Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12. Then the magnitudes of the forces are

The correct option is 13 N, 5N.

Let the magnitudes of the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force and $F_2$ is the larger force.

We are given the following conditions:

1. The sum of the magnitudes of the two forces is 18 N:
$F_1+F_2=18$  — (Equation 1)

2. The magnitude of the resultant force, $R$, is 12 N:
$R=12$

3. The resultant force vector $\overrightarrow{R}$ is perpendicular to the smaller force vector $\overrightarrow{F_1}$.

Since the resultant force $\overrightarrow{R}$ is the vector sum of $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$, and it is perpendicular to $\overrightarrow{F_1}$, we can form a right-angled triangle using vector addition. In this right-angled triangle:
• The base represents the smaller force $F_1$
• The perpendicular represents the resultant force $R$
• The hypotenuse represents the larger force $F_2$

By applying Pythagoras' theorem to this triangle, we have:
${F_2}^2={F_1}^2+R^2$

Rearranging the equation to solve for the difference of the squares of the forces:
${F_2}^2-{F_1}^2=R^2$

Substitute the given value of $R=12$:
${F_2}^2-{F_1}^2={12}^2$

${F_2}^2-{F_1}^2=144$

Using the algebraic identity $a^2-b^2=(a-b)(a+b)$, we can rewrite this as:
$(F_2-F_1)(F_2+F_1)=144$

Substitute the value of $F_2+F_1=18$ from Equation 1:
$(F_2-F_1)\left(18\right)=144$

$F_2-F_1=\frac{144}{18}$

$F_2-F_1=8$  — (Equation 2)

Now, we solve the system of linear equations (Equation 1 and Equation 2):
Add Equation 1 and Equation 2:
$(F_2+F_1)+(F_2-F_1)=18+8$

$2F_2=26$

$F_2=13\text{N}$

Substitute $F_2=13$ into Equation 1 to find $F_1$:
$13+F_1=18$

$F_1=18-13=5\text{N}$

Thus, the magnitudes of the forces are 13 N and 5 N.