Chapter 10.2, Problem 65E

(a)

To determine

To find: Explain why $w$ can be represented by the arrow in accompanied diagram.

Answer to Problem 65E

By the triangle law of vector $w$ can be represented by the arrow in accompanied diagram.

Explanation of Solution

Given information:

Given that and $\overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v}$ .

  

Calculation:

In the in diagram red line represents the resultant vector. By the triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. The sum of two vector is represented by tail to head representation. In the given diagram vector $u$ and $v$ are from tail to tail so the vector $w$ is represented as $\overrightarrow{w}=\overrightarrow{v}-\overrightarrow{u}$ in the diagram by the arrow.

To determine

To find: Explain why $|w{|}^2=|u{|}^2+|v{|}^2-2|u\left|\right|v|\cos \theta$

Answer to Problem 65E

The scalar product of two vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ of magnitude $\left|\overrightarrow{u}\right|$ and $\left|\overrightarrow{v}\right|$ is given as $\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta$ , where $\theta$ represents the angle between the vectors $\overrightarrow{u}$ .

Explanation of Solution

Given information:

Given that and $\overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v}$ .

  $\begin{array}{l}{\overrightarrow{w}}^2=\left(\overrightarrow{u}-\overrightarrow{v}\right)\cdot \left(\overrightarrow{u}-\overrightarrow{v}\right)\\ |w{|}^2=\left(\overrightarrow{u}-\overrightarrow{v}\right)\cdot \overrightarrow{u}-\left(\overrightarrow{u}-\overrightarrow{v}\right)\cdot \overrightarrow{v}\\ |w{|}^2=\overrightarrow{u}\cdot \overrightarrow{u}-\overrightarrow{v}\cdot \overrightarrow{u}-\overrightarrow{u}\cdot \overrightarrow{v}+\overrightarrow{v}\cdot \stackrel{\rightharpoonup }{v}\\ |w{|}^2=|\overrightarrow{u}{|}^2-2\overrightarrow{u}\cdot \overrightarrow{v}+|\overrightarrow{v}{|}^2\end{array}$

The scalar product of two vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ of magnitude $\left|\overrightarrow{u}\right|$ and $\left|\overrightarrow{v}\right|$ is given as $\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta$ , where $\theta$ represents the angle between the vectors $\overrightarrow{u}$ .

  $|w{|}^2=|\overrightarrow{u}{|}^2-2\left|\overrightarrow{u}\right|\cdot \left|\overrightarrow{v}\right|\cos \theta +|\overrightarrow{v}{|}^2$

(c)

To determine

To find: Find the component of $\overrightarrow{w}$ then prove $|u{|}^2+|v{|}^2-|w{|}^2=2\left(u_1{v}_1+u_2{v}_2\right)$ .

Answer to Problem 65E

The component of $\overrightarrow{w}$ is $\overrightarrow{w}=\left(u_1-v_1\right)\widehat{i}+\left(u_2-v_2\right)$ .

Explanation of Solution

Given information:

Given that and $\overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v}$ .

Calculation:

The components of $\overrightarrow{u}$ and $\overrightarrow{v}$ are given substitute these component in $\overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v}$ gives:

  $\begin{array}{l}\overrightarrow{w}=u_1\widehat{i}+u_2\widehat{j}-\left(v_1\widehat{i}+v_2\widehat{j}\right)\\ \overrightarrow{w}=\left(u_1-v_1\right)\widehat{i}+\left(u_2-v_2\right)\widehat{j}\end{array}$

This is the component form of $\overrightarrow{w}$

Recall the magnitude of a vector is given as:

  $\begin{array}{l}\overrightarrow{u}=u_1\widehat{i}+u_2\widehat{j}\\ \left|\overrightarrow{u}\right|=\sqrt{u_1^2+u_2^2}\\ |\overrightarrow{u}{|}^2=u_1^2+u_2^2\end{array}$

