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The line \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1} intersects the curve xy=c^{2},z=0  if c is equal to :

Option: 1

\pm 1


Option: 2

\pm \frac{1}{3}


Option: 3

\pm \sqrt{5}


Option: 4

none of these


Answers (1)

best_answer

 

Intersection of line and plane -

Let the line

\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} plane

a_{1}x+b_{1}y+c_{1}z+d=0 intersect at P

to find P assume general point on line as \left ( x_{1}+\lambda a_{1}y_{1} +\lambda b_{1}z_{1}+\lambda c_{1}\right )

now put it in plane to find \lambda,

a_{1}\left ( x_{1}+\lambda a \right )+b_{1}\left ( y_{1}+\lambda b \right )+c_{1}\left ( z_{1}+\lambda c \right )+d=0

-

 

 

 

Any pt. on line is(3\lambda + 2, 2 \lambda-1, 1 -\lambda )

but it lies on the curve xy = c^{2} \: and\: z = 0

\Rightarrow (3\lambda +2)(2\lambda -1)=c^{2}\: and\: 1-\lambda =0

\Rightarrow (3\lambda +2)(2\lambda -1)=c^{2}\: and\:\lambda =1

\Rightarrow c^{2}=5\; \; \; \Rightarrow\; \; \; c=\pm \sqrt{5}

Posted by

mansi

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