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Answer please! - Three Dimensional Geometry - JEE Main-9

If Q(0,-1,-3) is the image of the point P in the plane 3x-y+4z=2

and R is the point (3,-1,-2), then the area ( in sq. units) of \bigtriangleup PQR  is :

  • Option 1)

    2\sqrt{13}

  • Option 2)

    \frac{\sqrt{91}}{4}

  • Option 3)

    \frac{\sqrt{91}}{2}

  • Option 4)

    \frac{\sqrt{65}}{2}

 
Answers (1)
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V Vakul

Given equation of plane is 3x-y+4z=2 

Q ( 0 , -1, -3 ) is image of P 

Point R is lie on plane 

DQ=\frac{|1-12-2|}{\sqrt{9+1+16}}=\frac{13}{\sqrt{26}}

RQ=\sqrt{(3-0)^{2}+(-1+1)^{2}+(-2+3)^{2}}=\sqrt{10}

\Delta QRD is right angled triangle

So,

RD=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}

Area of \Delta PQR=2\times (\frac{1}{2}\times \sqrt{\frac{13}{2}}\times \sqrt{\frac{7}{2}})=\frac{\sqrt{91}}{2}

correct option (3) 


Option 1)

2\sqrt{13}

Option 2)

\frac{\sqrt{91}}{4}

Option 3)

\frac{\sqrt{91}}{2}

Option 4)

\frac{\sqrt{65}}{2}

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