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 segment
with 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.
C.
B.
D.
Over Lesson 8-4
Which of the following represents 3x – 5y + z
if x = 2, –7, 1, y = –5, 0, 3, and
z = –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
Simplify.
Solve for θ.
The measure of the angle
between u and v is
about 168.6°.
Answer: 168.6°
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.
u = 6, –1, –2
and
v = –1, –4, 2
Determinant of
a 3 × 3 matrix
Determinants of
2 × 2 matrices
Find the Cross Product of Two Vectors
Simplify.
Component
form
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
Find the Cross Product of Two Vectors
(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.
Answer: –10, –10, –25;
u × v • u = –10, –10, –25 • 6, –1, –2
= – 60 + 10 + 50 = 0
u × v • v = –10, –10, –25 • –1, –4, 2
= 10 + 40 – 50 = 0
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
Determinant
of a
3 × 3 matrix
Determinants of 2 × 2
matrices
Component form
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.
Answer: 9.9 N • m parallel to the positive y-axis
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
Determinants of 2 × 2
matrices
Step 2 Find the magnitude of u × v.
Magnitude of a vector
in space
Simplify.
Area of a Parallelogram in Space
The area of the parallelogram shown is
about 34.9 square units.
Answer:
or
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
Volume of a Parallelepiped
Find the volume of the parallelepiped with adjacent
edges t = –3i + 3j + 2k, u = –3i – 4j + 2k, and
v = 5i – 4j – k.
t = –3i + 3j + 2k ,
u = –3i – 4j + 2k and
v = 5i – 4j – k
Determinant of a 3 × 3
matrix
Volume of a Parallelepiped
Determinants of 2 × 2
matrices
Simplify.
The volume of the parallelepiped shown below is
| t ● (u × v)| or 49 cubic units.
Answer: 49 cubic units
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