Week 17, Day 3

As we continued our review, Harrison presented on lesson 7.6: Linear Programming. In this lesson, we learned how to perform linear programming. A 2D linear programming problem consists of a linear objective function and a system of linear inequalities called constraints. The objective function gives the quantity that is to be maximized or minimized, and the constraints determine the set of feasible solutions.

The following are steps to solving a linear programming problem:

1) Sketch the region corresponding to the system of constraints.

2) Find the vertices of the region.

3) Test the objective function, z = ax + by, at each of the vertices and select the values of the variables that optimize the objective function. For a bounded region, both a maximum and minimum will exist. For an unbound region, if an optimized solution exists, it will occur at a vertex. 

Week 17, Day 2

Today, Grace presented on section 10.6: Polar Coordinates. As opposed to the traditional x-y coordinate system, we use the polar coordinate system. To form this system in the plane, fix a point O, called the pole or origin, and construct from O an initial ray called the polar axis. Then each point P in the plane can be assigned polar coordinates (r, theta) as follows. (1) r = directed distance from O to P (2) Theta = directed angle, counterclockwise from polar axis to segment OP. 

Example 1: Plotting Points in the Polar Coordinate System

a) The point (r, theta) = (2, pi/3) lies two units from the pole on the terminal side 

of the angle theta = pi/3, as shown in the following image. 

b) The point (r, theta) = (3, -pi/6) lies three units from the pole on the terminal side 

of the angle theta = -pi/6, as shown in the following image. 

c) The point (r, theta) = (3, 11pi/6) coincides with the point (3, -pi/6), as shown in the following image. 

In rectangular coordinates, each point (x,y) has a unique representation. This is not true for polar coordinates. For instance, the coordinate (r, theta) and (r, 2pi + theta) represent the same point. Another way to obtain multiple reoresentations of a point is to use negative values for r. Because r is a directed distance, the coordinates (r, theta) and (-r, theta + pi) represent the same point. In general, the point (r, theta) can be represented as (r, theta) = (r, theta +/- 2npi) or (r, theta) = (-r, theta +/- (2n+1)pi) where n os any interger. Moreover, the pole is represented by (0, theta), where theta is any angle. 

Example 2: Multiple Representation of Points

Plot the point (3, -3pi/4) and find three additional polar representations of this point, using -2pi < theta < 2pi. The point is shown in the following image. 

Three other representations are as follows.

(3, -3pi/4 + 2pi) = (3, 5pi/4): Add 2pi to theta.

(-3, -3pi/4 - pi) = (-3, -7pi/4): Replace r by -r, then subtract pi from theta.

(-3, -3pi/4 + pi) = (-3, pi/4): Replace r by -r, then add pi to theta.

Coordination Conversion

To establish the relationship between polar and rectangular coordinates, let the polar axis coincide with the positive x-axis and the pole with the origin. Because (x,y) lies on a circle of radius r, it follows that r^2 = x^2 + y^2. Moreover, for r > 0, the defintions of the trignometric functions imply that tan theta = y/x, cos theta = x/r, and sin theta = y/r. You can show that the same relationships hold for r > 0. 

The polar coordinates (r, theta) are related to the rectangular coordinates (x,y) as follows:

x = rcos theta and tan theta = y/x

y = rsin theta 

r^2 = x^2 + y^2

Example 3: Polar-Rectangular Conversion

Convert the points (a) (2, pi) and (b) ( (3)^1/2, pi/6) to rectangular coordinates.

(a) For the point (r, theta) = (2, pi), you have

x = rcos theta = 2 cos pi = -2 and y = rsin theta = 2 sin pi = 0.

The rectangular coordinates are (x,y) = (-2,0).

(b) For the point (r, theta) = ( (3)^1/2, pi/6), you have

x = (3)^1/2 cos pi/6 = (3)^1/2 ( (3)^1/2 / 2) = 3/2 and

y = (3)^1/2 sin pi/6 = (3)^1/2 (1/2) = (3)^1/2 / 2.

The rectangular coordinates are (x,y) = ( 3/3, (3)^1/2).

Example 5: Converting Polar Equations to Rectangular Form

Describe the graph of each polar equation and find the corresponding rectangular equation.

