Question

A line $l$ is passing through the origin is perpendicular to the lines
$\begin{align}
  & {{l}_{1}}:\left( 3+t \right)\widehat{i}+\left( -1+2t \right)\widehat{j}+\left( 4+2t \right)\widehat{k},-\infty < t < \infty \\
 & {{l}_{2}}:\left( 3+2s \right)\widehat{i}+\left( 3+2s \right)\widehat{j}+\left( 2+s \right)\widehat{k},-\infty < t < \infty \\
\end{align}$
Then the coordinates of the points on ${{l}_{2}}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and ${{l}_{1}}$ is
A. \[\left( \dfrac{7}{3},\dfrac{7}{3},\dfrac{5}{3} \right)\]
B. \[\left( -1,-1,0 \right)\]
C. \[\left( 1,1,1 \right)\]
D. \[\left( \dfrac{7}{9},\dfrac{7}{9},\dfrac{8}{9} \right)\]

Answer 1

Hint: We first find the simplified form of the lines. Then we find the equation of the line $l$. We assume the intersection point and the required point and then from the formula of distance we find the required possible points.

Complete step by step solution:
We can rewrite the lines
$\begin{align}
  & {{l}_{1}}:\left( 3+t \right)\widehat{i}+\left( -1+2t \right)\widehat{j}+\left( 4+2t \right)\widehat{k},-\infty < t < \infty \\
 & {{l}_{2}}:\left( 3+2s \right)\widehat{i}+\left( 3+2s \right)\widehat{j}+\left( 2+s \right)\widehat{k},-\infty < t < \infty \\
\end{align}$ as
$\begin{align}
  & {{l}_{1}}:\left( 3\widehat{i}-\widehat{j}+4\widehat{k} \right)+t\left( \widehat{i}+2\widehat{j}+2\widehat{k} \right),-\infty < t < \infty \\
 & {{l}_{2}}:\left( 3\widehat{i}+3\widehat{j}+2\widehat{k} \right)+s\left( 2\widehat{i}+2\widehat{j}+\widehat{k} \right),-\infty < t < \infty \\
\end{align}$
The first line goes through the point $\left( 3,-1,4 \right)$ and the direction ratio is $1,2,2$.
The simplified form of the line will be ${{l}_{1}}:\dfrac{x-3}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{2}$.
Similarly, the second line goes through the point $\left( 3,3,2 \right)$ and the direction ratio is $2,2,1$.
The simplified form of the line will be ${{l}_{2}}:\dfrac{x-3}{2}=\dfrac{y-3}{2}=\dfrac{z-2}{1}$.
The line $l$ is passing through the origin $\left( 0,0,0 \right)$ and is perpendicular to the lines ${{l}_{1}}$ and ${{l}_{2}}$.
Let the direction ratios of the line $l$ be $a,b,c$.
If two lines are perpendicular then the sum of multiplication of respective direction ratios of those two lines will be 0.
The lines $l,{{l}_{1}},{{l}_{2}}$ have direction ratios $a,b,c$, $1,2,2$, $2,2,1$ respectively.
This gives $a+2b+2c=0$ and $2a+2b+c=0$.
The simplification of the equation in the form of ratio can give us
\[\dfrac{a}{2\times 1-2\times 2}=\dfrac{-b}{1\times 1-2\times 2}=\dfrac{c}{2\times 1-2\times 2}\] which gives \[\dfrac{a}{2}=\dfrac{b}{-3}=\dfrac{c}{2}\].
The direction ratios of the line $l$ is $2,-3,2$. It is passing through origin
Therefore, the equation is \[\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{2}\].
Let the point of intersection of $l$ and ${{l}_{1}}$ be ${{r}_{1}}=\dfrac{x-3}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{2}$. The point can be written as \[\left( {{r}_{1}}+3,2{{r}_{1}}-1,2{{r}_{1}}+4 \right)\]. This point is on the line $l:\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{2}$.
We get \[\dfrac{{{r}_{1}}+3}{2}=\dfrac{2{{r}_{1}}-1}{-3}=\dfrac{2{{r}_{1}}+4}{2}\].
The solution is \[{{r}_{1}}+3=2{{r}_{1}}+4\Rightarrow {{r}_{1}}=-1\].
The intersection point is \[\left( 2,-3,2 \right)\].
Let the point on ${{l}_{2}}$ be ${{r}_{2}}=\dfrac{x-3}{2}=\dfrac{y-3}{2}=\dfrac{z-2}{1}$. The point can be written as \[\left( 2{{r}_{2}}+3,2{{r}_{2}}+3,{{r}_{2}}+2 \right)\]. Now we find distance from \[\left( 2,-3,2 \right)\] which is equal to $\sqrt{17}$.
We get \[\sqrt{{{\left( 2{{r}_{2}}+3-2 \right)}^{2}}+{{\left( 2{{r}_{2}}+3+3 \right)}^{2}}+{{\left( {{r}_{2}}+2-2 \right)}^{2}}}=\sqrt{17}\].
The simplification gives
\[\begin{align}
  & \sqrt{{{\left( 2{{r}_{2}}+3-2 \right)}^{2}}+{{\left( 2{{r}_{2}}+3+3 \right)}^{2}}+{{\left( {{r}_{2}}+2-2 \right)}^{2}}}=\sqrt{17} \\
 & \Rightarrow 9{{r}_{2}}^{2}+28{{r}_{2}}+20=0 \\
 & \Rightarrow {{r}_{2}}=-2,-\dfrac{10}{9} \\
\end{align}\]
Putting value, we get \[\left( 2{{r}_{2}}+3,2{{r}_{2}}+3,{{r}_{2}}+2 \right)=\left( -1,-1,0 \right),\left( \dfrac{7}{9},\dfrac{7}{9},\dfrac{8}{9} \right)\].
The correct options are B and D.

Note: We need to be careful about the points we are taking in finding the intersection as the intersection point satisfies both the lines. We can also form the directions ratios directly in the formula instead of assuming the point to calculate the final solution.