If the line \frac{x-2}{3}-\frac{y+1}{2}=\frac{z-1}{-1} intersects the plane 2x+3y-z+13=0 at a point P and the plane 3x+y+4z=16 at a point Q, then PQ is equal to : 

 

 

  • Option 1)

    2\sqrt{7}   

  • Option 2)

     \sqrt{14}     

  • Option 3)

     2\sqrt{14}       

  • Option 4)

    14

 

Answers (1)
V Vakul

\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda

x=3\lambda +2,\: y=2\lambda -1,z=-\lambda +1

Intersection with plane 2x+3y-z+13=0

2\left ( 3\lambda +2 \right )+3\left ( 2\lambda -1 \right )-\left ( -\lambda +1 \right )+13=0

\lambda =-1

\therefore P\left ( -1,-3,-2 \right )

Intersection with plane

3x+y+4z=16

3\left ( 3\lambda +2 \right )+\left ( 2\lambda -1 \right )+4\left ( -\lambda +1 \right )=16

\lambda =1

Q=\left ( 5,1,0 \right )

PQ=\sqrt{\left ( -1-5 \right )^{2}+\left ( -3-1 \right )^{2}+\left ( -2-0 \right )^{2}}

         =\sqrt{6^{2}+4^{2}+2^{2}}

        =\sqrt{56}

        =2\sqrt{14}


Option 1)

2\sqrt{7}   

Option 2)

 \sqrt{14}     

Option 3)

 2\sqrt{14}       

Option 4)

14

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