Chapter 2, Problem 423RE

For the following exercises, determine whether the statement is true or false. Justify the answer with a proof or a counterexample.

423. For vectors $\text{a}$ and $\text{b}$ and any given scalar $c$ , $c\left(\text{a}\cdot \text{b}\right)=\left(c\text{a}\right)\cdot \text{b}$ .

To determine

Determine whether the statement is true or false. Justify the answer with a proof or a counterexample.

Explanation of Solution

Given information:

For vector $af$ and $b$ any given scalar $c$, $c\left(a\cdot b\right)=\left(ca\right)\cdot b$

From the given information we know that for vector $a$ and $b$ any given scalar $c$, $c\left(a\cdot b\right)=\left(ca\right)\cdot b$.

We consider $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$.

Then we can find dot product as follows.

$\begin{array}{l}a\cdot b=a_1{b}_1+a_2{b}_2+a_3{b}_3\\ \Rightarrow c\left(a\cdot b\right)=c\left(a_1{b}_1+a_2{b}_2+a_3{b}_3\right)\\ Bydistributivepropertyweget,\\ \Rightarrow c\left(a\cdot b\right)=c\left(a_1{b}_1\right)+c\left(a_2{b}_2\right)+c\left(a_3{b}_3\right)\\ Byassociativepropertyweget,\\ \Rightarrow c\left(a\cdot b\right)=\left(c{a}_1\right)b_1+\left(c{a}_2\right)b_2+\left(c{a}_3\right)b_3 (1)\end{array}$

By the definition of scalar multiplication we can figure out the vector $ca$.

$\begin{array}{l}ca=c\langle a_1,a_2,a_3\rangle \\ \Rightarrow ca=\langle c{a}_1,c{a}_2,c{a}_3\rangle \end{array}$

Therefore the dot product $\left(ca\right)\cdot b$ can be figure out as follows.

$\left(ca\right)\cdot b=\left(c{a}_1\right)b_1+\left(c{a}_2\right)b_2+\left(c{a}_3\right)b_3 (2)$

From (1) and (2) we can say that,

$c\left(a\cdot b\right)=\left(ca\right)\cdot b$

Hence the statement “For vector $a$ and $b$ any given scalar $c$, $c\left(a\cdot b\right)=\left(ca\right)\cdot b$ ” is true.