1. Sec 7.5 – Advanced Vector Operations Name:

The dot product of two vectors can be calculated two ways:

풖⃗ ∙ 풗⃗ = ‖풖⃗‖‖풗⃗‖ 퐜퐨퐬(휽) The same value can also be determined by finding the sum of the products of each corresponding element.

풖⃗ ∙ 풗⃗ = 풖ퟏ풗ퟏ + 풖ퟐ풗ퟐ + ⋯+ 풖풏풗풏

For example, let 푢⃗:⟨−4,6⟩ and 푣⃗:⟨3,2⟩ as shown at the right.

푢⃗ ∙ 푣⃗ = (−4)(3) + (6)(2) = −12 + 12 = 0

Notice that the if the dot product of two non-zero vectors is 0 then the two vectors are perpendicular. This could be proven to some extent for a 2-dimensional vectors without too much difficulty if we can assume that perpendicular lines have negative reciprocal slopes.

Consider vector 푎⃗:⟨푥,푦⟩ would have a slope of and a vector that is perpendicular must then have a slope

of − where c is some constant, which would suggest that a perpendicular vector would be of the form

푏⃗:⟨−푐푦, 푐푥⟩ which would have a dot product of 푎⃗ ∙ 푏⃗ = (푥)(−푐푦) + (푦)(푐푥) = −푐푥푦 + 푐푥푦 = 0 1. Find the dot product of the following set of vectors, graph the vectors, and determine which are perpendicular.

a. 푝⃗:⟨−4,3⟩ and

푞⃗:⟨4,5⟩

b. 푚⃗:⟨−8,2⟩ and

푛⃗:⟨1, 4⟩

c. 푢⃗:⟨−4,−3,2⟩ and 푣⃗:⟨−3,2,−3⟩

M. Winking © Unit 7-5 pg. 138

Magnitude of vector 푢⃗

Magnitude of vector 푣⃗

Cosine of the angle,, between the two vectors

2. Answer the following using the dot product. a. Consider the vectors 푝⃗:⟨−6,4⟩ and

푞⃗:⟨6,푎⟩. Using the dot product, what value of ‘a’ would ensure the vectors are perpendicular?

b. Consider the vectors 푚⃗:⟨2, 3,−1⟩ and 푛⃗:⟨5,푏, 푐⟩. Using the dot product, determine 2 sets of values that would ensure the vectors are perpendicular. (Can you determine a general solution to ensure the vectors are perpendicular?)

The cross product of two 3-d vectors can be calculated two ways:

풖⃗ × 풗⃗ = (‖풖⃗‖‖풗⃗‖ 퐬퐢퐧(휽)) ∙ 풏 The same value can also be determined by finding the determinant of the matrix created by

풖⃗ × 풗⃗ = 풅풆풕⃗ ⃗ 풌⃗풖ퟏ 풖ퟐ 풖ퟑ풗ퟏ 풗ퟐ 풗ퟑ

For example, let 풖⃗:⟨ퟏ,−ퟐ,ퟏ⟩ and 풗⃗:⟨ퟐ,−ퟏ,ퟏ⟩ as shown at the right.

푢⃗ × 푣⃗ = 푑푒푡횤⃗ 횥⃗ 푘⃗1 −2 12 −1 1

=횤⃗ 횥⃗ 푘⃗1 −2 12 −1 1

횤⃗ 횥⃗1 −22 −1

This cross product creates a new vector that is perpendicular to both of the original vectors simultaneously which we could verify with the dot product:

푢⃗ ∙ 푐⃗ = ⟨1,−2, 1⟩ ∙ ⟨−1, 1,3⟩ = (1)(−1) + (−2)(1) + (1)(3) = −1 + −2 + 3 = 0

푣⃗ ∙ 푐⃗ = ⟨2,−1, 1⟩ ∙ ⟨−1, 1,3⟩ = (2)(−1) + (−1)(1) + (1)(3) = −2 + −1 + 3 = 0

Since the dot products are each zero that suggests that each of the two sets of vectors are perpendicular but the dot product of 푢⃗ ∙ 푣⃗ = 5 so the original vectors 푢⃗ and 푣⃗ are not perpendicular

Magnitude of vector 푢⃗

Magnitude of vector 푣⃗

Sine of the angle,, between the two vectors

The unit vector at right angles to both vectors

−2횤⃗+ 2횥⃗ − 1푘⃗ − 1횥⃗ − 1횤⃗ − 4푘⃗ = −1횤⃗ + 1횥⃗ + 3푘⃗ = ⟨−1,1,3⟩ = 푐⃗

M. Winking © Unit 7-5 pg. 139

3. Find the cross product of the following set of vectors a. 푚⃗:⟨3,−4,2⟩ and 푛⃗:⟨5, 2, 1⟩

b. 푝⃗:⟨3,−1,2⟩ and 푞⃗:⟨−6, 2,−4⟩

4. Consider the vectors 푢⃗:⟨3,2,2⟩ and 푣⃗:⟨4, 1, 2⟩ a. Determine 푢⃗ × 푣⃗

b. Determine 푣⃗ × 푢⃗

c. Is the cross product commutative?

The angle between two vectors can be calculated using the Law of Cosines:

풄ퟐ = 풂ퟐ + 풃ퟐ − ퟐ풂풃풄풐풔(푪)

M. Winking © Unit 7-5 pg. 140

The formula can be extended to higher dimensions as well. Using the formula complete the problems below.

풄풐풔(흑) = 풖⃗ ∙ 풗⃗

‖풖⃗‖‖풗⃗‖

5. Graph the following vectors and find the angle between the following set of vectors

a. 푎⃗:⟨4,−2⟩ and 푏⃗:⟨−1,4⟩

b. 푚⃗:⟨−6,−4⟩ and 푛⃗:⟨−2,7⟩

c. 푢⃗:⟨2,−2, 3⟩ and 푣⃗:⟨−3, 2, 1⟩

M. Winking © Unit 7-5 pg. 141