Chapter 14.2, Problem 4CE

a.

To determine

Find out the sum of each given pair.

Answer to Problem 4CE

The solution is,

  $\begin{array}{l}u=ai+bj=11.2349i+5.4796j\\ v=a_{1}i+b_{1}j=-7.3536i+15.7698j\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\Vert u\Vert =12.5,\alpha =26\\ \Vert v\Vert =17.4,\beta =115\end{array}$

Calculation:

Using the trigonometric relation,

  $\begin{array}{l}\Vert u\Vert =\frac{a}{\cos \alpha }=\frac{b}{\sin \alpha }\\ \Vert v\Vert =\frac{a}{\cos \beta }=\frac{b}{\sin \beta }\end{array}$

Calculate the direction vector,

  $\begin{array}{l}a=\Vert u\Vert \cos \alpha =11.2349\\ b=\Vert u\Vert \sin \alpha =5.4796\end{array}$

Write into the vector from,

  $u=ai+bj=11.2349i+5.4796j$

Repeat the above process for v vector and get.

  $v=a_{1}i+b_{1}j=-7.3536i+15.7698j$

program:

clc
clear
close all
u1=12.5; v1=17.4;
alpha=26;
beta=115;
a=u1.*cosd(alpha);
b=u1.*sind(alpha);
a1=v1.*cosd(beta);
b1=v1.*sind(beta);

Query:

• First, define the given normal vector u.
• Then calculate the directional vector.
• Convert into vector form.
• Then repeat the same process for v.

b.

To determine

Find out the sum of each given pair.

Answer to Problem 4CE

The solution is,

  $\begin{array}{l}u=ai+bj=6.7905i-1.8195j\\ v=a_{1}i+b_{1}j=1.9169i+9.0185j\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\Vert u\Vert =7.03,\alpha =-15\\ \Vert v\Vert =9.22,\beta =78\end{array}$

Calculation:

Using the trigonometric relation,

  $\begin{array}{l}\Vert u\Vert =\frac{a}{\cos \alpha }=\frac{b}{\sin \alpha }\\ \Vert v\Vert =\frac{a}{\cos \beta }=\frac{b}{\sin \beta }\end{array}$

Calculate the direction vector,

  $\begin{array}{l}a=\Vert u\Vert \cos \alpha =6.7905\\ b=\Vert u\Vert \sin \alpha =-1.8195\end{array}$

Write into the vector from,

  $u=ai+bj=6.7905i-1.8195j$

Repeat the above process for v vector and get.

  $v=a_{1}i+b_{1}j=1.9169i+9.0185j$

program:

clc
clear
close all
u1=12.5; v1=17.4;
alpha=26;
beta=115;
a=u1.*cosd(alpha);
b=u1.*sind(alpha);
a1=v1.*cosd(beta);
b1=v1.*sind(beta);

Query:

• First, define the given normal vector u.
• Then calculate the directional vector.
• Convert into vector form.
• Then repeat the same process for v.

c.

To determine

Find out the sum of each given pair.

Answer to Problem 4CE

The solution is,

  $\begin{array}{l}u=ai+bj=-33.2281i+41.0333j\\ v=a_{1}i+b_{1}j=-35.0304i-25.4511j\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\Vert u\Vert =52.8,\alpha =129\\ \Vert v\Vert =43.3,\beta =216\end{array}$

Calculation:

Using the trigonometric relation,

  $\begin{array}{l}\Vert u\Vert =\frac{a}{\cos \alpha }=\frac{b}{\sin \alpha }\\ \Vert v\Vert =\frac{a}{\cos \beta }=\frac{b}{\sin \beta }\end{array}$

Calculate the direction vector,

  $\begin{array}{l}a=\Vert u\Vert \cos \alpha =-33.2281\\ b=\Vert u\Vert \sin \alpha =41.0333\end{array}$

Write into the vector from,

  $u=ai+bj=-33.2281i+41.0333j$

Repeat the above process for v vector and get.

  $v=a_{1}i+b_{1}j=-35.0304i-25.4511j$

program:

clc
clear
close all
u1=52.8; v1=43.3;
alpha=129;
beta=216;
a=u1.*cosd(alpha);
b=u1.*sind(alpha);
a1=v1.*cosd(beta);
b1=v1.*sind(beta);

Query:

• First, define the given normal vector u.
• Then calculate the directional vector.
• Convert into vector form.
• Then repeat the same process for v.