How do I find the modulus of a sum of vectors which is part of a triangle embedded in a quarter a circle?

Question

The problem is as follows:

In the figure from below, calculate the modulus of $\vec{x}+\vec{y}$. $P$ is tangential point. Show the answer in terms of $R$.

The alternatives are as follows:

$\begin{array}{ll} 1.&1R\\ 2.&0.41R\\ 3.&0.59R\\ 4.&1.41R\\ 5.&2.12R\\ \end{array}$

The only thing which I was able to spot here was to establish that

$x=\frac{(R+a)\sqrt{2}}{2}+a$

$y=\frac{(R+a)\sqrt{2}}{2}+a$

But this doesn't seem very convincing to me. How exactly can I use the vector decomposition in this set of vectors?.

Answer 1

Let ${\vec x}$ start from $A$ and ${\vec y}$ start from $B$. Let the right angle be at $O$.

$${\vec x}=\vec {AO}+\vec {OP}$$

$${\vec y}=\vec {BO}+\vec {OP}$$

$$\vec x+\vec y=\vec {AO}+\vec {BO} +2\vec {OP}$$ $$\vec x+\vec y=-\sqrt 2\vec {OP}+2\vec {OP}$$ $$|\vec x+\vec y|=(2-\sqrt 2)R$$ $$|\vec x+\vec y|=0.59R$$

Answer 2

Let $\vec{x} = \vec{XP}$, $\vec{y} = \vec{YP} $ and write the vectors $\vec{X}=(0,R)$, $\vec{Y}=(R,0)$ and $\vec{P}=(\frac R{\sqrt2}, \frac R{\sqrt2})$. Then,

$$\vec{x} = \vec{P} -\vec{X}=(\frac R{\sqrt2}, \frac R{\sqrt2}-R),\>\>\>\>\>\>\> \vec{y} = \vec{P} -\vec{Y}=(\frac R{\sqrt2}-R, \frac R{\sqrt2}) $$

Evaluate

$$\vec{x}+\vec{y} = (\sqrt2 R-R, \sqrt2 R-R)$$

Thus, the modulus is $\sqrt{2(\sqrt2 R-R)^2}=(2-\sqrt2)R=0.59R$.