Chapter 8.4, Problem 39E

To determine

To find: the value of $k$ .

Answer to Problem 39E

  $-\frac{19}{29}$

Explanation of Solution

Given information:

  $\begin{array}{l}\overrightarrow{v}=\left(1,2\right)\\ \overrightarrow{w}=\left(-1,2\right)\\ \overrightarrow{u}=\left(5,12\right)\\ \left(k\overrightarrow{v}+\overrightarrow{w}\right)\perp \overrightarrow{u}\end{array}$

Calculation:

Since $\left(k\overrightarrow{v}+\overrightarrow{w}\right)\perp \overrightarrow{u}$ , so their dot product will be zero. Evaluate the value of $k$ as shown below.

  $\begin{array}{c}\left(k\overrightarrow{v}+\overrightarrow{w}\right)\cdot \overrightarrow{u}=0\\ \left(k\left(1,2\right)+\left(-1,2\right)\right)\cdot \left(5,12\right)=0\\ \left(\left(k,2k\right)+\left(-1,2\right)\right)\cdot \left(5,12\right)=0\\ \left(k-1,2k+2\right)\cdot \left(5,12\right)=0\\ 5\left(k-1\right)+12\left(2k+2\right)=0\left[\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2\right]\\ 5k-5+24k+24=0\\ 29k+19=0\\ k=-\frac{19}{29}\end{array}$

Thus, the value of $k$ is $-\frac{19}{29}$ .