The vertices B and C of a \Delta ABC   lie on the line , \frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}   such that BC=5  units. Then the area ( in sq. units ) of this triangle , given that the point A(1,-1,2), is :

  • Option 1)

    5\sqrt{17}

  • Option 2)

    2\sqrt{34}

  • Option 3)

    6

  • Option 4)

    \sqrt{34}

 

Answers (1)

Area = \frac{1}{2} BC \times AD

let\:\:D=(3r-2,1,4r)\:\:on\:\:BC

 

DR's\:\:of \:\:AD=(3r-3,\:\:1+1,4r-2)

Since Ad is perpendicular to BC

\\(3r-3).3+2(0)+4.(4r-2)=0\\\\\\\:9r-9+0+16r-8=0

\\25r-17=0\\\\\\\:r=\frac{17}{25}

\\AD=\frac{2}{5}\cdot \sqrt{34}\\\\\\\:Area=\frac{1}{2}\cdot\frac{2}{5}\sqrt{34}\times 5=\sqrt{34}\:\:sq\:\:\:unit

 

 

 


Option 1)

5\sqrt{17}

Option 2)

2\sqrt{34}

Option 3)

6

Option 4)

\sqrt{34}

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