,2,2) is 45?, then

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Theorems--Theorem 3.3.1

• Let u and v be vectors in 2- or 3-space.
• v v v2 that is, v2 (v v)½
• If the vectors u and v are nonzero and ? is the
  angle between them, then
• ? is acute if and only if u v gt 0
• ? is obtuse if and only if u v lt 0
• ? ?/2 if and only if u v 0

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Theorems-- Theorem 3.3.2 (Properties of the Dot
Product)

• If u, v and w are vectors in 2- or 3-space, and k
  is a scalar, then
• u v v u
• u (v w) u v u w
• k(u v) (ku) v u (kv)
• v v gt 0 if v ? 0, and v v 0 if v 0

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Orthogonal Vectors

• Definition
• Perpendicular vectors are also called orthogonal
  vectors.
• Two nonzero vectors are orthogonal if and only if
  their dot product is zero.
• To indicate that u and v are orthogonal vectors
  we write
• u?v.

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Orthogonal Vectors

• Example
• Show that in 2-space the nonzero vector n (a,b)
  is perpendicular to the line ax by c 0.

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An Orthogonal Projection

• To "decompose" a vector u into a sum of two
  terms, one parallel to a specified nonzero vector
  a and the other perpendicular to a.
• We have w2 u w1 and w1 w2 w1 (u w1)
  u
• The vector w1 is called the orthogonal projection
  of u on a or sometimes the vector component of u
  along a, and denoted by projau

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An Orthogonal Projection

• The vector w2 is called the vector component of u
  orthogonal to a, and denoted by w2 u projau

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Theorem 3.3.3

• If u and a are vectors in 2-space or 3-space and
  if a ? 0, then

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Example

• Let u(2,-1,3) and a(4,-1,2). Find the vector
  component of u along a and the vector component
  of u orthogonal to a.
• Solution
• Thus, the vector component of u along a is
• And the vector component of u orthogonal to a is
  

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Example

• Find a formula for the distance D between point
  p0(x0,y0) and the line axbyc0
• Solution
• Let Q(x1,y1) be any point on the line and
  position the vector n(a,b) so that its initial
  point is at Q.
• Thus, vector n is perpendicular to the line, and
  the distance D is equal to the length of the
  orthogonal projection of on n

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Example

• So that
• Since the point Q(x1,y1) lies on the line, its
  coordinates satisfy the equation of the line, so
• ax1by1c0 or c-ax1-by1
• Substituting this expression and yielding

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Cross Product

• Definition
• If u (u1, u2, u3) and v (v1, v2, v3) are
  vectors in 3-space, then the cross product u ? v
  is the vector defined byor in determinant
  notation
• Example
• Find u ? v, where u (1, 2, -2) and v (3, 0,
  1).

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Theorems

• Theorem 3.4.1 (Relationships Involving Cross
  Product and Dot Product)
• If u, v and w are vectors in 3-space, then
• u (u ? v) 0
• v (u ? v) 0
• u ? v u2v2 (u v)2 (Lagranges
  identity)
• u ? (v ? w) (u w) v (u v)
  w (relationship between cross dot product)
• (u ? v) ? w (u w) v (v w) u (relationship
  between cross dot product)

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Theorems

• Theorem 3.4.2 (Properties of Cross Product)
• If u, v and w are any vectors in 3-space and k is
  any scalar, then
• u ? v - (v ? u)
• u ? (v w) u ? v u ? w
• (u v) ? w u ? v v ? w
• k(u ? v) (ku) ? v u ? (kv)
• u ? 0 0 ? u 0
• u ? u 0

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Standard Unit Vectors

• The vectors
• i (1,0,0), j (0,1,0), k (0,0,1)
• have length 1 and lie along the coordinate axes.
  They are called the standard unit vectors in
  3-space.
• Every vector v (v1, v2, v3) in 3-space is
  expressible in terms of i, j, k since we can
  write
• v (v1, v2, v3) v1(1,0,0) v2 (0,1,0) v3
  (0,0,1)
• v1i v2j v3k

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Standard Unit Vectors

• For example, (2, -3, 4) 2i 3j 4k
• Note that
• i ? i 0, j ? j 0, k ? k 0
• i ? j k, j ? k i, k ? i j
• j ? i -k, k ? j -i, i ? k -j

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Cross Product

• A cross product can be represented symbolically
  in the form of 3?3 determinant
• Geometric interpretation of cross product
• From Lagranges identity, we have

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Area of a Parallelogram

• Theorem 3.4.3 (Area of a Parallelogram)
• If u and v are vectors in 3-space, then u ? v
  is equal to the area of the parallelogram
  determined by u and v.
• Example
• Find the area of the triangle determined by the
  point (2,2,0), (-1,0,2), and (0,4,3).

