Question

Let $P$ be a point in the first octant , whose image $Q$ is in the plane (that is the line segment $PQ$ is perpendicular to the plane $x+y=3$ and the midpoint of $PQ$ lies on the plane $x+y=3$) lies on the $z$-axis. Let the distance of $P$ from $x$-axis be 5. If $R$ is the image of $P$in the $xy$-plane then what is the length of $PR$ \[\]

Answer 1

Hint: Use the equation perpendicular line containing two points that are images of each other to find out the coordinate $P$. Then use the fact that two points that are image of each other are equidistant from the reflecting plane.\[\]

Complete step-by-step answer:

We can see from the image that the point $P$is reflected onto $Q$ by the plane $x+y=3$. If $P$ is an image of $Q$ then $Q$ is also an image of $P$. Let us denote the coordinates as $P\left( {{x}_{p}},{{y}_{p}},{{z}_{p}} \right)$ and $Q\left( {{x}_{q}},{{y}_{q}},{{z}_{q}} \right)$. We know that the perpendicular line containing $P$ and $Q$ where $P$ is the image of $Q$ with respect to the plane $ax+by+cz+d=0$ is given by
\[\dfrac{{{x}_{p}}-{{x}_{q}}}{a}=\dfrac{{{y}_{p}}-{{y}_{q}}}{b}=\dfrac{{{z}_{p}}-{{z}_{q}}}{c}=-2\left( \dfrac{a{{x}_{q}}+b{{y}_{q}}+c{{z}_{q}}+d}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right)\]

As given in the body of the solution that the point lies $Q$ on the $z$-axis. So ${{x}_{p}}=0$ and ${{y}_{p}}=0$. The given equation of the plane is $x+y=3$. So $a=1,b=1,c=0,d=-3$. Putting these values in above equation
\[\begin{align}
  & \dfrac{{{x}_{p}}-0}{1}=\dfrac{{{y}_{p}}-0}{1}=\dfrac{{{z}_{p}}-{{z}_{q}}}{0}=-2\left( \dfrac{1\cdot 0+1\cdot 0+0{{z}_{q}}-3}{{{1}^{2}}+{{1}^{2}}+{{0}^{2}}} \right) \\
 & \Rightarrow {{x}_{p}}={{y}_{p}}=3,{{z}_{p}}={{z}_{q}} \\
\end{align}\]

So the coordinates $P$ are obtained as $\left( 3,3,{{z}_{q}} \right)$. It is also given that the point $P$ is at distance 5 from $x$-axis. The distance of any point is given by the formula $d=\sqrt{{{y}^{2}}+{{z}^{2}}}$. So ,
\[\begin{align}
  & 5=\sqrt{{{3}^{2}}+{{z}_{q}}^{2}} \\
 & \Rightarrow {{z}_{q}}=\sqrt{{{5}^{2}}-{{3}^{2}}}=4 \\
\end{align}\]
We have rejected the negative result because ${{z}_{q}}$ is a distance. Now the coordinates $P$ are obtained as $\left( 3,3,4 \right).$ We also know that the distance of any point from the $xy$-plane is equal to the $z$ coordinate of the point. So the distance $P\left( 3,3,4 \right)$ from the $xy$-plane is 4. \[\]
It is given that $R$ is the image of $P$ in $xy$-plane. Then R and $P$ are equidistant from $xy$-plane. Hence the length $PR=2\times 4=8$.\[\]

Note: The question tests the application distance formulas and reflection points. It is good to remember these formulas for example the point$\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ to plane $ax+by+cz+d=0$distance formula \[r=\left| \dfrac{ax+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|\] and the line perpendicular to the reflecting plane $\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}=-2\left( \dfrac{a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right)$ for shorter calculation.