Question

The equation of the line of shortest distance between the lines $${\displaystyle \frac{x+4}{4}}={\displaystyle \frac{y-2}{-2}}={\displaystyle \frac{z-3}{0}}$$ and $${\displaystyle \frac{x-5}{5}}={\displaystyle \frac{y-3}{3}}={\displaystyle \frac{z}{0}}$$, is
$$\begin{array}{rl}& (A){\displaystyle \frac{x+4}{0}}={\displaystyle \frac{y-2}{0}}={\displaystyle \frac{z-3}{1}}\\ & (B){\displaystyle \frac{x-5}{0}}={\displaystyle \frac{y-3}{0}}={\displaystyle \frac{z}{1}}\\ & (C){\displaystyle \frac{x}{0}}={\displaystyle \frac{y}{0}}={\displaystyle \frac{z-3}{1}}\\ & (D)\text{ None of these}\end{array}$$

Answer 1

Hint: We needed to remember that the line of shortest distance between lines $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ is $$L_3:{\displaystyle \frac{x-x_3}{a_3}}={\displaystyle \frac{y-y_3}{b_3}}={\displaystyle \frac{z-z_3}{c_3}}$$ which must be perpendicular and also passes through the both $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$. By this point, we can solve the problem. Now we should equate $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ to a constant $$\lambda$$. In the similar way, we should equate $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ to a constant $$\mu$$. Now we should find the line passing through these two points. Let us assume this line as $$L_3:{\displaystyle \frac{x-x_3}{a_3}}={\displaystyle \frac{y-y_3}{b_3}}={\displaystyle \frac{z-z_3}{c_3}}$$. This line should be perpendicular to both $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ This will give us the $$L_3:{\displaystyle \frac{x-x_3}{a_3}}={\displaystyle \frac{y-y_3}{b_3}}={\displaystyle \frac{z-z_3}{c_3}}$$. Now by using the concept of sum of product of directional ratios of perpendicular will be zero, we can find the values of both $$\lambda$$ and $$\mu$$.

