Question

If $\stackrel{\rightharpoonup }{v}=2\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j}$and $\stackrel{\rightharpoonup }{w}=x\stackrel{\rightharpoonup }{i}+3\stackrel{\rightharpoonup }{j}$, find all numbers x for which $\left|\stackrel{\rightharpoonup }{v}+\stackrel{\rightharpoonup }{w}\right|=5$.

Short answer

There will be two such values of x which are$\sqrt{21}-2and-\sqrt{21}-2$

Step-by-step solution

Step. 1 Given

$\stackrel{\rightharpoonup }{v}=2\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j}$and$\stackrel{\rightharpoonup }{w}=x\stackrel{\rightharpoonup }{i}+3\stackrel{\rightharpoonup }{j}$

Step. 2 Magnitude of a vector

Firstly, we need to find the vector $\stackrel{\rightharpoonup }{v}+\stackrel{\rightharpoonup }{w}$,

so,

$$\begin{gathered} \stackrel{\rightharpoonup }{v}+\stackrel{\rightharpoonup }{w}=(2\stackrel{\rightharpoonup }{i}-\stackrel{\rightharpoonup }{j})+(x\stackrel{\rightharpoonup }{i}+3\stackrel{\rightharpoonup }{j}) \\ \stackrel{\rightharpoonup }{v}+\stackrel{\rightharpoonup }{w}=(2\stackrel{\rightharpoonup }{i}+x\stackrel{\rightharpoonup }{i})+(-\stackrel{\rightharpoonup }{j}+3\stackrel{\rightharpoonup }{j}) \\ \stackrel{\rightharpoonup }{v}+\stackrel{\rightharpoonup }{w}=(x+2)\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j} \end{gathered}$$

we get the vector whose magnitude is given i.e., 5

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

Step. 3 Final solution

Now we have one equation in one variable. Solve it for x,

$\sqrt{{(x+2)}^2+4}=5$,

Squaring both sides we get,

$$\begin{gathered} {(x+2)}^2+4=25 \\ {(x+2)}^2=21 \end{gathered}$$,

Taking square root both sides we get,

$$\begin{gathered} (x+2)=\pm \sqrt{21}, \\ x=\pm \sqrt{21}-2, \end{gathered}$$

So,

$x=\sqrt{21}-2orx=-\sqrt{21}-2$