Similarly magnitude of $\overrightarrow{v}$ is given as:

  $\begin{array}{l}\overrightarrow{v}=v_1\widehat{i}+v_2\widehat{j}\\ \left|\overrightarrow{v}\right|=\sqrt{v_1^2+v_2^2}\\ |\overrightarrow{v}{|}^2=v_1^2+v_2^2\end{array}$

Now the magnitude of $\overrightarrow{w}$ is given as:

  $|\overrightarrow{w}{|}^2={\left(u_1-v_1\right)}^2+{\left(u_2-v_2\right)}^2$

Substitute the values of $|u{|}^2,|v{|}^2,|w{|}^2$ in $|u{|}^2+|v{|}^2-|w{|}^2=2\left(u_1{v}_1+u_2{v}_2\right)$

  $\begin{array}{l}|u{|}^2+|v{|}^2-|w{|}^2=u_1^2+u_2^2+v_1^2+v_2^2-[{\left(u_1-v_1\right)}^2+{\left(u_2-v_2\right)}^2]\\ |u{|}^2+|v{|}^2-|w{|}^2=u_1^2+u_2^2+v_1^2+v_2^2-\left[u_1^2+v_1^2-2u_1{v}_1+u_2^2+v_2^2-2u_2{v}_2\right]\\ |u{|}^2+|v{|}^2-|w{|}^2=u_1^2+u_2^2+v_1^2+v_2^2-u_1^2-v_1^2+2u_1{v}_1-u_2^2-v_2^2+2u_2{v}_2\\ |u{|}^2+|v{|}^2-|w{|}^2=2\left(u_1{v}_1+u_2{v}_2\right)\end{array}$

Hence proved.

(d)

To determine

To find: Prove that $\overrightarrow{u}\cdot \overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\cos \theta$

Explanation of Solution

Given information:

Given that and $\overrightarrow{w}=\overrightarrow{u}-\overrightarrow{v}$ . $\theta$ is the angle between $\overrightarrow{u},\overrightarrow{v}$ .

Calculation:

In the given figure $\overrightarrow{w}$ represents the side $\overrightarrow{u}-\overrightarrow{v}$ .

  ${\left|\overrightarrow{u}-\overrightarrow{v}\right|}^2=|\overrightarrow{u}{|}^2+|\overrightarrow{v}{|}^2-2\left|\overrightarrow{u}\right|\cdot \left|\overrightarrow{v}\right|\cos \theta$

The magnitude of the vector is given as $|\overrightarrow{u}{|}^2-\overrightarrow{u}\cdot \overrightarrow{u}$ .

  ${\left|\overrightarrow{u}-\overrightarrow{v}\right|}^2$ can be written as $\left(\overrightarrow{u}-\overrightarrow{v}\right)\cdot \left(\overrightarrow{u}-\overrightarrow{v}\right)$ .

  $\begin{array}{l}\left(\overrightarrow{u}-\overrightarrow{v}\right)\cdot \left(\overrightarrow{u}-\overrightarrow{v}\right)=|\overrightarrow{u}{|}^2+|\overrightarrow{v}{|}^2-2\left|\overrightarrow{u}\right|\cdot \left|\overrightarrow{v}\right|\cos \theta \\ |\overrightarrow{u}{|}^2-2\overrightarrow{u}\cdot \overrightarrow{v}+|\overrightarrow{v}{|}^2=|\overrightarrow{u}{|}^2+|\overrightarrow{v}{|}^2-2\left|\overrightarrow{u}\right|\cdot \left|\overrightarrow{v}\right|\cos \theta \\ -2\overrightarrow{u}\cdot \overrightarrow{v}=-2\left|\overrightarrow{u}\right|\cdot \left|\overrightarrow{v}\right|\cos \theta \\ \overrightarrow{u}\cdot \overrightarrow{v}=\left|\overrightarrow{u}\right|\cdot \left|\overrightarrow{v}\right|\cos \theta \end{array}$

Hence proved.