Chapter 6.2, Problem 57E

Solution

i.

To prove: The vector property $0\cdot u=0$ using component form.

Given:

An expression $0\cdot u=0$ .

Calculation:

Suppose that $0=〈0,0〉$ and $u=〈u_1,u_2〉$ .

  $\begin{array}{l}0\cdot u=〈0,0〉\cdot 〈u_1,u_2〉\\ 0\cdot u=0\left(u_1\right)+0\left(u_2\right)\\ 0\cdot u=0+0\\ 0\cdot u=0\end{array}$

Conclusion:

Hence, It is proved that $0\cdot u=0$ .

ii.

To prove: The vector property $u\cdot \left(v+w\right)=u\cdot v+u\cdot w$ using component form.

Given:

An expression $u\cdot \left(v+w\right)=u\cdot v+u\cdot w$ .

Calculation:

Suppose that $u=〈u_1,u_2〉,v=〈v_1,v_2〉$ and $w=〈w_1,w_2〉$ .

Substitute the assumed values in the expression $u\cdot \left(v+w\right)=u\cdot v+u\cdot w$ .

  $\begin{array}{c}〈u_1,u_2〉\cdot \left(〈v_1,v_2〉+〈w_1,w_2〉\right)=〈u_1,u_2〉\cdot 〈v_1,v_2〉+〈u_1,u_2〉\cdot 〈w_1,w_2〉\\ 〈u_1,u_2〉\cdot \left(〈v_1+w_1,v_2+w_2〉\right)=u_1\cdot v_1+u_1\cdot w_1+u_2\cdot v_2+u_2\cdot w_2\\ u_1\left(v_1+w_1\right)+u_2\left(v_2+w_2\right)=u_1{v}_1+u_1{w}_1+u_2{v}_2+u_2{w}_2\\ u_1{v}_1+u_1{w}_1+u_2{v}_2+u_2{w}_2=u_1{v}_1+u_1{w}_1+u_2{v}_2+u_2{w}_2\end{array}$

The left-hand side is equal to right hand side. So,

  $u\cdot \left(v+w\right)=u\cdot v+u\cdot w$

Conclusion:

Hence, it is proved that $u\cdot \left(v+w\right)=u\cdot v+u\cdot w$ .

iii.

To prove: The vector property $\left(u+v\right)\cdot w=u\cdot w+v\cdot w$ using component form.

Given:

An expression $\left(u+v\right)\cdot w=u\cdot w+v\cdot w$ .

Calculation:

Suppose that $u=〈u_1,u_2〉,v=〈v_1,v_2〉$ and $w=〈w_1,w_2〉$ .

Substitute the assumed values in the expression $\left(u+v\right)\cdot w=u\cdot w+v\cdot w$ .

  $\begin{array}{c}\left(〈u_1,u_2〉+〈v_1,v_2〉\right)\cdot 〈w_1,w_2〉=〈u_1,u_2〉\cdot 〈w_1,w_2〉+〈v_1,v_2〉\cdot 〈w_1,w_2〉\\ 〈u_1+v_1,u_2+v_2〉\cdot 〈w_1,w_2〉=u_1\cdot w_1+u_2\cdot w_2+v_1\cdot w_1+v_2\cdot w_2\\ u_1{w}_1+v_1{w}_1+u_2{w}_2+v_2{w}_2=u_1{w}_1+v_1{w}_1+u_2{w}_2+v_2{w}_2\end{array}$

The left-hand side is equal to right hand side. So,

  $\left(u+v\right)\cdot w=u\cdot w+v\cdot w$

Conclusion:

Hence, it is proved that $\left(u+v\right)\cdot w=u\cdot w+v\cdot w$ .

iv.

To prove: The vector property $\left(u+v\right)\cdot w=u\cdot w+v\cdot w$ using component form.

Given:

An expression $\left(cu\right)\cdot v=u\cdot \left(cv\right)=c\left(u\cdot v\right)$ .

Calculation:

Suppose that $u=〈u_1,u_2〉$ and $v=〈v_1,v_2〉$ .

Substitute the assumed values in the expression $\left(cu\right)\cdot v=u\cdot \left(cv\right)=c\left(u\cdot v\right)$ .

  $\begin{array}{c}\left(c〈u_1,u_2〉\right)\cdot 〈v_1,v_2〉=〈u_1,u_2〉\cdot \left(c〈v_1,v_2〉\right)=c\left(〈u_1,u_2〉\cdot 〈v_1,v_2〉\right)\\ 〈c{u}_1,c{u}_2〉\cdot 〈v_1,v_2〉=〈u_1,u_2〉\cdot 〈c{v}_1,c{v}_2〉=c\left(u_1{v}_1+u_2{v}_2\right)\\ c{u}_1{v}_1+c{u}_2{v}_2=c{v}_1{u}_1+c{v}_2{u}_2=c{u}_1{v}_1+c{u}_2{v}_2\end{array}$

All sides are equal. So, $\left(cu\right)\cdot v=u\cdot \left(cv\right)=c\left(u\cdot v\right)$ .

Conclusion:

Hence, it is proved that $\left(cu\right)\cdot v=u\cdot \left(cv\right)=c\left(u\cdot v\right)$ .