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  • If two distinct points lie on the line of intersection of the planes <img alt="\mathrm{-x+2y-z=

If two distinct points \mathrm{Q,R} lie on the line of intersection of the planes \mathrm{-x+2y-z= 0\; and\; 3x-5y+2z= 0\; and\; PQ= PR= \sqrt{18}} where the point \mathrm{P} is \mathrm{\left ( 1,-2,3 \right ),} then the area of the triangle \mathrm{PQR} is equla to

Option: 1

\frac{2}{3}\sqrt{38}


Option: 2

\frac{4}{3}\sqrt{38}


Option: 3

\frac{8}{3}\sqrt{38}


Option: 4

\sqrt{\frac{152}{3}}


Answers (1)

best_answer

\mathrm{Let\: z= k}

\mathrm{\begin{gathered} -x+2y= k\: ---(1) \\ 3x-5y= -2k\: ---(2)\\ \\ \end{gathered}}
\mathrm{\frac{(1) \times 3 \Rightarrow-3 x+6 y=3 k\, --- (3)}{(2) +(3) \Rightarrow y=k ; \quad x=k}}

Also both the lines pass through \mathrm{(0,0,0)} So theie line of intersection is
\mathrm{\frac{x}{1}=\frac{y}{1}=\frac{z}{1}=k.}

Let the Foot of Perpendicular from \mathrm{P} to Line \mathrm{QR} is

\mathrm{S(k, k, k) . \quad P(1,-2,3)}

\mathrm{\overrightarrow{P S}=(k-1, k+2, k-3) . \quad \overrightarrow{P S} \cdot }\vec{b}=0
\mathrm{\Rightarrow k-1+k+2+k-3=0 \Rightarrow k=2 / 3 .}

\mathrm{S\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right), \quad|\overrightarrow{P S}| =\sqrt{\frac{1}{9}+\frac{64}{9}+\frac{49}{9}}=\frac{\sqrt{114}}{3} .}
\mathrm{\text { RS }=Q S =\sqrt{18-\frac{114}{9}}=\sqrt{\frac{48}{9}}=\frac{4}{\sqrt{3}}.}

\mathrm{Area\, of\: \Delta PQR= 2\times\frac{1}{2}\times RS \times PS= \frac{\sqrt{114}}{3}\times\frac{4}{\sqrt{3}}= \frac{4}{3}\sqrt{38}}
Option (B)



 

Posted by

himanshu.meshram

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