Let the original force be $\vec{F}$. This force is resolved into two components, $\vec{P}$ and $\vec{Q}$. Therefore, we can write: $$\vec{F} = \vec{P} + \vec{Q}$$ According to the problem: 1. The magnitude of one component is equal to the magnitude of the force: $$P = F$$ 2. This component is perpendicular to the force: $$\vec{P} \perp \vec{F}$$ We need to find the magnitude and direction of the other component, $\vec{Q}$. ### **1. Magnitude of the other component ($\vec{Q}$):** From the vector addition, we can express $\vec{Q}$ as: $$\vec{Q} = \vec{F} - \vec{P}$$ Since $\vec{P}$ is perpendicular to $\vec{F}$, the angle between them is $90^\circ$. The magnitude of the vector difference $\vec{Q}$ is: $$Q = \sqrt{F^2 + P^2 - 2FP \cos 90^\circ}$$ Since $\cos 90^\circ = 0$ and $P = F$, we substitute these values into the equation: $$ \begin{aligned} Q &= \sqrt{F^2 + F^2 - 0} \\ &= \sqrt{2F^2} \\ &= \sqrt{2}F \end{aligned} $$ Thus, the magnitude of the other component is $\sqrt{2}F$. ### **2. Direction of the other component ($\vec{Q}$):** Let $\theta$ be the angle between the other component $\vec{Q}$ and the original force $\vec{F}$. Using the definition of the dot product: $$\vec{Q} \cdot \vec{F} = QF \cos \theta$$ We can also find $\vec{Q} \cdot \vec{F}$ by substituting $\vec{Q} = \vec{F} - \vec{P}$: $$ \begin{aligned} \vec{Q} \cdot \vec{F} &= (\vec{F} - \vec{P}) \cdot \vec{F} \\ &= \vec{F} \cdot \vec{F} - \vec{P} \cdot \vec{F} \end{aligned} $$ Since $\vec{P}$ is perpendicular to $\vec{F}$, their dot product is zero ($\vec{P} \cdot \vec{F} = 0$). Also, $\vec{F} \cdot \vec{F} = F^2$. Therefore: $$\vec{Q} \cdot \vec{F} = F^2$$ Equating the two expressions for the dot product: $$ \begin{aligned} QF \cos \theta &= F^2 \\ \Rightarrow \cos \theta &= \frac{F^2}{QF} \\ \Rightarrow \cos \theta &= \frac{F}{Q} \end{aligned} $$ Substituting $Q = \sqrt{2}F$: $$ \begin{aligned} \cos \theta &= \frac{F}{\sqrt{2}F} \\ \Rightarrow \cos \theta &= \frac{1}{\sqrt{2}} \\ \Rightarrow \theta &= 45^\circ \end{aligned} $$ Thus, the other component makes an angle of $45^\circ$ with the original force $\vec{F}$ (on the opposite side of the first component $\vec{P}$).