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  • Let be the plane passing through the intersectio of th planes  <img alt="\mathrm{\vec{r} \cdot(\

Let \mathrm{P} be the plane passing through the intersectio of th planes  \mathrm{\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})=5 \text { and } \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=3,} and the point \mathrm{ (2,1,-2)}. Let the position vectors of the points \mathrm{ X \: and \: Y} be \mathrm{ \hat{i}-2 \hat{j}+4 \hat{k} \text { and } 5 \hat{i}-\hat{j}+2 \hat{k}} respectively. Then the points 

Option: 1

\mathrm{X\: and\: X+Y} are on the same side of \mathrm{P}


Option: 2

\mathrm{Y\: and\: Y-X} are on the opposite side of \mathrm{P}


Option: 3

\mathrm{X\: and\: Y} are on the opposite side of \mathrm{P}


Option: 4

\mathrm{X+Y\: and\: X-Y} are on the same side of \mathrm{P}


Answers (1)

best_answer

\mathrm{P_{1}:\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})=5}\\

\mathrm{x+3 y-z=5}\\

\mathrm{P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=3}\\

\mathrm{2 x-y+z=3}

Equation of a plane passing through the intersection of \mathrm{P_{1}\: and\: P_{2}}

\mathrm{P : \quad P_{1}+k P_{2}=0}\\

\mathrm{P:(x+3 y-z-5)+k(2 x-y+z-3)=0}\\

\mathrm{\because P}\\ is passing through \mathrm{(2,1 ;-2)}\\

\mathrm{\therefore P:(2+3+2-5)+k(4-1-2-3)=0}

\mathrm{\because \quad 2-2 k=0 \quad k=1} \\

\mathrm{\quad 3 x+2 y-8=0} \\

\mathrm {X=(1,-2,4)} \\\mathrm {X=(1,-2,4)} \\

\mathrm{Y=(5,-1,2)}

\mathrm{\therefore } \mathrm{X\: and \: Y } are on opposite sides of \mathrm{P}

Hence the correct answer is option 3.

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