(a) r = 2 

The graph of the polar equation r = 2 consists of all points that are two units from the pole. In other words, this graph is a circle centered at the origin with a radius of 2. You can confirm this by converting to rectangular form, using the relationship r^2 = x^2 + y^2. 

r = 2 is the polar equation. 

r ^2 = 2^2 

x^2 + y^2 = 2^2 is the rectangular equation. 

Week 17, Day 1

This week we continued our review of topics that will be tested on our final. 
Cristian Vera presented lesson 9.5: Binomial Theorem and Expansion.
When you write out the coefficients for a binomial that is raised to a power, you are expanding a binomial. There are two methods of expansion, both of which provide binomial coefficients. The first is Binomial Theorem, which states that in the expansion of 

(x + y)^n: (x + y) ^n = x^n + (nx^n-1)y +...+ (n)C(r) (x^n-r)y^r +...+ nxy^n-1 + y^n.The coefficient of (x^n-r)y^r is given by a procedure known as combination: (n)C(r) = n! / (n-r)!r!. The second is Pascal's Triangle, in which the first andlast number in each row is 1. Every other number in each row is formed by adding the two numbers immediately above the number. A basic illustration of Pascal's triangle is:

To expand a binomial, it is easiest to use Pascal's Triangle as shown below in an example I got from my original post on the topic:

Because the exponent of the binomial is 6, you would use the 6th row of the triangle. Using the given coefficients, write each appropriate term, and simplify. 

Week 16, Day 3

Today we reviewed the topic of vectors, as explained in Darron's presentation.

Physical forces and velocities are not confined to the plane, therefore it is natural to extend the concept of vectors from two-dimensional space to three-dimensional space. In space, vectors are denoted by ordered triples v = <v1, v2, v3>, otherwise known as component form. The zero vector is denoted by 0 = < 0, 0, 0 >. Using the unit vectors i = < 1, 0, 0 > in the direction of the positive z-axis, the standard unit vector notation for v is v = v1i + v2j + v3k, otherwise known as unit vector form. If v is represented by the directed line segment from P(p1, p2, p3) to Q(q1, q2, q3), the component form of v is produced by subtracting the coordinates of the initial point from the coordinates of the terminal point, v ='<v1, v2, v3> = <q1-p1, q2-p2, q3-p3>. 

Vectors in Space:

1. Two vectors are equal if and only if their corresponding components are equal. 

2. The length of u = <u1, u2, u3> is ||u|| = (u1^2 + u2^2 + u3^2)^1/2 = vector length. 

3. A unit vector u in the direction of v is given by u = v / ||v|| = unit vector.

4. The sum of u = <u1, u2, u3> and v = <v1, v2, v3> is u + v = <u1 + v1, u2 + v2, u3 + ||v3>.

5. The scalar multiple of the real number c and u = <u1, u2, u3> is cu = <cu1, cu2, cu3>.

6. The dot product of u = <u1, u2, u3> and v = <v1, v2, v3> is u x v = u1v1 + u2v2 + u3v3.

Example 1: Finding the Component Form of a Vector

Find the component form and length of the vector v having initial point (3,4,2) amd terminal point (3,6,4). Then find a unit vector in the direction of v. 

The component form of v is:

v = <3-3, 6-4, 4-2> = <0,2,2>

This implies that the vector's length is 

||v|| = (0^2 + 2^2 + 2^2)^1/2 = (8)^1/2 = 2(2)^1/2.

The unit vector in the direction of v is:

u = v / ||v|| = 1 / 2(2)^1/2 <0,2,2> = <0, 1/(2)^1/2, 1/(2)^1/2>.

Example 2: Finding the Dot Product of Two Vectors

Find the dot product of <0,3,-2> and <4,-2,3>. 

<0,3,-2> x <4,-2,3> = 0(4) + 3(-2) + (-2)(3) = 0-6-6 = -12.

The angle between two nonzero vectors is the angle theta, 0 <(=) theta <(=) pi, between its respective standard position vectors. Thus, if theta is the angle between two nonzero vectors u and v, then cos(theta) =u x v / ||u|| ||v||. If the dot product of two nonzero vectors is zero, the angle between the vectors is 90 degrees. Such vectors are called orthogonal. For instance, the standard unit vectors i,j, and k are orthogonal to each other. 

