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Cross product practice problems

u × v by the determinant, and the area of the parallelogram it spans. 70 problems with worked solutions; in 70 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 9.23

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, 2, -2 \rangle,\ \mathbf{v} = \langle 2, -4, -4 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-16\\-8\\0\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 8 \sqrt{5} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -16, -8, 0 \rangle,\quad \text{area} = 8 \sqrt{5} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.25

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, 2, 3 \rangle,\ \mathbf{v} = \langle -4, -3, -1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}7\\-10\\2\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 3 \sqrt{17} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 7, -10, 2 \rangle,\quad \text{area} = 3 \sqrt{17} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.61

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, 2, 2 \rangle,\ \mathbf{v} = \langle -2, -1, -3 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-4\\-13\\7\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 3 \sqrt{26} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -4, -13, 7 \rangle,\quad \text{area} = 3 \sqrt{26} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.65

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 3, 3 \rangle,\ \mathbf{v} = \langle 4, 2, 0 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-6\\12\\-16\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 2 \sqrt{109} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -6, 12, -16 \rangle,\quad \text{area} = 2 \sqrt{109} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.69

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, 1, 2 \rangle,\ \mathbf{v} = \langle 4, 1, -4 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-6\\12\\-3\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 3 \sqrt{21} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -6, 12, -3 \rangle,\quad \text{area} = 3 \sqrt{21} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.72

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 1, 1 \rangle,\ \mathbf{v} = \langle -4, 4, 2 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-2\\-10\\16\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 6 \sqrt{10} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -2, -10, 16 \rangle,\quad \text{area} = 6 \sqrt{10} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.103

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, 3, -2 \rangle,\ \mathbf{v} = \langle 0, 3, 0 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}6\\0\\-9\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 3 \sqrt{13} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 6, 0, -9 \rangle,\quad \text{area} = 3 \sqrt{13} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.127

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, -4, 2 \rangle,\ \mathbf{v} = \langle 1, -3, 0 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}6\\2\\16\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 2 \sqrt{74} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 6, 2, 16 \rangle,\quad \text{area} = 2 \sqrt{74} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.129

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, -1, 2 \rangle,\ \mathbf{v} = \langle 4, 4, -1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-7\\7\\0\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 7 \sqrt{2} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -7, 7, 0 \rangle,\quad \text{area} = 7 \sqrt{2} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.130

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, -4, -4 \rangle,\ \mathbf{v} = \langle 2, 4, 0 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}16\\-8\\4\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 4 \sqrt{21} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 16, -8, 4 \rangle,\quad \text{area} = 4 \sqrt{21} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.151

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 2, -2 \rangle,\ \mathbf{v} = \langle -3, 0, 3 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}6\\12\\6\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 6 \sqrt{6} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 6, 12, 6 \rangle,\quad \text{area} = 6 \sqrt{6} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.196

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, -3, -3 \rangle,\ \mathbf{v} = \langle 0, 2, -4 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}18\\-16\\-8\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 2 \sqrt{161} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 18, -16, -8 \rangle,\quad \text{area} = 2 \sqrt{161} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.197

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, -3, -1 \rangle,\ \mathbf{v} = \langle 2, 4, -1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}7\\0\\14\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 7 \sqrt{5} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 7, 0, 14 \rangle,\quad \text{area} = 7 \sqrt{5} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.198

medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 2, 2 \rangle,\ \mathbf{v} = \langle 3, 4, 0 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-8\\6\\6\end{matrix}\right] \]
    The components.Reviewed
  3. \[ 2 \sqrt{34} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -8, 6, 6 \rangle,\quad \text{area} = 2 \sqrt{34} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.18

easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 4, -3 \rangle,\ \mathbf{v} = \langle -1, -3, 2 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-1\\-3\\-5\end{matrix}\right] \]
    The components.Reviewed
  3. \[ \sqrt{35} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -1, -3, -5 \rangle,\quad \text{area} = \sqrt{35} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.30

easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, 0, 2 \rangle,\ \mathbf{v} = \langle -2, -1, 2 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}2\\2\\3\end{matrix}\right] \]
    The components.Reviewed
  3. \[ \sqrt{17} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 2, 2, 3 \rangle,\quad \text{area} = \sqrt{17} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.63

easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 2, -4 \rangle,\ \mathbf{v} = \langle -4, 2, 1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}10\\13\\14\end{matrix}\right] \]
    The components.Reviewed
  3. \[ \sqrt{465} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle 10, 13, 14 \rangle,\quad \text{area} = \sqrt{465} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.67

easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 0, 3 \rangle,\ \mathbf{v} = \langle 3, 1, -1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-3\\7\\-2\end{matrix}\right] \]
    The components.Reviewed
  3. \[ \sqrt{62} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -3, 7, -2 \rangle,\quad \text{area} = \sqrt{62} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.68

easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 0, -2, 1 \rangle,\ \mathbf{v} = \langle 4, 3, 1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-5\\4\\8\end{matrix}\right] \]
    The components.Reviewed
  3. \[ \sqrt{105} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -5, 4, 8 \rangle,\quad \text{area} = \sqrt{105} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 9.124

easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, -3, -2 \rangle,\ \mathbf{v} = \langle -3, -2, 1 \rangle \), and the area of the parallelogram they span.
  1. Expand the determinant with rows i j k, u, v.
    Reviewed
  2. \[ \left[\begin{matrix}-7\\3\\-15\end{matrix}\right] \]
    The components.Reviewed
  3. \[ \sqrt{283} \]
    Area = |u × v|.Reviewed
Answer \( \mathbf{u}\times\mathbf{v} = \langle -7, 3, -15 \rangle,\quad \text{area} = \sqrt{283} \)

✓ Nihil obstat Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More cross product practice

Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, 0, 1 \rangle,\ \mathbf{v} = \langle -2, 4, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.16easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, -2, 3 \rangle,\ \mathbf{v} = \langle 4, 0, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.17easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, -1, -2 \rangle,\ \mathbf{v} = \langle -1, 2, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.21easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 3, -1 \rangle,\ \mathbf{v} = \langle 0, -1, 0 \rangle \), and the area of the parallelogram they span.
Problem 9.22easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, 1, 1 \rangle,\ \mathbf{v} = \langle -1, -4, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.24easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, 3, 0 \rangle,\ \mathbf{v} = \langle 4, -1, 3 \rangle \), and the area of the parallelogram they span.
Problem 9.62easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 0, -4, -2 \rangle,\ \mathbf{v} = \langle 2, -2, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.64easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, -3, 0 \rangle,\ \mathbf{v} = \langle 3, 1, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.70easy✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, 3, 0 \rangle,\ \mathbf{v} = \langle 0, 2, 3 \rangle \), and the area of the parallelogram they span.
Problem 9.75easy✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, -3, -2 \rangle,\ \mathbf{v} = \langle 2, -1, 2 \rangle \), and the area of the parallelogram they span.
Problem 9.100easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, -2, 1 \rangle,\ \mathbf{v} = \langle 0, -4, 3 \rangle \), and the area of the parallelogram they span.
Problem 9.101easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 0, 1 \rangle,\ \mathbf{v} = \langle 0, -1, 0 \rangle \), and the area of the parallelogram they span.
Problem 9.102easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, -3, 0 \rangle,\ \mathbf{v} = \langle -2, 2, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.104easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, 0, 1 \rangle,\ \mathbf{v} = \langle -1, 0, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.105easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 3, -1 \rangle,\ \mathbf{v} = \langle -1, 1, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.125easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, 0, -3 \rangle,\ \mathbf{v} = \langle -1, -4, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.148easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, -4, -1 \rangle,\ \mathbf{v} = \langle 0, -1, 0 \rangle \), and the area of the parallelogram they span.
Problem 9.150easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 2, -1 \rangle,\ \mathbf{v} = \langle 1, -2, -4 \rangle \), and the area of the parallelogram they span.
Problem 9.152easy✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 0, -2, 1 \rangle,\ \mathbf{v} = \langle 1, 2, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.154easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, -2, 0 \rangle,\ \mathbf{v} = \langle -1, 1, 0 \rangle \), and the area of the parallelogram they span.
Problem 9.171easy✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, 1, 0 \rangle,\ \mathbf{v} = \langle 3, 1, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.172easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 3, -4 \rangle,\ \mathbf{v} = \langle 4, 1, 3 \rangle \), and the area of the parallelogram they span.
Problem 9.173easy✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, -4, -4 \rangle,\ \mathbf{v} = \langle -2, 1, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.174easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, -1, -4 \rangle,\ \mathbf{v} = \langle 4, 2, 3 \rangle \), and the area of the parallelogram they span.
Problem 9.175easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, 1, 1 \rangle,\ \mathbf{v} = \langle 3, 0, -4 \rangle \), and the area of the parallelogram they span.
Problem 9.176easyReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, -1, -2 \rangle,\ \mathbf{v} = \langle -2, 1, 4 \rangle \), and the area of the parallelogram they span.
Problem 9.200easy✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, 0, -4 \rangle,\ \mathbf{v} = \langle 3, -1, 4 \rangle \), and the area of the parallelogram they span.
Problem 9.202easyNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, -3, 2 \rangle,\ \mathbf{v} = \langle 2, 1, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.19mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 0, 0, 1 \rangle,\ \mathbf{v} = \langle 2, 2, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.20mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, -3, 3 \rangle,\ \mathbf{v} = \langle -1, -4, 4 \rangle \), and the area of the parallelogram they span.
Problem 9.26mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, -2, 2 \rangle,\ \mathbf{v} = \langle 0, 1, 0 \rangle \), and the area of the parallelogram they span.
Problem 9.27mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, -2, 4 \rangle,\ \mathbf{v} = \langle -4, 4, 2 \rangle \), and the area of the parallelogram they span.
Problem 9.28mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, -1, -4 \rangle,\ \mathbf{v} = \langle -1, -2, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.29mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, 2, 1 \rangle,\ \mathbf{v} = \langle 3, 0, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.66mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, 2, 2 \rangle,\ \mathbf{v} = \langle -4, -3, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.71mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, 4, 3 \rangle,\ \mathbf{v} = \langle -3, -4, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.73medium✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, 3, 1 \rangle,\ \mathbf{v} = \langle 4, -1, -2 \rangle \), and the area of the parallelogram they span.
Problem 9.74mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 3, 2 \rangle,\ \mathbf{v} = \langle 2, 1, -4 \rangle \), and the area of the parallelogram they span.
Problem 9.99mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, -3, -4 \rangle,\ \mathbf{v} = \langle 4, 2, 0 \rangle \), and the area of the parallelogram they span.
Problem 9.106mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, 0, 1 \rangle,\ \mathbf{v} = \langle -2, 2, -2 \rangle \), and the area of the parallelogram they span.
Problem 9.123mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, -4, -4 \rangle,\ \mathbf{v} = \langle -3, 3, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.126mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, 0, -3 \rangle,\ \mathbf{v} = \langle 3, -3, -3 \rangle \), and the area of the parallelogram they span.
Problem 9.128mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, 1, 0 \rangle,\ \mathbf{v} = \langle -2, 1, 4 \rangle \), and the area of the parallelogram they span.
Problem 9.147mediumNot checked
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, 2, 4 \rangle,\ \mathbf{v} = \langle -4, 2, -1 \rangle \), and the area of the parallelogram they span.
Problem 9.149mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -1, -2, 4 \rangle,\ \mathbf{v} = \langle -2, -4, 2 \rangle \), and the area of the parallelogram they span.
Problem 9.153mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 1, -3, 2 \rangle,\ \mathbf{v} = \langle 1, -1, 4 \rangle \), and the area of the parallelogram they span.
Problem 9.177mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 2, 0, 4 \rangle,\ \mathbf{v} = \langle -4, -2, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.178mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 4, -1, 1 \rangle,\ \mathbf{v} = \langle 0, -1, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.195mediumReviewed
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -3, 3, -3 \rangle,\ \mathbf{v} = \langle 3, 4, 4 \rangle \), and the area of the parallelogram they span.
Problem 9.199medium✓ Nihil obstat
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -2, 4, 0 \rangle,\ \mathbf{v} = \langle 0, 3, 1 \rangle \), and the area of the parallelogram they span.
Problem 9.201medium✓ Nihil obstat