Question

Vector $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$which is directed along an x axis, which to be added to

vector $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$, which has a magnitude of 7.0 m.The sum is a third vector that

is directed along the y axis, with a magnitude that is 3.0 times that of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$. What is that magnitude of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$?

Short answer

Magnitude of$\stackrel{\rightharpoonup }{A}=2.2m$

Step-by-step solution

To understand the concept

Using the vector algebra and simple mathematical operations, we can find the unknown vectors. herethe vector addition is to be used to find the magnitude of vector $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$.

Vector$\stackrel{\rightharpoonup }{A}$can be written as

$\stackrel{\rightharpoonup }{A}=x\hat{i}+y\hat{j}$ (1)

With magnitude

$\left|\stackrel{\rightharpoonup }{A}\right|=\sqrt{x^2+y^2}$ (2)

Given

$$\begin{gathered} \left|\stackrel{\rightharpoonup }{\mathrm{B}}\right|=7.0\mathrm{m}\left(3\right) \\ \left|\stackrel{\rightharpoonup }{\mathrm{B}}+\overrightarrow{\mathrm{A}}\right|=3\times \left|\overrightarrow{\mathrm{A}}\right|\left(4\right) \end{gathered}$$

To find the magnitude of vector

Now,

$$\begin{gathered} \overrightarrow{\mathrm{B}}={\mathrm{B}}_{\mathrm{x}}\hat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{y}}\hat{\mathrm{j}} \\ \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}}=\left({\mathrm{B}}_{\mathrm{x}}\hat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{y}}\hat{\mathrm{j}}\right)+\mathrm{A}\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}}=({\mathrm{B}}_{\mathrm{x}}+\mathrm{A})\hat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\hat{\mathrm{j}} \end{gathered}$$

as the resultant is directed along y axis, so the x component will be zero.

$$\begin{gathered} {\mathrm{B}}_{\mathrm{x}}+\mathrm{A}=0 \\ {\mathrm{B}}_{\mathrm{x}}=-\mathrm{A} \\ {\mathrm{B}}_{\mathrm{y}}=3\mathrm{A} \\ \overrightarrow{\mathrm{B}}={\mathrm{B}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\hat{\mathrm{j}} \end{gathered}$$

Using equation (ii),

$$\begin{gathered} 7=\sqrt{{(-\mathrm{A})}^2+{\left(3\mathrm{A}\right)}^2} \\ 7=\mathrm{A}\times \sqrt{10} \\ \mathrm{A}=\frac{7}{\sqrt{10}} \\ \mathrm{A}=2.2\mathrm{m} \end{gathered}$$