Example 3: Finding the Angle Between Two Vectors

Find the angle between u = <1,0,2> and v = <3,1,0>.

cos(theta) = u x v / ||u|| ||v|| = <1,0,2> x <3,1,0> / ||<1,0,2>|| ||<3,1,0>|| = 3 / (50)^1/2

This implies that the angle between the two vectors is 

theta = arccos 3 / (50)^1/2 = 64.9 degrees. 

Parallel Vectors:

In general two nonzero vectors u and v are parallel if there is some scalar c such that u = cv. 

For example, the vectors shown below u, v, and w are parallel because u = 2v and w = -v. 

Example 4: Parallel Vectors

Vector w has initial point (1,-2,0) and terminal point (3,2,1). 

Which of the following vectors is parallel to w? 

a) u = <4,8,2> 

b) v = <4,8,4> 

Begin by writing w in component form.

w = <3,-1, 2 -(-2), 1-0> = <2,4,1>

a) The vector u is parallel to w because

u = <4,8,2> = 2<2,4,1> = 2w

b) In this case, you need to find a scalar c such that <4,8,4> = c<2,4,1>.

However, equating corresponding components produces c = 2 for the first two components and c = 4 for the third. Hence, the equation has no solution, and the vectors are not parallel. 

Week 16, Day 2

As we continued our review of the semester, we dove into the concept of matrices.
In her presentation, Jasmine explained how to perform inverse matrices.

A^-1 is called the inverse of A.

Show that B is the inverse of A where, 

To show that B is the inverse of A, show that AB = I = BA, as follows.

If a matrix A has an inverse, A is called invertible or nonsingular; otherwise, A is called singular. A nonsquare matrix cannot have an inverse. The following are example problems I took from my original post on the topic demonstrating how to solve for inverse matrices (one is singular and one is not). 

Week 16, Day 1

Today we began our review of this semester's material.
One of our topics, as Axie explained in her presentation, was partial fractions.

In this lesson, we learned how to perform partial fraction decomposition, which involves writing a rational expression as the sum of two or more rational expressions. For instance, (x+7) / (x^2-x-6) can be written as the sum of two fractions with linear denominators. That is, (x+7) / (x^2-x-6) = (2 / x-3) + (-1 / x+2). Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side. The following are steps to carry out partial fraction decomposition: 

1. Multiply by LC

2. Distribute

3. Collect like terms

4. Factor out the variable

5. Equate the coefficients

6. Solve the system of equations

7. Write the answer as a partial fraction

When the denominator of a partial fraction is, for example, (x+1)^2, it is called a repeated factor.

When the denominator of a partial fraction is, for example, (x^2+1), it is called a quadratic factor. 

Here are a few examples:

Example 1: Distinct Linear Factors

5/(x^2+x-6)--> 5/(x+3)(x-2) = A/(x+3) + B/(x-2)

1. 5 = A(x-2) + B(x+3)

2. 5 = Ax - 2A + Bx + 3B

3. 5 = Ax + Bx - 2A + 3B 

4. 5 = (A + B)x - 2A + 3B 

5. 0 = A + B, 5 = -2A + 3B

6. 5 = 5B--> B = 1, A = -1

7. -1/(x+3) + 1/(x-2)

Example 2:Repeated Linear Factors

2x - 3/(x-1)^2 

In this case, a factor (x-1) repeats, and thus the steps to solve it are as follows: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Equate the coefficients

5. Solve the system of equations

6. Write the answer as a partial fraction

2x - 3/(x-1)^2 = A/(x-1) + B/(x-1)^2 Every exponent gets a factor!

1. 2x-3 = A(x-1) + B

2. 2x-3 = Ax - A + B

3. 2x-3 = Ax + (-A + B)

4. A = 2, -3 = -A + B

5. -3 + 2 = B = -1

6. 2/(x-1) - 1/(x-1)^2

Example 3: Distinct Linear Factors

4x^2 + 2x - 1 / x^2(x +1) = A/x + B/x^2 + C/x+1

1. 4x^2 + 2x - 1 = Ax(x+1) + B(x+1) + Cx^2

2. 4x^2 + 2x - 1 = Ax^2 + Ax + Bx + B + Cx^2

3. 4x^2 + 2x - 1 = Ax^2 + Cx^2 + Ax + Bx + B

4. 4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B

5. A + C = 4, A + B = 2, B = -1

6. A = 3, B = -1, C = 1

7. 3/x -1/x^2 + 1/x+1

Example 4: Distinct Linear and Quadratic Factors

x^2 - 1 / x(x^2 + 1) 

