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Question: Find the images of the points \(\left( 5,-3,1 \right)\) in the plane \(2x-2y-3z=10\) ....

Find the images of the points (5,−3,1)\left( 5,-3,1 \right) in the plane 2x−2y−3z=102x-2y-3z=10 .

Explanation

Solution

To solve this problem we will assume the image of the given point as P(x1,y1,z1)P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and then we will use the concept of midpoint of the line joining P and the given point Q. We first find the co-ordinates of the point on the line (joining P and the given point Q) which also lies on the plane. Upon finding that point we use the concept of midpoints to determine the co-ordinates of the image point.

Complete step by step answer:
We are given the point say Q as (5,−3,1)\left( 5,-3,1 \right) and a plane 2x−2y−3z=102x-2y-3z=10
To solve the problem, we first assume the point P(x1,y1,z1)P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) as the image of the given point on the given plane.
We know that equation of plane (ax+by+cz=k)\left( ax+by+cz=k \right) when a line PQ where P has co-ordinates (p,q,r)\left( p,q,r \right) and Q has co-ordinates (x,y,z)\left( x,y,z \right) is given by
p−xa=q−yb=r−zc=k\dfrac{p-x}{a}=\dfrac{q-y}{b}=\dfrac{r-z}{c}=k
Substituting the given values in the above equation we get
⇒5−x2=−3−y−2=1−z−3=k\Rightarrow \dfrac{5-x}{2}=\dfrac{-3-y}{-2}=\dfrac{1-z}{-3}=k
Hence, a point O lying on the plane as well as the line will have co-ordinates
⇒x=5−2k\Rightarrow x=5-2k , y=2k−3y=2k-3 , z=3k+1z=3k+1
So, the point is O[(5−2k),(2k−3),(3k+1)]O\left[ \left( 5-2k \right),\left( 2k-3 \right),\left( 3k+1 \right) \right]
Now, we substitute the above co-ordinates in the given equation of the plane as shown below
⇒2(5−2k)−2(2k−3)−3(3k+1)=10\Rightarrow 2\left( 5-2k \right)-2\left( 2k-3 \right)-3\left( 3k+1 \right)=10
Further simplifying as
⇒10−4k−4k+6−9k−3=10 ⇒k=317 \begin{aligned} & \Rightarrow 10-4k-4k+6-9k-3=10 \\\ & \Rightarrow k=\dfrac{3}{17} \\\ \end{aligned}
Hence, co-ordinates of the point O are
⇒[(5−2⋅317),(2⋅317−3),(3⋅317+1)] ⇒[(7917),(−4517),(2617)] \begin{aligned} & \Rightarrow \left[ \left( 5-2\cdot \dfrac{3}{17} \right),\left( 2\cdot \dfrac{3}{17}-3 \right),\left( 3\cdot \dfrac{3}{17}+1 \right) \right] \\\ & \Rightarrow \left[ \left( \dfrac{79}{17} \right),\left( \dfrac{-45}{17} \right),\left( \dfrac{26}{17} \right) \right] \\\ \end{aligned}
Since, the above point O is the midpoint of PQ we can apply the midpoint formula as shown below

& \dfrac{{{x}_{1}}+5}{2}=\dfrac{79}{17} \\\ & \Rightarrow {{x}_{1}}=\dfrac{73}{17} \\\ \end{aligned}$$ $$\begin{aligned} & \dfrac{{{y}_{1}}-3}{2}=-\dfrac{45}{17} \\\ & \Rightarrow {{y}_{1}}=-\dfrac{39}{17} \\\ \end{aligned}$$ $$\begin{aligned} & \dfrac{{{z}_{1}}+1}{2}=\dfrac{26}{17} \\\ & \Rightarrow {{z}_{1}}=\dfrac{35}{17} \\\ \end{aligned}$$ Therefore, the image of the given point is $\left( \dfrac{73}{17},-\dfrac{39}{17},\dfrac{35}{17} \right)$ . ![](https://www.vedantu.com/question-sets/8303c679-dae9-4053-939d-b10b7b66c8df6912681319442348786.png) **Note:** We must be very careful while doing these types of calculations as very little number of mistakes will lead us to an inappropriate solution. Also, we must keep in mind that the points and the equations of the plane must be defined properly to avoid any type of mistakes.