Question

Two vectors ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ each of magnitude are inclined to each other such that their resultant is to $\sqrt{3}F$ then the resultant of ${\overrightarrow{F}}_1$ and $-{\overrightarrow{F}}_2$ is

• A. $F$
• B. $\sqrt{2}F$
• C. $\sqrt{3}F$
• D. 2F

Answer 1

The correct option is A

$F$

Step 1: Given data

First vector is $F_1$.

Second vector is $F_2$.

Step 2: To find

Resultant of the two vectors.

Step 3: Formula used

$\left|\overrightarrow{R}\right|=\sqrt{{\overrightarrow{F}}_1^2+{\overrightarrow{F}}_2^2+2\overrightarrow{F_1{F}_2}cos}\theta$

Where, the resultant of vector is $\left|\overrightarrow{R}\right|$, angle between the vector is $\theta$.

Step 4: Calculate the resultant of the vector

Consider that ${\overrightarrow{F}}_1=F$ and ${\overrightarrow{F}}_2=F$.

$$\begin{gathered} \sqrt{3}F=\sqrt{F^2+F^2+2F\times F\cos \theta } \\ \sqrt{3}F=\sqrt{2F^2\times 2{\cos }^2\frac{\theta }{2}} \\ \sqrt{3}F=2F\cos \frac{\theta }{2} \\ \cos \theta ={60}^{\circ } \end{gathered}$$

For resultant of vector ${\overrightarrow{F}}_1$ and $-{\overrightarrow{F}}_2$.

$\left|R_1\right|=\sqrt{{\overrightarrow{F}}_1^2+\left(-{\overrightarrow{F}}_2^2\right)+2{\overrightarrow{F}}_1\left(-{\overrightarrow{F}}_2\right)\cos \theta }$

Here, the resultant of the given vector is $R_1$.

consider that ${\overrightarrow{F}}_1=F$ and $-{\overrightarrow{F}}_2=-F$

$$\begin{gathered} \left|R_1\right|=\sqrt{F^2+F^2+2F\times \left(-F\right)\cos {60}^{\circ }} \\ =\sqrt{2F^2-2F^2\times \frac{1}{2}} \\ =\sqrt{2F^2-F^2} \\ =F \end{gathered}$$

Hence, Option (A) is correct.