Chapter 1, Problem 1.57P

Compute the dot product A - B for each of the following cases. Identify the units of each product.

$\begin{array}{l}\text{A}=12\text{j}+8\text{k}\text{ft}\text{B}=4\text{i}-2\text{j}+3\text{k}\text{ft}\\ \text{A}=5\text{i}+3\text{j}\text{m}\text{B}=7\text{i}-12\text{k}\text{N}\\ \text{A}=3\text{j}+2\text{j-k}\text{m}\text{B}=-6\text{i}+2\text{j}-8\text{k}\text{m}\end{array}$

To determine

(a)

The dot product $A·B$

Answer to Problem 1.57P

The dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(0\right)\left(4\right)+\left(12\right)\left(-2\right)+\left(8\right)\left(3\right)=0f{t}^2$

Explanation of Solution

Given that:

  $\overrightarrow{A}=12j+8k$ ft

  $\overrightarrow{B}=4i-2j+3k$ ft

Concept: - We know that the dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(A_{x}i+A_{y}j+A_{z}k\right)·\left(B_{x}i+B_{y}j+B_{z}k\right)$

  $=A_x{B}_x+A_y{B}_y+A_z{B}_z$

Calculation: -

The dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(12j+8k\right)·\left(4i-2j+3k\right)$

  $\overrightarrow{A}·\overrightarrow{B}=\left(0\right)\left(4\right)+\left(12\right)\left(-2\right)+\left(8\right)\left(3\right)=0f{t}^2$

To determine

(b)

The dot product $A·B$

Answer to Problem 1.57P

The dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(5\right)\left(7\right)+\left(3\right)\left(0\right)+\left(0\right)\left(12\right)=35Nm$

Explanation of Solution

Given that:

  $\overrightarrow{A}=5i+3j$ N

  $\overrightarrow{B}=7i+12k$ m

Concept: - We know that the dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(A_{x}i+A_{y}j+A_{z}k\right)·\left(B_{x}i+B_{y}j+B_{z}k\right)$

  $=A_x{B}_x+A_y{B}_y+A_z{B}_z$

Calculation: -

The dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(5i+3j\right)·\left(7i+12k\right)$

  $\overrightarrow{A}·\overrightarrow{B}=\left(5\right)\left(7\right)+\left(3\right)\left(0\right)+\left(0\right)\left(12\right)=35Nm$

To determine

(c)

The dot product $A·B$

Answer to Problem 1.57P

The dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(3\right)\left(-6\right)+\left(2\right)\left(2\right)+\left(-1\right)\left(-8\right)=-6m^2$

Explanation of Solution

Given that: -

  $\overrightarrow{A}=3i+2j-k$ m

  $\overrightarrow{B}=-6i+2j-8k$ m

Concept: - We know that the dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(A_{x}i+A_{y}j+A_{z}k\right)·\left(B_{x}i+B_{y}j+B_{z}k\right)$

  $=A_x{B}_x+A_y{B}_y+A_z{B}_z$

Calculation: -

The dot product $A·B$ is given by:

  $\overrightarrow{A}·\overrightarrow{B}=\left(3i+2j-k\right)·\left(-6i+2j-8k\right)$

  $\overrightarrow{A}·\overrightarrow{B}=\left(3\right)\left(-6\right)+\left(2\right)\left(2\right)+\left(-1\right)\left(-8\right)=-6m^2$