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Find the ratio in which the point $ P(5,4,-6) $ divides the line segment joining the points $ A(3,2,-4) $ and $ B(9,8,-10) $ . Also, find the harmonic conjugate of $ P $ .

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Hint: In order to solve this question, we have to apply the section formula which is if point $ P(x,y,z) $ divides the line segment joining the points $ A({{x}_{1}},{{y}_{1}},{{z}_{1}}) $ and $ B({{x}_{2}},{{y}_{2}},{{z}_{2}}) $ in the ratio $ m:n $ , then
 $ \therefore x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} $ , $ y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} $ and $ z=\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} $
After applying the section formula, get the ratio in which $ P $ divides $ AB $ . With that ratio we have to find the harmonic conjugate of $ P $ and then get the final answer.

Complete step-by-step answer:
Let’s suppose that point $ P $ divides the line segment $ AB $ in the ratio $ k:1 $ .

In order to find the value of $ k $ we have to apply a section formula.
According to section formula if point $ P(x,y,z) $ divides the line segment joining the points $ A({{x}_{1}},{{y}_{1}},{{z}_{1}}) $ and $ B({{x}_{2}},{{y}_{2}},{{z}_{2}}) $ in the ratio $ k:1 $ , then
 $ \therefore x=\dfrac{k{{x}_{2}}+{{x}_{1}}}{k+1} $ , $ y=\dfrac{k{{y}_{2}}+{{y}_{1}}}{k+1} $ and $ z=\dfrac{k{{z}_{2}}+{{z}_{1}}}{k+1} $
Using the same logic, we have $ P(5,4,-6) $ , $ A(3,2,-4) $ and $ B(9,8,-10) $ , applying the section formula, we get
 $ \Rightarrow 5=\dfrac{9k+3}{k+1} $
 $ \Rightarrow 5k+5=9k+3 $
 $ \Rightarrow 4k=2 $
 $ \Rightarrow k=\dfrac{2}{4}=\dfrac{1}{2} $
Hence, $ P $ divides the line segment $ AB $ in the ratio $ \dfrac{1}{2}:1 $ , or we can say that $ 1:2 $ .
Now, let’s suppose $ Q(a,b,c) $ is the harmonic conjugate of $ P $ .
If $ P $ and $ Q $ are the harmonic conjugate, that means the ratio in which $ P $ divides the line segment $ AB $ internally $ Q $ also have to divide it in that ratio externally.
So, now $ Q $ divides the line segment $ AB $ in the ratio $ 1:2 $ externally.
Externally division means if point $ Q(x,y,z) $ divides the line segment joining the points $ A({{x}_{1}},{{y}_{1}},{{z}_{1}}) $ and $ B({{x}_{2}},{{y}_{2}},{{z}_{2}}) $ externally in the ratio $ m:n $ , then
 $ \therefore x=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n} $ , $ y=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n} $ and $ z=\dfrac{m{{z}_{2}}-n{{z}_{1}}}{m-n} $
Using the same concept, we get
 $ \Rightarrow a=\dfrac{9\times 1-2\times 3}{1-2}=\dfrac{9-6}{-1}=-3 $
 $ \Rightarrow b=\dfrac{8\times 1-2\times 2}{1-2}=\dfrac{8-4}{-1}=-4 $
 $ \Rightarrow c=\dfrac{-10\times 1-2\times -4}{1-2}=\dfrac{-10+8}{-1}=2 $
Hence, $ Q(-3,-4,2) $ .
So, the harmonic conjugate of $ P $ is $ (-3,-4,2) $ .

Note: In this question we just have to understand the difference between external and internal division of a line segment. First we have to apply internal division in order to find the ratio and then apply the external division formula to get the harmonic conjugate. Students often confuse external and internal division so apply the formulas carefully.
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