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Triple Product

• Definition
• If u, v and w are vectors in 3-space, then u (v
  ? w) is called the scalar triple product of u, v
  and w.
• Remarks
• The symbol (u v) ? w make no sense.
• u (v ? w) w (u ? v) v (w ? u)

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Theorem 3.4.4

• The absolute value of the determinant
• is equal to the area of the parallelogram in
  2-space determined by the vectors u (u1, u2),
  and v (v1, v2),

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Theorem 3.4.4

• The absolute value of the determinant
• is equal to the volume of the parallelepiped
  in 3-space determined by the vectors u (u1, u2,
  u3), v (v1, v2, v3), and w (w1, w2, w3),

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Remark
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Theorem 3.4.5

• If the vectors u (u1, u2, u3), v (v1, v2,
  v3), and w (w1, w2, w3) have the same initial
  point, then they lie in the same plane if and
  only if

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Planes in 3-Space

• One can specify a plane in 3-space by giving its
  inclination and specifying one of its points.
• A convenient method for a plane is to specify a
  nonzero vector, called a normal, that is
  perpendicular to the plane.
• The point-normal form of the equation of a plane
• n (a, b, c)
• a(x-x0) b(y-y0) c(z-z0) 0

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Example

• Find an equation of the plane passing through the
  point (3,-1,7) and perpendicular to the vector
  n(4,2,-5).
• Solution
• A point-normal form is
• 4(x-3)2(y1)-5(z-7)0
• By multiplying out and collecting terms,
• 4x2y-5z250

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Theorem 3.5.1

• If a, b, c, and d are constants and a, b, and c
  are not all zero, then the graph of the equation
• ax by cz d 0
• is a plane having the vector n (a, b, c) as a
  normal.
• Remark
• The above equation is a linear equation in x, y,
  and z it is called the general form of the
  equation of a plane.

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Theorem 3.5.1

• Theorem 3.5.2 (Distance between a Point and a
  Plane)
• The distance D between a point P0(x0,y0,z0) and
  the plane ax by cz d 0 is

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The Solution of a System in 3-Space
39
Example

• Find the equation of the plane passing through
  the points P1(1,2,-1), P2(2,3,1) and P3(3,-1,2).
• Solution
• Since the three points lie in the plane, their
  coordinates must satisfy the general equation
  axbyczd0 of the plane. Thus,
• a2b-cd0
• 2a3bcd0
• 3a-b2c0

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Example

• Solving this system gives a-9t/16, b-t/16, dt.
  Letting t-16 yields the desired equation
• 9xy-5z-160

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Line in 3-Space

• Suppose that l is the line in 3-space through the
  point P0(x0,y0,z0) and parallel to the nonzero
  vector v (a, b, c).
• l consists precisely of those points P0(x0,y0,z0)
  for which the vector is parallel to v,
  that is, for which there is a scalar t such that
• Parametric equations for l

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Example

• Parametric equations of a line
• The line through the point (1,2,-3) and parallel
  to the vector v(4,5,-7) has parametric equations
• x14t
• y25t
• z-3-7t (-8lttlt8)

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Example (Intersection of a Line and the xy-Plane)

• Find parametric equations for the line l passing
  through the points P1(2,4,-1) and P2(5,0,7).
• Where does the line intersect the xy-plane?
• Solution
• Since the vector is parallel to
  l and P1(2,4,-1) lies on l, the line l is given
  by
• x23t, y4-4t, z-18t
• The line intersects the xy-plane at the point
  where z-18t0, that is, where t1/8.
  Substituting this value of t yields
• (x,y,z)(19/8, 7/2, 0)

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Example (Line of Intersection of Two Planes)

• Find parametric equations for the line of
  intersection of the planes
• 3x2y-4z-60 and x-3y-2z-40
• Solution
• The line of intersection consists of all points
  (x,y,z) that satisfy the two equations in the
  system
• 3x2y-4z-60
• x-3y-2z-40
• Solving this system gives
• x26/1116t/11,
• y-6/11-2t/11,
• zt.

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Example

• A line parallel to a given vector
• The equation
• (x,y,z)(-2,0,3)t(4,-7,1)
• is the vector equation of the line through the
  point (-2,0,3) that is parallel to the vector
  v(4,-7,1)

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Theorem 3.5.2 (Distance Between a Point and a
Plane)

• The distance D between a point P0(x0,y0,z0) and
  the plane ax by cz d 0 is

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Example (Distance Between a Pont and a Plane)

• Find the distance D between the point (1,-4,-3)
  and the plane 2x-3y6z-1.
• Solution
• Applying above theorem,

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Example (Distance Between Parallel Planes)

• The planes x2y-2z3 and 2x4y-4z7
• Are parallel since their normals, (1,2,-2) and
  (2,4,-4) are parallel vectors. Find the distance
  between these planes.
• Solutions
• To find the distances between the planes, one may
  select an arbitrary point in one of the planes
  and compute its distances to the other plane.
  Thus point (3,0,0) is chosen from plane x2y-2z3
  and the distance between this point to plane
  2x4y-4z7 is