Question

If $P=\left(-3,1\right)andQ=\left(x,4\right)$, find all numbers x such that the vector represented by $\stackrel{\rightharpoonup }{PQ}$has length 5.

Short answer

There will be two values for x which are -7 and 1.

Step-by-step solution

Step. 1 Given

We have coordinates of two points,

$P\equiv \left(-3,1\right)andQ\equiv \left(x,4\right)$

and, length of the vector

$\stackrel{\rightharpoonup }{PQ}is5$

i.e.,

$\left|\stackrel{\rightharpoonup }{PQ}\right|=5$

Step. 2 Formula to find length of a vector

As we all know,

The length of the vector having coordinates $\left(x_1,y_1\right)and\left(x_2,y_2\right)$is

$\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}units$

Now, $$\begin{gathered} \stackrel{\rightharpoonup }{PQ}=\left(x-\left(-3\right)\right)\stackrel{\rightharpoonup }{i}+\left(4-1\right)\stackrel{\rightharpoonup }{j} \\ \stackrel{\rightharpoonup }{PQ}=\left(x+3\right)\stackrel{\rightharpoonup }{i}+3\stackrel{\rightharpoonup }{j} \end{gathered}$$,

Step. 3 Calculating the value of x

Now we have,

$$\begin{gathered} \left|\stackrel{\rightharpoonup }{PQ}\right|=5 \\ \sqrt{{\left(x+3\right)}^2+{\left(3\right)}^2}=5 \end{gathered}$$,

Squaring both sides we have,

$$\begin{gathered} {\left(x+3\right)}^2+9=25, \\ {\left(x+3\right)}^2=16, \end{gathered}$$

Now taking square root both sides we get,

$$\begin{gathered} \left(x+3\right)=\pm 4, \\ So, \\ x=\pm 4-3, \\ i.e., \\ x=1or-7 \end{gathered}$$