Proving Q R in Vector Mathematics

In vector mathematics, the resultant of two forces P and Q is denoted as R. When Q is doubled, the new resultant becomes perpendicular to P. This article aims to demonstrate through mathematical proof that in such a scenario, the magnitude of the new resultant R is equal to Q. We will use vector properties and trigonometric identities to derive this conclusion.

Understanding the Problem

The problem states that the resultant of vectors P and Q is R. Mathematically, the magnitude of R can be described by:
$$R^2 P^2 Q^2 2PQcosalpha$$
Here, alpha is the angle between P and Q.
Now, the new resultant, when Q is doubled, is perpendicular to P. The condition for two vectors to be perpendicular is that their dot product is zero:

$$vec{P} cdot vec{2Q} 0$$

Deriving the Relationship

Given that the new resultant is perpendicular to P when Q is doubled, let's denote the new resultant as S. Therefore, S is perpendicular to P and we have:

$$cos 90^circ frac{vec{P} cdot vec{2Q}}{|vec{P}| |vec{2Q}|} frac{vec{P} cdot vec{2Q}}{P cdot 2Q} 0$$

This implies:

$$vec{P} cdot vec{2Q} 0$$

Which further implies:

$$vec{P} cdot vec{Q} 0$$

Now, to find the relationship between R and Q, let's square R using the original equation:

$$R^2 P^2 Q^2 2PQcosalpha$$

Since P and 2Q are perpendicular, P and Q must have a specific relationship:

$$vec{P} cdot vec{Q} 0 Rightarrow Q sinalpha -P/2 Rightarrow cosalpha -P/2Q$$

Substitute this value into the equation for R squared:

$$R^2 P^2 Q^2 2PQ left(-frac{P}{2Q}right) P^2 Q^2 - P^2 Q^2$$

Therefore:

$$R^2 Q^2 Rightarrow R |Q|$$

Conclusion

We have shown that when Q is doubled and the new resultant is perpendicular to P, the magnitude of the resultant R is indeed equal to the magnitude of Q. This conclusion is based on vector properties and the condition of perpendicularity.