Chapter 9.4, Problem 56AYU

To determine

To find: all the number xfor which the vectorrepresented by $\overrightarrow{PQ}=5$ .

Answer to Problem 56AYU

  $x=7,-1$ .

Explanation of Solution

Given:

The given point is $P=\left(-3,1\right)\text{and }Q=\left(x,4\right)$

And the condition is $\overrightarrow{PQ}=5$ .

Calculation:

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

The given point is $P=\left(-3,1\right)\text{and }Q=\left(x,4\right)$

Thus,

  $\begin{array}{l}PQ=Q-P=\left(x,4\right)-\left(3,5\right)=\left(x+3,3\right)\\ \therefore \Vert PQ\Vert =\sqrt{{\left(x+3\right)}^2+3^2}=\sqrt{x^2+6x+9+9}=\sqrt{x^2+6x+18}\end{array}$

Since it is given that $\overrightarrow{PQ}=5$ .

Therefore,

  $\begin{array}{l}\sqrt{x^2+6x+18}=5\{\because PQ=5\}\\ \therefore x^2+6x+18=25\left\{squaringbothsides\right\}\\ \Rightarrow x^2+6x-7=0\\ \therefore \left(x-7\right)\left(x+1\right)=0\left\{factor\right\}\end{array}$

Either $\left(x-7\right)=0$ or $\left(x+1\right)=0$ ,

When, $\left(x-7\right)=0\Rightarrow x=7$

when $\left(x+1\right)=0\Rightarrow x=-1$

Hence, all the number x for which the vector represented by $\overrightarrow{PQ}=5$ is $x=7,-1$