Chapter 9.4, Problem 55AYU

To determine

To find: all the number xfor which the vector $\Vert \text{v}+\text{w}\Vert =5$ .

Answer to Problem 55AYU

  $x=-2\pm \sqrt{21}$ .

Explanation of Solution

Given:

The given vector is $\text{v}=2\text{i}+\text{j}\text{and }-x\text{i}-3\text{j}$

And the condition is $\Vert \text{v}+\text{w}\Vert =5$ .

Calculation:

As the magnitude of the vector $\text{v}=a\text{i}+b\text{j}=\langle a,b\rangle \text{is }\Vert \text{v}\Vert =\sqrt{a^2+b^2}$ .

Since it is given that $\Vert \text{v}+\text{w}\Vert =5$ .

Therefore,

  $\begin{array}{l}\sqrt{x^2+4x+8}=5\{\because \text{v}=5\}\\ \therefore x^2+4x+8=25\left\{squaringbothsides\right\}\\ \Rightarrow x^2+4x-17=0\\ x=\frac{-4\pm \sqrt{16-\left(-68\right)}}{2\times 1}\left\{quadratic\text{formula}\text{apply}\right\}\\ x=\frac{-4\pm \sqrt{84}}{2}\\ x=\frac{-4\pm 2\sqrt{21}}{2}=-2\pm \sqrt{21}\end{array}$

Hence, all the number x for which the vector $\Vert \text{v}+\text{w}\Vert =5$ is $x=-2\pm \sqrt{21}$