Solve the problem.
-Find a field $\mathrm { G } = \mathrm { P } ( \mathrm { x } , \mathrm { y } ) \mathbf { i } + \mathrm { Q } ( \mathrm { x } , \mathrm { y } ) \mathbf { j }$ in the $x y - p l a n e$ with the property that at any point $( a , b ) \neq ( 0,0 ) , \mathrm { G }$ is a unit vector pointing away from the origin.
A) $- \frac { x i + y j } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$
B) $x i + y j$
C) $\frac { x i + y j } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$
D) $\frac { x i - y j } { \sqrt { x ^ { 2 } - y ^ { 2 } } }$

Question

Quizplus · Accepted Answer

Let $\mathrm{G} = \mathrm{P}(x,y)\mathbf{i} + \mathrm{Q}(x,y)\mathbf{j}$ be a unit vector pointing away from the origin. Then, we have
\begin{align*}
\mathrm{G}\cdot\mathrm{G} &= \left(\mathrm{P}(x,y)\mathbf{i} + \mathrm{Q}(x,y)\mathbf{j}\right)\cdot\left(\mathrm{P}(x,y)\mathbf{i} + \mathrm{Q}(x,y)\mathbf{j}\right)\
&= \mathrm{P}^2(x,y) + \mathrm{Q}^2(x,y)\
&= 1,
\end{align*}
since $\mathrm{G}$ is a unit vector. Thus, we have
$$\mathrm{P}(x,y)^2 + \mathrm{Q}(x,y)^2 = 1$$
for all $(x,y)\neq (0,0)$.
Given the choices, we can try $\mathrm{G} = \frac{x\mathbf{i} + y\mathbf{j}}{\sqrt{x^2+y^2}}$. Then, we have
\begin{align*}
\mathrm{G}\cdot\mathrm{G} &= \left(\frac{x\mathbf{i}+y\mathbf{j}}{\sqrt{x^2+y^2}}\right)\cdot\left(\frac{x\mathbf{i}+y\mathbf{j}}{\sqrt{x^2+y^2}}\right)\
&= \frac{x^2+y^2}{x^2+y^2}\
&= 1,
\end{align*}
as desired. Therefore, the best choice is C.

Solve the Problem G=P(x,y)i+Q(x,y)j\mathrm { G } = \mathrm { P } ( \mathrm { x } , \mathrm { y } ) \mathbf { i } + \mathrm { Q } ( \mathrm { x } , \mathrm { y } ) \mathbf { j }G=P(x,y)i+Q(x,y)j

Solve the Problem $\mathrm { G } = \mathrm { P } ( \mathrm { x } , \mathrm { y } ) \mathbf { i } + \mathrm { Q } ( \mathrm { x } , \mathrm { y } ) \mathbf { j }$