In this case, the denominator contains an irreducible quadratic factor, so the set up is as follows:

x^2 - 1 / x(x^2 + 1) = A/x + Bx + C / x^2 + 1

1. x^2 - 1 = A(x^2 + 1) + (Bx + C)x

2. x^2 - 1 = Ax^2 + A + Bx^2 + Cx

3. x^2 - 1 = Ax^2 + Bx^2 + Cx + A

4. x^2 - 1 = (A + B)x^2 + Cx + A

5. A + B = 1, C = 0, A = -1

6. A = -1, B = 2, C = 0

7. -1/x + 2x/ x^2 + 1

When the exponent of the numerator is greater than or equal to the exponent of the denominator, use long division. Once you have found the quotient, take the remainder and put it over the original denominator, using it as your new partial fraction. 

Mathematical Imagery

The connection between mathematics and art goes back thousands of years. Mathematics has been used in the design of Gothic cathedrals, Rose windows, oriental rugs, mosaics, and tilings. Geometric forms were fundamental to the cubists and many abstract expressionists, and award-winning sculptors have used topology as the basis for their pieces. Dutch artist M.C. Escher represented infinity, Möbius bands, tessellations, deformations, reflections, Platonic solids, spirals, symmetry, and the hyperbolic plane in his works. Mathematicians and artists continue to create stunning works in all media. To further explore the visualization of mathematics, refer to origami, computer-generated landscapes, tesselations, fractals, and anamorphic art, which I have shown below. Hope you enjoy! 

Week 15, Day 2

Today we saw how direct substitution and operations with limits can be used to evaluate limits of certain functions, such as polynomial functions and rational functions with nonzero denominators.

Limits of Polynomial and Rational Functions:

1. If p is a polynomial function and c is a real number, then lim x-->c p(x) = p(c). 

2. If r is a rational function given by r(x) = p(x) / q(x), and c is a real number such that q(c) does not equal 0, then lim x--> c r(x) = r(c) = p(c) / q(c), q(c) does not equal 0.

Methods of Evaluating Limits:

1. Direct substitution (plug-in).

2. Cancellation technique (factor/cancel).

3. Rationalization technique (multiply radicals by conjugate).

Once you reach the intermediate form (0/0), use either the cancellation or rationalization method to evaluate the limit. If it contains a radical, use the rationalization technique. Finally, plug the value of x 

in the equation using direct substitution. This will give you the value of the limit. 

Example 1: Evaluating Limits By Direct Substitution

Find each of the following limits. 

a) lim x-->-1 (x^2 + x - 6) 

b) lim x-->-1 (x^2 + x - 6 / x + 3)

The first function is a polynomial function and the second is a rational function (with a nonzero denominator at x = -1). Thus, you can evaluate the limits by direct substitution. 

a) lim x-->-1 (x^2 + x - 6) = (-1)^2 + (-1) - 6 = -6

b) lim x-->-1 (x^2 + x - 6) / x + 3 = -6 / -1 + 3 = -3

Example 2: Evaluating Limits By Cancellation

Find the limit: lim x--> -3 (x^2 + x - 6) / x + 3.

Begin by factoring the numerator and canceling any common factors.

lim x--> -3 (x^2 + x - 6) / x + 3 (Factor numerator)

lim x--> -3 (x - 2)(x + 3) / x + 3 (Cancel common factor)

lim x--> -3 (x - 2) (Simplify)

lim = 5 (Direct substitution) 

Find the limit: lim x--> 1 (x - 1) / x^3 - x^2 + x - 1

Begin by substituting x = 1 into the numerator and denominators.

1 - 1 = 0 (Numerator is 0 when x = 1)

(1)^3 - (1)^2 + 1 - 1 = 0 (Denominator is 0 when x = 1)

Because both the numerator and denominator are zero when x = 1, direct substitution will not yield the limit. To fins the limit, you should factor the numerator and denominator, cancel any common factors, and then try direct substitution again. 

lim x--> 1 (x - 1) / x^3 - x^2 + x - 1 =

lim x--> 1 (x - 1) / (x - 1)(x^2 + 1) (Factor denominator)

lim x--> 1 (x - 1) / (x - 1)(x^2 + 1) (Cancel common factor)

lim x--> 1 (1) / (x^2 + 1) (Simplify)

(1) / (1)^2 + 1 (Substitute)

lim = 1 / 2 (Simplify)

Example 3: Evaluating Limits By Rationalization 

Find the limit: (x + 1)^1/2 - 1 / x

By direct substitution, you obtain the indeterminate form 0 / 0.

lim x-->0 (x + 1)^1/2 - 1 / x = 0 / 0

In this casem you can rewrite the fraction by rationalizing the numerator.

