Questions: The scalar product

Author

Ritwik Anand

Summary
A selection of questions for the study guide on the scalar product

Before attempting these questions, it is highly recommended that you read Guide: The scalar product, as well as Guide: Introduction to quadratic equations.

Q1

Find the scalar product of \(\mathbf{a}\) and \(\mathbf{b}\).

1.1. \(\mathbf{a}= \begin{pmatrix}6\\3\\4\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}1\\4\\2\end{pmatrix}\)

1.2. \(\mathbf{a}= \begin{pmatrix}10\\-7\\4\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}3\\-5\\13\end{pmatrix}\)

1.3. \(\mathbf{a}= \begin{pmatrix}-44\\-12\\3\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}61\\-25\\93\end{pmatrix}\)

1.4. \(\mathbf{a}= \begin{pmatrix}54\\38\\0\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}32\\-55\\13\end{pmatrix}\)

1.5. \(\mathbf{a}= 2\mathbf{i} + 7\mathbf{j} + \mathbf{k}\) and \(\mathbf{b}= 6\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}\)

1.6. \(\mathbf{a}= -3\mathbf{i} + 10\mathbf{j} - 8\mathbf{k}\) and \(\mathbf{b}= \mathbf{i} - 12\mathbf{j} + 9\mathbf{k}\)

1.7. \(\mathbf{a}= 17\mathbf{j} + 23\mathbf{k}\) and \(\mathbf{b}= 6\mathbf{i} - 23\mathbf{j} - 8\mathbf{k}\)

1.8. \(\mathbf{a}= \mathbf{i}\) and \(\mathbf{b}= \mathbf{j}\).

What can you say about the result of 1.8.? Can you deduce similar conclusions for the scalar product of different combinations of the vectors \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\)?

Q2

Using the geometric definition of the scalar products, find the smallest angle \(\theta\) in between \(\mathbf{a}\) and \(\mathbf{b}\) in degrees. If your answer is not a whole number, give your answer to an accuracy of one decimal place.

2.1. \(\mathbf{a}= \begin{pmatrix}-5\\2\\-3\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}2\\-2\\11\end{pmatrix}\)

2.2. \(\mathbf{a}= \begin{pmatrix}1\\1\\1\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}1\\-1\\1\end{pmatrix}\)

2.3. \(\mathbf{a}= \begin{pmatrix}-8\\1\\-4\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}-1\\-5\\7\end{pmatrix}\)

2.4. \(\mathbf{a}= \begin{pmatrix}1.2\\-1.4\\-3.1\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}-5.4\\9.7\\-7.5\end{pmatrix}\)

2.5. \(\mathbf{a}= \begin{pmatrix}45\\65\\54\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}-19\\-58\\71\end{pmatrix}\)

2.6. \(\mathbf{a}= \begin{pmatrix}1\\0\\0\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}0\\0\\1\end{pmatrix}\)

2.7. \(\mathbf{a}= \begin{pmatrix}-1\\-2\\3\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}4\\-5\\6\end{pmatrix}\)

2.8. \(\mathbf{a}= \begin{pmatrix}-17\\3\\8\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}12\\-19\\-16\end{pmatrix}\)

Q3

Find the value(s) of \(\lambda\) for which \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.

3.1. \(\mathbf{a}= \begin{pmatrix}2\\4\\7\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}1\\\lambda\\-2\end{pmatrix}\)

3.2. \(\mathbf{a}= \begin{pmatrix}0\\1\\\lambda\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}1\\2\\3\end{pmatrix}\)

3.3. \(\mathbf{a}= \begin{pmatrix}9\\-2\\11\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}\lambda\\-\lambda\\3\end{pmatrix}\)

3.4. \(\mathbf{a}= \begin{pmatrix}\lambda\\6\\1\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}\lambda\\\lambda\\8\end{pmatrix}\)

3.5. \(\mathbf{a}= \begin{pmatrix}-2\lambda^2\\4\\14\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}3\\2\lambda\\1\end{pmatrix}\)

3.6. \(\mathbf{a}= \begin{pmatrix}-5\\9\\2\lambda\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}\lambda\\-2\\\lambda\end{pmatrix}\)

3.7. \(\mathbf{a}= \begin{pmatrix}-7\\4\\2\lambda\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}2\lambda\\1\\6\lambda\end{pmatrix}\)

3.8. \(\mathbf{a}= \begin{pmatrix}-25\\-\lambda^2\\-2\end{pmatrix}\) and \(\mathbf{b}= \begin{pmatrix}3\lambda\\-11\\7\end{pmatrix}\)


After attempting the questions above, please click this link to find the answers.


Version history and licensing

v1.0: initial version created 08/23 by Ritwik Anand as part of a University of St Andrews STEP project.

  • v1.1: edited 05/24 by tdhc.

This work is licensed under CC BY-NC-SA 4.0.

Feedback

Your feedback is appreciated and useful. Feel free to leave a comment here.