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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  3-D Geometry

The point on the line $\overrightarrow r=(-2\hat i-\hat j+3\hat k)+\lambda(3\hat i+2\hat j+2\hat k)$ which is at a distance of $3\sqrt 2$ units from the point $(1,2,3)$ is ?

$\begin{array}{1 1} (a)\:(\large\frac{17}{56}.\frac{17}{43},\frac{17}{111})\:\qquad\:(b)\:(\large\frac{56}{17}.\frac{43}{17},-\frac{111}{17})\:\qquad\:(c)\:(-2,-1,3)\:\:\qquad\:(d)\:(-\large\frac{53}{17},\frac{43}{17},\frac{111}{17}) \end{array} $

1 Answer

Given line is $\large\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$$=\lambda$
Any point on the line in terms of $\lambda$ is $A(3\lambda-2,2\lambda-1,2\lambda+3)$
Given point $P(1,2,3)$
Given: $AP=3\sqrt 2$
$\Rightarrow\:\sqrt {(3\lambda-3)^2+(2\lambda-3)^2+(2\lambda)^2}=3\sqrt 2$
$\Rightarrow\:17\lambda^2-30\lambda+18=18$
$\Rightarrow\:\lambda=0\:\:or\:\:\large\frac{30}{17}$
$\therefore A(-2,-1,3)$

 

answered Jan 1, 2014 by rvidyagovindarajan_1
 

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