Animated Solution for Mathematics - Vector Algebra: The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are

Imagine you are standing on a flat, open field with a heavy block. You pull with force $\vec{P}$, and your friend pulls with force $\vec{Q}$. The block moves in a direction determined by the resultant force $\vec{R}$.

The problem states that the resultant force $\vec{R}$ acts at a perfect $90^{\circ }$ angle to your force $\vec{P}$. This geometric constraint is the key that unlocks the entire puzzle.

When we add vectors using the triangle law, we place the tail of $\vec{Q}$ at the head of $\vec{P}$. The vector that connects the start of $\vec{P}$ to the end of $\vec{Q}$ is the resultant $\vec{R}$.

Because $\vec{R}$ is perpendicular to $\vec{P}$, these two vectors form the legs of a right-angled triangle. The vector $\vec{Q}$ acts as the hypotenuse connecting the head of $\vec{P}$ to the head of $\vec{R}$.

With the right-angled triangle identified, we apply the Pythagorean theorem: the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this configuration, the relationship is:

We are given that the resultant $R=12$ N. Substituting this value, our equation becomes:

We now have a system of two equations. First, the sum of the magnitudes is given as:

Second, our geometric derivation provides the difference of squares:

We utilize the algebraic identity $Q^2-P^2=(Q-P)(Q+P)$ to simplify the calculation. Substituting the known sum:

Dividing both sides by $18$, we find:

We now solve the simple linear system:

Adding these two equations yields $2Q=26$, which results in $Q=13$ N. Subtracting the equations yields $2P=10$, resulting in $P=5$ N.

This problem demonstrates that when physics appears complex, the right visualization—in this case, a simple right-angled triangle—can turn a difficult challenge into a straightforward algebraic solution.