Chapter 6.2, Problem 44E

Solution

Find the value of vector $v$ .

The vector $v$ is either $〈3,-1〉$ or $〈-1.47,-2.79〉$ .

Given:

  $u=〈-2,5〉,u\cdot v=-11,\&{\left|v\right|}^2=10$

Calculation:

Let the vector $v=〈a,b〉$ .

Then, the modulus square of the vector $v$ is:

  $\begin{array}{c}{\left|v\right|}^2=10\\ {\left(\sqrt{a^2+b^2}\right)}^2=10\\ a^2+b^2=10\mathrm{...}\left(i\right)\end{array}$

The dot product of vector $u$ and $v$ is:

  $\begin{array}{l}u\cdot v=〈-2,5〉.〈a,b〉\\ u\cdot v=-2a+5b\end{array}$

Substitute the value of dot product:

  $-2a+5b=-11$

Find the value of variable $a$ in terms of variable $b$ :

  $\begin{array}{c}-2a+5b=-11\\ 5b+11=2a\\ a=\frac{5b+11}{2}\end{array}$

Substitute the value of variable $a$ in equation $\left(i\right)$ :

  $\begin{array}{c}{\left(\frac{5b+11}{2}\right)}^2+b^2=10\\ \frac{{\left(5b+11\right)}^2}{4}+b^2=10\\ \frac{25b^2+110b+121}{4}+b^2=10\\ \frac{25b^2+110b+121+4b^2}{4}=10\\ 29b^2+110b+121=40\\ 29b^2+110b+81=0\\ b=-1\&-2.79\end{array}$

The value of variable $a$ for $b=-1$ :

  $\begin{array}{c}a=\frac{5\left(-1\right)+11}{2}\\ a=3\end{array}$

The value of variable $a$ for $b=-2.79$ :

  $\begin{array}{c}a=\frac{5\left(-2.79\right)+11}{2}\\ a\approx -1.47\end{array}$

Hence, the vector $v$ is either $〈3,-1〉$ or $〈-1.47,-2.79〉$ .