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Question: A line \(l\) is passing through the origin is perpendicular to the lines \(\begin{aligned} & ...

A line ll is passing through the origin is perpendicular to the lines
l1:(3+t)i^+(−1+2t)j^+(4+2t)k^,−∞<t<∞ l2:(3+2s)i^+(3+2s)j^+(2+s)k^,−∞<t<∞ \begin{aligned} & {{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{aligned}
Then the coordinates of the points on l2{{l}_{2}} at a distance of 17\sqrt{17} from the point of intersection of ll and l1{{l}_{1}} is
A. (73,73,53)\left( \dfrac{7}{3},\dfrac{7}{3},\dfrac{5}{3} \right)
B. (−1,−1,0)\left( -1,-1,0 \right)
C. (1,1,1)\left( 1,1,1 \right)
D. (79,79,89)\left( \dfrac{7}{9},\dfrac{7}{9},\dfrac{8}{9} \right)

Explanation

Solution

We first find the simplified form of the lines. Then we find the equation of the line ll. 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
l1:(3+t)i^+(−1+2t)j^+(4+2t)k^,−∞<t<∞ l2:(3+2s)i^+(3+2s)j^+(2+s)k^,−∞<t<∞ \begin{aligned} & {{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{aligned} as
l1:(3i^−j^+4k^)+t(i^+2j^+2k^),−∞<t<∞ l2:(3i^+3j^+2k^)+s(2i^+2j^+k^),−∞<t<∞ \begin{aligned} & {{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{aligned}
The first line goes through the point (3,−1,4)\left( 3,-1,4 \right) and the direction ratio is 1,2,21,2,2.
The simplified form of the line will be l1:x−31=y+12=z−42{{l}_{1}}:\dfrac{x-3}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{2}.
Similarly, the second line goes through the point (3,3,2)\left( 3,3,2 \right) and the direction ratio is 2,2,12,2,1.
The simplified form of the line will be l2:x−32=y−32=z−21{{l}_{2}}:\dfrac{x-3}{2}=\dfrac{y-3}{2}=\dfrac{z-2}{1}.
The line ll is passing through the origin (0,0,0)\left( 0,0,0 \right) and is perpendicular to the lines l1{{l}_{1}} and l2{{l}_{2}}.
Let the direction ratios of the line ll be a,b,ca,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,l1,l2l,{{l}_{1}},{{l}_{2}} have direction ratios a,b,ca,b,c, 1,2,21,2,2, 2,2,12,2,1 respectively.
This gives a+2b+2c=0a+2b+2c=0 and 2a+2b+c=02a+2b+c=0.
The simplification of the equation in the form of ratio can give us
a2×1−2×2=−b1×1−2×2=c2×1−2×2\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 a2=b−3=c2\dfrac{a}{2}=\dfrac{b}{-3}=\dfrac{c}{2}.
The direction ratios of the line ll is 2,−3,22,-3,2. It is passing through origin
Therefore, the equation is x2=y−3=z2\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{2}.
Let the point of intersection of ll and l1{{l}_{1}} be r1=x−31=y+12=z−42{{r}_{1}}=\dfrac{x-3}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{2}. The point can be written as (r1+3,2r1−1,2r1+4)\left( {{r}_{1}}+3,2{{r}_{1}}-1,2{{r}_{1}}+4 \right). This point is on the line l:x2=y−3=z2l:\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{2}.
We get r1+32=2r1−1−3=2r1+42\dfrac{{{r}_{1}}+3}{2}=\dfrac{2{{r}_{1}}-1}{-3}=\dfrac{2{{r}_{1}}+4}{2}.
The solution is r1+3=2r1+4⇒r1=−1{{r}_{1}}+3=2{{r}_{1}}+4\Rightarrow {{r}_{1}}=-1.
The intersection point is (2,−3,2)\left( 2,-3,2 \right).
Let the point on l2{{l}_{2}} be r2=x−32=y−32=z−21{{r}_{2}}=\dfrac{x-3}{2}=\dfrac{y-3}{2}=\dfrac{z-2}{1}. The point can be written as (2r2+3,2r2+3,r2+2)\left( 2{{r}_{2}}+3,2{{r}_{2}}+3,{{r}_{2}}+2 \right). Now we find distance from (2,−3,2)\left( 2,-3,2 \right) which is equal to 17\sqrt{17}.
We get (2r2+3−2)2+(2r2+3+3)2+(r2+2−2)2=17\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

& \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{aligned}$$ Putting value, we get $$\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.