Discussion Forum

A line l passing through the origin is perpendicular to the lines

$l_{1}:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},-\infty<t<\infty$

$l_{2}:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},-\infty<t<\infty$

Then, the coordinate (s)  of the point(s)  on l₂ at a distance of   $\sqrt{17}$ from the point of intersection of l and l₁ is (are)

Answer:

Explanation:

Concept involved

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 l₁ and l₂

$\therefore$  its DR's are cross product of l₁  and l₂

Now, to find a point on l₂ 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 

               $\begin{bmatrix}\hat{i} & \hat{j} &\hat{k}\\1 & 2& 2 \\2 &2&1 \end{bmatrix}$

$=-2\hat{i}+3\hat{j}-2 \hat{k}$

$\therefore$    $l:\frac{x}{-2}=\frac{y}{3}=\frac{z}{-2}=k_{1},k_{2}$

 Now, A(-2k₁,3k₁,-2k₁)

 and   B (-2k₂,3k₂,-2k₂)

 since , A lies on l₁

 $\therefore$   $(-2k_{1})\hat{i}+(3k_{1})\hat{j}-2k_{1})\hat{k}$

                    = $(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$

   $\Rightarrow$    $  3+t=-2k_{1},-1+2t=3k_{1}$

   4+2t=-2k₁

 $\therefore$   k₁=-1

$\Rightarrow  $    A(2,-3,2)

Let any point on l₂ (3+2s,3+2s,2+5)

 $\Rightarrow  $            $\sqrt{(2-3+2s)^{2}+(-3-3-2s)^{2}+(2+2+s)^{2}}=\sqrt{17}$

$\Rightarrow  $   $9s^{2}+28s+37=17$

$\Rightarrow  $   $9s^{2}+28s+20=0$

$\Rightarrow  $  $9s^{2}+18s+10s+20=0$

$\therefore$ (9s+10)(s+2) =0

$\Rightarrow  $     s=-2,-10/9

Hence (-1,-1,0) and   $(\frac{7}{9},\frac{7}{9},\frac{8}{9})$ are required points