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  4. A line l passing through the origin is perpendicular to the lines l1:(3+t) hat i +(-1+2 t) hat j +(4+2 t) hat k ,-∞ < t < ∞ l2:(3+2 s) hat i +(3+2 s) hat j +(2+s) hat k ,-∞ < s < ∞ Then, the coordinate(s) of the point(s) on l2 at a distance of √17 from the point of intersection of l and l1 is (are)

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Q. A line $l$ passing through the origin is perpendicular to the lines
$l_{1}:(3+t) \hat{ i }+(-1+2 t) \hat{ j }+(4+2 t) \hat{ k },-\infty < t < \infty$
$l_{2}:(3+2 s) \hat{ i }+(3+2 s) \hat{ j }+(2+s) \hat{ k },-\infty < s < \infty$
Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_{1}$ is (are)

JEE AdvancedJEE Advanced 2013Three Dimensional Geometry

Solution:

PLAN Equation of straight line is
$l: \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
Since, $l$ is perpendicular to $I_{1}$ and $I_{2}$.
So, its DR's are cross-product of $l_{1}$ and $I_{2}$.
Now, to find a point on $I_{2}$ whose distance is given, assume a point and find its distance to obtain point.
Let $l: \frac{x-0}{a}=\frac{y-0}{b}=\frac{z-0}{c}$
which is perpendicular to
$ l_{1}:(3 \hat{ i }-\hat{ j }+4 \hat{ k })+t(\hat{ i }+2 \hat{ j }+2 \hat{ k })$
$ l_{2}:(3 \hat{ i }+3 \hat{ j }+2 \hat{ k })+s(2 \hat{ i }+2 \hat{ j }+\hat{ k })$
$\therefore $ DR's of $ l$ is $\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\1 & 2 & 2 \\2 & 2 & 1\end{vmatrix}=-2 \hat{ i }+3 \hat{ j }-2 \hat{ k }$
$l: \frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=k_{1}, k_{2}$
Now, $A\left(-2 k_{1}, 3 k_{1},-2 k_{1}\right)$ and $B\left(-2 k_{2}, 3 k_{2},-2 k_{2}\right)$.
Since, $A$ lies on $l_{1}$.
$\therefore\left(-2 k_{1}\right) \hat{ i }+\left(3 k_{1}\right) \hat{ j }-\left(2 . k_{1}\right) \hat{ k }=(3+t) \hat{ i }-(-1+2 t) \hat{ j }$
$+(4+2 t) \hat{ k }$
$\therefore 3+t=-2 k_{1},-1+2 t=3 k_{1}, 4+2 t=-2 k_{1}$
$\therefore h_{1}=-1$
$\Rightarrow A(2,-3,2)$
Let any point on $l_{2}(3+2 s, 3+2 s, 2+s 1$
$\sqrt{(2-3-2 s)^{2}+(-3-3-2 s)^{2}+(2-2-s)^{2}}=\sqrt{17}$
$\Rightarrow 9 s^{2}+28 s+37=17$
$\Rightarrow 9 s^{2}+28 s+20=0$
$\Rightarrow 9 s^{2}+18 s+10 s+20 =0 $
$\Rightarrow (9 s+10)(s+2) =0 $
$\therefore s =-2, \frac{-10}{9}$
Hence, $(-1,-1,0)$ and $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$ are required points.