Five-Minute Check (over Lesson 8-4)

Then/Now

New Vocabulary

Key Concept:Dot Product and Orthogonal Vectors in Space

Example 1:Find the Dot Product to Determine Orthogonal Vectors in Space

Example 2:Angle Between Two Vectors in Space

Key Concept:Cross Product of Vectors in Space

Example 3:Find the Cross Product of Two Vectors

Example 4:Real-World Example: Torque Using Cross Product

Example 5:Area of a Parallelogram in Space

Key Concept:Triple Scalar Product

Example 6:Volume of a Parallelepiped

Over Lesson 8-4

Find the length and the midpoint of the segmentwith endpoints at (−2, 3, 4) and (6, 1, −5).

A. 8.31;

B. 10.04;

C. 12.21;

D. 12.21;

Over Lesson 8-4

Locate and graph v = (–3, 4, 2).

A.

B.

C.

D.

Over Lesson 8-4

Which of the following represents 3x – 5y + zif x = 2, –7, 1, y = –5, 0, 3, andz = –1, 6, –4?

A. –23, –15, 14

B. 6, –1, –6

C. 30, −15, −16

D. 30, −15, 14

You found the dot product of two vectors in the plane. (Lesson 8-3)

• Find dot products of and angles between vectors in space.

• Find cross products of vectors in space and use cross products to find area and volume.

• cross product

• torque

• parallelepiped

• triple scalar product

Find the Dot Product to Determine Orthogonal Vectors in Space

A. Find the dot product of u and v for u = –1, 6, –3 and v = 3, –1, –3. Then determine if u and v are orthogonal.

u • v = –1(3) + 6(–1) + (–3)(–3)

= –3 + (–6) + 9 or 0

Since u • v = 0, u and v are orthogonal.

Answer: 0; orthogonal

Find the Dot Product to Determine Orthogonal Vectors in Space

B. Find the dot product of u and v for u = 2, 4, –6 and v = –3, 2, 4. Then determine if u and v are orthogonal.

u • v = 2(–3) + 4(2) + (–6)(4)

= –6 + 8 + (–24) or –22

Since u • v ≠ 0, u and v are not orthogonal.

Answer: –22; not orthogonal

Find the dot product of u = –4, 5, –1 and v = 3, –3, 1. Then determine if u and v are orthogonal.

A. – 28; orthogonal

B. – 28; not orthogonal

C. – 4; orthogonal

D. – 4; not orthogonal

Angle Between Two Vectors in Space

Find the angle θ between u = –4, –1, –3 and v = 7, 3, 4 to the nearest tenth of a degree.

Angle between two vectors

u = –4, –1, –3 and v = 7, 3, 4

Evaluate the dot product and magnitudes.

Angle Between Two Vectors in Space

The measure of the angle

between u and v is about 168.6°.

Answer: 168.6°

Simplify.

Solve for θ.

Find the angle between u = –2, 3, –1 and v = –4, –3, 4 to the nearest tenth of a degree.

A. 12.0°

B. 78.0°

C. 82.8°

D. 102.0°

Find the Cross Product of Two Vectors

Find the cross product of u = 6, –1, –2 and v = –1, –4, 2. Then show that u × v is orthogonal to both u and v.

Determinant of a 3 × 3 matrix

u = 6, –1, –2 and v = –1, –4, 2

Determinants of 2 × 2 matrices

Find the Cross Product of Two Vectors

To show that u × v is orthogonal to both u and v, find the dot product of u × v with u and u × v with v.

(u × v) • u

= –10, –10, –25 • 6, –1, –2 = –10(6) + (–10)(–1) + (–25)( –2)

= –60 + 10 + 50 or 0

Component form

Simplify.

Find the Cross Product of Two Vectors

Answer: –10, –10, –25; u × v • u = –10, –10, –25 • 6, –1, –2

= – 60 + 10 + 50 = 0u × v • v = –10, –10, –25 • –1, –4, 2

= 10 + 40 – 50 = 0

(u × v) • v

= –10, –10, –25 • –1, –4, 2= –10(–1) + (–10)(–4) + (–25)( 2)

= 10 + 40 + (–50) or 0

Because both dot products are zero, the vectors are orthogonal.

Find the cross product of u = 2, 3, –1 and v = –3, 1, 4.

A. u × v = 13, 5, 11

B. u × v = 13, –5, 11

C. u × v = 12, –5, 7

D. u × v = 13, 5, 11

Torque Using Cross Product

MACHINERY A mechanic uses a 0.4-meter long wrench to tighten a nut. Find the magnitude and direction of the torque about the nut if the force is 30 newtons straight down to the end of the handle when it is 35° above the positive x-axis.

Step 1 Graph each vector in standard position.

Torque Using Cross Product

Step 2 Determine the component form of each vector.

The component form of the vector representing the directed distance from the axis of rotation to the end of the handle can be found using the triangle in the figure below and trigonometry.

Torque Using Cross Product

Vector r is therefore 0.4 cos 35°, 0, 0.4 sin 35° or about 0.33, 0, 0.23. The vector representing the force applied to the end of the handle is 30 newtons straight down, so F = 0, 0, –30. Step 3 Use the cross product of these vectors to find the vector representing the torque about the nut.

T = r × F Torque Cross Product Formula

Cross product of r and F

Torque Using Cross Product

Determinants of 2 × 2 matrices

Component form

Determinant of a 3 × 3 matrix

Answer: 9.9 N • m parallel to the positive y-axis

Torque Using Cross Product

Step 4 Find the magnitude and direction of the torque vector.

The component form of the torque vector 0, 9.9, 0 tells us that the magnitude of the vector is about 9.9 newton-meters parallel to the positive y-axis as shown below.

MACHINERY A mechanic uses a 0.3-meter-long wrench to tighten a nut. Find the magnitude and direction of the torque about the nut if the force is 35 newtons straight down to the end of the handle when it is 40° below the positive x-axis as shown below.A. 10.5 N • m parallel

to the positive y-axis

B. 8.0 N • m parallel to the positive y-axis

C. 6.7 N • m parallel to the positive y-axis

D. 4.1 N • m parallel to the positive y-axis

Area of a Parallelogram in Space

Find the area of the parallelogram with adjacent sides u = –3i – 4j +2k and v = 5i – 4j – k.

Step 1 Find u × v.

u = –3i – 4j +2k and v = 5i – 4j – k

Determinant of a 3 × 3 matrix

Area of a Parallelogram in Space

Step 2 Find the magnitude of u × v.

Determinants of 2 × 2 matrices

Magnitude of a vector in space

Simplify.

Area of a Parallelogram in Space

The area of the parallelogram shown is or about 34.9 square units.

Answer:

Find the area of a parallelogram with sides u = 4i + 5j – 2k and v = i – j + 3k.

A. about 7.3 square units

B. about 10.0 square unit

C. about 20.8 square units

D. about 21.1 square units

Find the volume of the parallelepiped with adjacent edges t = –3i + 3j + 2k, u = –3i – 4j + 2k, and v = 5i – 4j – k.

Volume of a Parallelepiped

t = –3i + 3j + 2k , u = –3i – 4j + 2k and v = 5i – 4j – k

Determinant of a 3 × 3 matrix

The volume of the parallelepiped shown below is | t ● (u × v)| or 49 cubic units.

Answer: 49 cubic units

Volume of a Parallelepiped

Determinants of 2 × 2 matricesSimplify.

Find the volume of the parallelepiped with adjacent sides t = 2i – j + k, u = i + 2j – k, and v = 2i + 3j – k.

A. 2 cubic units

B. 6 cubic units

C. 8 cubic units

D. 14 cubic units