(x + 1)^1/2 - 1 / x  = ((x + 1)^1/2 - 1 / x)((x + 1)^1/2 - 1 / (x + 1)^1/2 + 1)) 

= (x + 1) - 1 / x((x + 1)^1/2 + 1)) (Multiply)

= x / x((x + 1)^1/2 + 1)) (Simplify)

= x / x((x + 1)^1/2 + 1)) (Cancel common factor)

= 1 / (x + 1)^1/2 + 1) (Simplify)

Now you can evaluate the limit by direct substitution.

lim x-->0 (x + 1) ^1/2 - 1 / x 

lim x--> 0 1 / (x + 1)^1/2 + 1 

= 1 / 1 + 1

= 1 / 2

You can reinforce your conclusion that the limit is 1/2 by constructing a table, or by sketching a graph.

One-Sided Limits

lim x--> c- f(x) = L (Limit from the left)

lim x--> c+ f(x) = L (Limit from the right)

The function approaches a different value from the left side of c than it approaches from the right of c.

Thus, a limit can fail to exist. This is a concept known as a one-sided limit.

Example 4: Evaluating One-Sided Limits

Find the limit as x--> 0 from the left and the limit as x--> 0 from the right for the function given by:

f(x) = |2x| / x.

From the graph of f, shown below, you can see that f(x) = -2 for all x < 0. Therefore, the limit frol the left is lim x--> 0 |2x| / x = -2 (Limit from the left). Because f(x) = 2 for all x > 0, the limit from the right is lim x--> 0 |2x| / x = 2 (Limit from the right). 

Existence of a Limit:

If f is a function and c and L are real numbers, then lim x--> c f(x) = L if and only if both the left and right limits are equal to L. 

Week 15, Day 1

Today we introduced the concept of limits. This is a fundamental aspect of calculus. 

Example 1: Finding a Rectangle of Maximum Area

You are given 24 inches of wire and are asked to form a rectangle 

whose area is as large as possible. What dimensions should the rectangle have?

Let w represent the width of the rectangle and let l represent the length of the rectangle. 

Because 2w + 2l = 24 (perimeter is 24), it follows that l = 12-w and that the area of the rectangle is 

A = wl --> Formula for area

A = w(12 - w) --> Substitute 12 - w for l

A = 12w - w^2 --> Simplify

Using this model for area, you can experiment with difficult values of w to see how to obtain the maximum area. After trying several values, it appears that the maximum area occurs when w = 6.

In limit terminology, we say that the limit of A as w approaches 6 is 36. This is written as:

lim A = 36

w--> 6

Example 2 : Finding a Limit That Can Be Reached

Use a table to estimate numerically the limit lim x--> 2 (3x - 2)

Let f(x) = 3x - 2. Then construct a table that shows values of f(x) when x is close to 2.

From the table, it appears that the closer x gets to 2, the closer f(x) gets to 4. Thus, you can estimate the limit to be 4. For this particular function, you can obtain the limit simply by substituting 2 for x to then obtain lim x--> 2 (3x - 2) = 3(2) - 2 = 4. 

Example 3: Finding a Limit That Cannot Be Reached

Use a table to estimate numerically the limit lim x-->1 (x^3 - x^2 + x - 1) / x - 1

Let f(x) = (x^3 - x^2 + x - 1) / (x - 1). Then construct a table that shows values of f(x) when x = 1.

From the table, it appears that the limit is 2. The graph below supports this conclusion. In this case, notice that you cannot obtain the limit simply by evaluating f(x) when x = 1.

Definition of a Limit:

If f(x) becomes arbitrarily close to a unique number L as x approaches 

c from either side, the limit of f(x) as x approaches c is L. This is written as lim x-->c f(x) = L. 

Conditions Under Which Limits Do Not Exist:

The limit of f(x) as x--> c does not exist if any of the following conditions is true.

1. f(x) approaches a different a different number from the right side of c than from the left side of c.

