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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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