Question

A line $l$ passing through the origin is perpendicular to the lines
$l_1:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},-\infty <t<\infty$
$l_2:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},-\infty <s<\infty$
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)

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

Answer 1

Answer: (B), (D)

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 $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 2& 1\end{array}\right|=-2\hat{i}+3\hat{j}-2\hat{k}$
$l:\frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=k_1,k_2$
Now, $A\left(-2k_1,3k_1,-2k_1\right)$ and $B\left(-2k_2,3k_2,-2k_2\right)$.
Since, $A$ lies on $l_1$.
$\therefore \left(-2k_1\right)\hat{i}+\left(3k_1\right)\hat{j}-\left(2.k_1\right)\hat{k}=(3+t)\hat{i}-(-1+2t)\hat{j}$
$+(4+2t)\hat{k}$
$\therefore 3+t=-2k_1,-1+2t=3k_1,4+2t=-2k_1$
$\therefore h_1=-1$
$\Rightarrow A(2,-3,2)$
Let any point on $l_2(3+2s,3+2s,2+s1$
$\sqrt{(2-3-2s{)}^2+(-3-3-2s{)}^2+(2-2-s{)}^2}=\sqrt{17}$
$\Rightarrow 9s^2+28s+37=17$
$\Rightarrow 9s^2+28s+20=0$
$\Rightarrow 9s^2+18s+10s+20=0$
$\Rightarrow (9s+10)(s+2)=0$
$\therefore s=-2,\frac{-10}{9}$
Hence, $(-1,-1,0)$ and $\left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right)$ are required points.