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Question: The image of the origin in the line \[\dfrac{x+1}{2}=\dfrac{y-2}{3}=\dfrac{z}{\sqrt{3}}\] is a. \[...

The image of the origin in the line x+12=y−23=z3\dfrac{x+1}{2}=\dfrac{y-2}{3}=\dfrac{z}{\sqrt{3}} is
a. (−1,112,32)\left( -1,\dfrac{11}{2},\dfrac{\sqrt{3}}{2} \right)
b. (3,−52,32)\left( 3,\dfrac{-5}{2},\dfrac{\sqrt{3}}{2} \right)
c. (−3,52,−32)\left( -3,\dfrac{5}{2},-\dfrac{\sqrt{3}}{2} \right)
d. (1,−112,32)\left( 1,-\dfrac{11}{2},\dfrac{\sqrt{3}}{2} \right)

Explanation

Solution

From the given question we have been asked to find the image of the origin in a given line. For solving this question we will assume that the image of a point in the line as some variable and we will use the perpendicular and midpoint concept and use the directional ratios concept and further solve the equations using basic mathematical operations like addition, division and find the solution.

Complete step by step solution:
Let us assume that the origin be P(0,0,0)P\left( 0,0,0 \right) and also we will assume that,
⇒x+12=y−23=z3=k\Rightarrow \dfrac{x+1}{2}=\dfrac{y-2}{3}=\dfrac{z}{\sqrt{3}}=k
So any point on this line is given as Q(2k−1,3k+2,3k)Q\left( 2k-1,3k+2,\sqrt{3}k \right)
Let the image of P in the line as M, so PM will be the perpendicular to the given line and Q will be the midpoint of PM.
Now, the directional ratios of PQ are, (2k−1,3k+2,3k)\left( 2k-1,3k+2,\sqrt{3}k \right)
Also PQ is perpendicular to the given line. So, when we multiply with directional cosines it will be equal to zero. So, we get the equation as follows.
⇒2(2k−1)+3(3k+2)+3(3k)=0\Rightarrow 2\left( 2k-1 \right)+3(3k+2)+\sqrt{3}(\sqrt{3}k)=0
Expanding and simplifying the above equation. We get the equation reduced as follows.
⇒4k−2+9k+6+3k=0\Rightarrow 4k-2+9k+6+3k=0
⇒16k+4=0\Rightarrow 16k+4=0
⇒k=−14\Rightarrow k=-\dfrac{1}{4}
Thus Q=(−32,54,−34)Q=\left( \dfrac{-3}{2},\dfrac{5}{4},\dfrac{-\sqrt{3}}{4} \right)
Since Q is the midpoint of PM, therefore coordinate of M is (−32,54,−34)\left( \dfrac{-3}{2},\dfrac{5}{4},\dfrac{-\sqrt{3}}{4} \right)
Therefore the image of the origin in the line x+12=y−23=z3\dfrac{x+1}{2}=\dfrac{y-2}{3}=\dfrac{z}{\sqrt{3}} is Q=(−3,52,−32)Q=\left( -3,\dfrac{5}{2},\dfrac{-\sqrt{3}}{2} \right).

Note: Students must do the calculations very carefully. Students should have good knowledge in 3d-geometry. Students should know the concept of directional ratios to solve the question. Students should not interchange the coordinates while writing it for Q and ultimately the direction ratios for line PQ.