Question

A line $l$ passing through the origin is perpendicular to the lines $ L_1 : (3+t) \widehat {i} +(1-2t) \widehat {j}+(4+2t)\widehat {k}, - \infty < t < \infty$ $ L_2 : (3+2s) \widehat {i} +(3+2s) \widehat {j}+(2+s)\widehat {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)

• A. $\bigg(\frac{7}{3},\frac{7}{3},\frac{5}{3}\bigg)$
• B. $(-1,-1,0)$
• C. $(1, 1, 1)$
• D. $\bigg(\frac{7}{9},\frac{7}{9},\frac{8}{9}\bigg)$

Answer 1

The Correct Option is D

PLAN Equation of straight line is
$\hspace20mm l: \frac{x-x_1}{a} = \frac{y-y_1}{b}=\frac{z-z_1}{c}$
Since, l is perpendicular to $l_1$ and c.
So, its DR's are cross - product of $l_1$ and $l_2$.
Now, to find a point on $l_2$ 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
$\ \ \ L_1 : (3\widehat {j}-\widehat {j} +4\widehat {k}) + t(\widehat {i} +2 \widehat {j}+2\widehat {k})$
$ \ \ \ L_2 : (3\widehat {i}-3\widehat {j} +2\widehat {k}) + s(2\widehat {i} +2 \widehat {j}+\widehat {k})$
$\therefore$ $DR's\ of\ l \ is \begin{vmatrix}
\widehat {i} &\widehat {j} &\widehat {k} \\
1 & 2 & 2 \\
2 & 2 & 1 \\
\end{vmatrix}= -2 \widehat {i}+3\widehat {j} +2\widehat {k}$
$ \hspace30mm l: \frac{x}{-2} = \frac{y}{3}=\frac{z}{-2}=k_1,k_2$

Now, $A(-2k_1,3k_1,-2k_1)$ and $B(-2k_2,3k_2,-2k_2).$
Since, A lies on $l_1$
$\therefore$ $(-2k_1)\widehat {i} + (3k_1)\widehat {j} - (2k_1)\widehat {k}=(3+t)\widehat {i} + (-1+2t) \widehat {j}$
$ \hspace70mm + (4 + 2t) \widehat {k}$
$\Rightarrow \ \ \ \ $$3+t = -2k_ 1,-1 + 2t = 3k_1, 4 + 2t = -2k_1$
$\therefore$ $\ \ \ \ k_ = -1$
$\Rightarrow$ A (2, -3, 2)
Let any point on $ l_2 (3+2s,3+2s,2+s)$
$ \ \ \sqrt{ (2-3-2s)^2 (-3 - 3 +2s)^2+(2-2-s)^2} = \sqrt{17}$
$\Rightarrow$ $ \hspace50mm 9s^2+28s+37 = 17$
$\Rightarrow$ $ \hspace50mm 9s^2+28s+20 = 0$
$\Rightarrow$ $\hspace20mm 9s^2+18s+10s+20=0$
$\Rightarrow$ $\hspace30mm (9s+10) (s+2)=0$
$\therefore$ $\hspace50mm s = -2, \frac{-10}{9}$
Hence, (-1,-1,0) and $\bigg(\frac{7}{9},\frac{7}{9},\frac{8}{9}\bigg)$ are required points.