Chapter 6.4, Problem 14E

To determine

To find $\text{3u}\cdot \text{v}$

Answer to Problem 14E

  $3\text{u}\cdot \text{v}=-18$ & it is a scalar.

Explanation of Solution

Given:

  $\text{u}=\langle 3,3\rangle ,\text{v}=\langle -4,2\rangle$

Concept used:

The dot product of $u=\langle u_1,u_2\rangle \text{and}v=\langle v_1,v_2\rangle$ is $u\cdot v=u_1{v}_1+u_2{v}_2$ .

Calculation:

  $\begin{array}{l}3\text{u}=3\langle 3,3\rangle \\ \Rightarrow 3\text{u}=\langle 3\left(3\right),3\left(3\right)\rangle \\ \Rightarrow 3\text{u}=\langle 9,9\rangle \\ \therefore 3\text{u}\cdot \text{v}=\langle 9,9\rangle \cdot \langle -4,2\rangle \\ =9\left(-4\right)+9\left(2\right)\\ =-36+18\\ =-18\end{array}$

Since dot product of two vectors is a scalar, therefore $3\text{u}\cdot \text{v}=-18$ is a scalar.