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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)

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

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

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)

$14$

Posted by

Vakul

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