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

Example 4: Oscillating Behavior

Discuss the existence of the limit.

lim x--> 0 sin (1/x) 

Let f(x) = sin(1/x). In the diagram above, you can see that as x approaches 0, f(x) oscillates between -1 and 1. Therefore, the limit does not exist because no matter how close you are to 0, it is possible to choose values of x(1) and x(2) such that sin (1/x(1)) = 1 and sin (1/x(2)) = -1, as indicated in the table. 

Is Math in Your Future?

One of the benefits of studying mathematics is the variety of career paths it offers. A 2009 study indicated that the top three professions in terms of income and other factors were math-oriented. Another recent survey stated that the top 15 highest-earning college degrees were in mathematics. The following are seven popular career choices: 

Teaching: The teaching of mathematics at the K-12th level is a high-demand field in which the need is expected to grow significantly as we approach the future. 

Actuarial Science: Actuarial science takes mathematics and statistics and applies them to finance and insurance. It includes a number of interrelating disciplines, including probability and statistics, finance, and economics. 

Computer Science: Computer science is the study of the theoretical foundations of information and computation and their implementation and application in computer systems. Mathematicians, with their training in logical and precise thinking, are highly prized in this field. 

Operations Research: Operations research is an interdisciplinary branch of mathematics which uses mathematical methods to arrive at optimal decisions to problems in maximizing or minimizing things like costs or profits. The eventual intention behind using Operations Research is to elicit a best possible solution to a problem mathematically, which improves or optimizes the performance of the system. 

Biomathematics: Mathematical biology or biomathematics is an interdisciplinary field of study. It models natural and biological processes using mathematical techniques and tools. Results have been applied to areas such as cellular neurobiology, epidemic modelling, and population genetics. 

Cryptography: Cryptography is the practice and study of hiding information. Cryptography is considered to be a branch of both mathematics and computer science. Not just for spies anymore, cryptography applications include the security of ATM cards and computer passwords. 

Finance: Finance is a field that studies and addresses the ways in which individuals, businesses, and organizations raise, allocate, and use monetary resources over time, taking into account the risks entailed in their projects. Mathematicians can build models to help explain and predict the behavior of financial markets. Several schools offer Master's degrees in Financial Mathematics. 

Week 14, Day 2

Today we learned about the cross product of two vectors. Let u = u1i + u2j + u3k and v = v1i + v2j + v3k  be vectors in space. The cross product of u and v is the vector. u x v = (u2v3 - u3v2)i - (u1v3 - u3v1)j + (u1v2 - u2v1)k. A convenient way to calculate u x v is to use the following determinant form with cofactor expansion. (This 3 x 3 determinant form is used simply to help remember the formula for the cross product--it is technically not a determinant because not all the entries of the corresponding matrix are real numbers). 

Example 1: Finding Cross Products

Given u = i + 2j + k and v = 3i + j + 2k, find the following:

a) u x v 

b) v x u

c) v x v

Algebraic Properties of the Cross Product

1) u x v = -(v x u)

2) u x (v + w) = (u x v) + (u x w)

3) c (u x v) = (cu) x v = u x (cv)

4) u x 0 = 0 x u = 0

5) u x u = 0

6) u x (v x w) = (u x v) x w

Geometric Properties of the Cross Product

This property indicates that the vectors u x v and v x u have equal lengths but opposite directions. 

Let u and v be nonzero vectors in space, and let theta be the angle between u and v. 

1) u x v is orthogonal to both u and v.

2) ||u x v|| = ||u|| ||v|| sin theta.

3) u x v = 0 if and only if u and v are scalar multiples.

4) ||u x v|| = area of parallelogram having u and v as adjacent sides.

Example 2: Using the Cross Product

Find a unit vector that is orthogonal to both

u = 3i - 4j + k and v = -3i + 6j

The Triple Scalar Product

For vectors u, v, and w in space, the dot product of u and v x w 

is called the triple scalar product of u, v, and w. 

If the vectors u, v, and w do not lie in the same plane, the triple scalar product u x (v x w) can be used to determine the volume of the parallelpiped with u, v, and w as adjacent edges. 

Geometric Property of Triple Scalar Product:

Example 4: Volume by the Triple Scalar Product

Find the volume of the parallelpiped having

u = 3i - 5j + k, v = 2j - 2k, and w = 3i + j + k as adjacent edges.