Complete step-by-step solution:
From the given information, let us assume $$L_1:{\displaystyle \frac{x+4}{4}}={\displaystyle \frac{y-2}{-2}}={\displaystyle \frac{z-3}{0}}$$ and $$L_2:{\displaystyle \frac{x-5}{5}}={\displaystyle \frac{y-3}{3}}={\displaystyle \frac{z}{0}}$$.
Let $$L_1:{\displaystyle \frac{x+4}{4}}={\displaystyle \frac{y-2}{-2}}={\displaystyle \frac{z-3}{0}}=\lambda .....(1)$$
From the equation (1) we get
$${\displaystyle \frac{x+4}{4}}=\lambda$$
By using cross multiplication, we get
$$\Rightarrow x=4\lambda -4......(2)$$
In the same way, from equation (1) we get
$${\displaystyle \frac{y-2}{-2}}=\lambda$$
By using cross multiplication, we get
$$\Rightarrow y=-2\lambda +2......(3)$$
In the same way, from equation (3) we get
$${\displaystyle \frac{z-3}{0}}=\lambda$$
By using cross multiplication, we get
$$\Rightarrow z=3......(4)$$
From equation (2), (3) and (4) let us assume a point on $$L_1:{\displaystyle \frac{x+4}{4}}={\displaystyle \frac{y-2}{-2}}={\displaystyle \frac{z-3}{0}}$$is $$A(4\lambda -4,-2\lambda +2,3)$$.
Let $$L_2:{\displaystyle \frac{x-5}{5}}={\displaystyle \frac{y-3}{3}}={\displaystyle \frac{z}{0}}=\mu ....(5)$$
From the equation (5) we get
$${\displaystyle \frac{x-5}{5}}=\mu$$
By using cross multiplication, we get
$$\Rightarrow x=5\mu +5.....(6)$$
In the same way, from equation (5) we get
$${\displaystyle \frac{y-3}{3}}=\mu$$
By using cross multiplication, we get
$$\Rightarrow y=3\mu +3.....(7)$$
In the same way, from equation (5) we get
$${\displaystyle \frac{z}{0}}=\mu$$
By using cross multiplication, we get
$$\Rightarrow z=0......(8)$$
From equation (5), (6) and (7) let us assume a point on $$L_2:{\displaystyle \frac{x-5}{5}}={\displaystyle \frac{y-3}{3}}={\displaystyle \frac{z}{0}}=\mu ....(5)$$ is $$B(5\mu +5,3\mu +3,0)$$.
We know that the equation of line passing through $$A(x_1,y_1,z_1)$$ and $$B(x_2,y_2,z_2)$$ is $${\displaystyle \frac{x-x_1}{x_2-x_1}}={\displaystyle \frac{y-y_1}{y_2-y_1}}={\displaystyle \frac{z-z_1}{z_2-z_1}}$$
Now we can find the equation of line passing through $$A(4\lambda -4,-2\lambda +2,3)$$ and $$B(5\mu +5,3\mu +3,0)$$ is $$L_3:{\displaystyle \frac{x-(4\lambda -4)}{(5\mu +5)-(4\lambda -4)}}={\displaystyle \frac{y-(-2\lambda +2)}{(3\mu +3)-(-2\lambda +2)}}={\displaystyle \frac{z-3}{0-3}}$$
$$\Rightarrow L_3:{\displaystyle \frac{x-4\lambda +4}{(5\mu -4\lambda +9)}}={\displaystyle \frac{y+2\lambda -2}{(2\lambda +3\mu +1)}}={\displaystyle \frac{z-3}{-3}}$$
We know that the line of shortest distance between lines $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ is $$L_3:{\displaystyle \frac{x-x_3}{a_3}}={\displaystyle \frac{y-y_3}{b_3}}={\displaystyle \frac{z-z_3}{c_3}}$$which must be perpendicular to both }$$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$.

From the above condition, we get
$$L_3:{\displaystyle \frac{x-4\lambda +4}{(5\mu -4\lambda +9)}}={\displaystyle \frac{y+2\lambda -2}{(2\lambda +3\mu +1)}}={\displaystyle \frac{z-3}{-3}}$$is perpendicular to both $$L_1:{\displaystyle \frac{x+4}{4}}={\displaystyle \frac{y-2}{-2}}={\displaystyle \frac{z-3}{0}}$$ and $$L_2:{\displaystyle \frac{x-5}{5}}={\displaystyle \frac{y-3}{3}}={\displaystyle \frac{z}{0}}$$
We know that a line $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ is said to be perpendicular to $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ , if $$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$$.
So, Line$$L_3:{\displaystyle \frac{x-4\lambda +4}{(5\mu -4\lambda +9)}}={\displaystyle \frac{y+2\lambda -2}{(2\lambda +3\mu +1)}}={\displaystyle \frac{z-3}{-3}}$$ should be perpendicular to $$L_1:{\displaystyle \frac{x+4}{4}}={\displaystyle \frac{y-2}{-2}}={\displaystyle \frac{z-3}{0}}$$.
We get $$a_1=5\mu -4\lambda +9,b_1=(2\lambda +3\mu +1),c_1=-3$$ and $$a_2=4,b_2=-2,c_2=0$$.
$$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$$
$$\Rightarrow 4(5\mu -4\lambda +9)+(-2)(2\lambda +3\mu +1)+3(0)=0$$
$$\Rightarrow 20\mu -16\lambda +36-4\lambda -6\mu -2=0$$
$$\Rightarrow -20\lambda +14\mu +34=0$$
$$\Rightarrow 20\lambda -14\mu -34=0$$
$$\Rightarrow 10\lambda -7\mu -17=0.....(9)$$
In the similar manner, we know that line$$L_3:{\displaystyle \frac{x-4\lambda +4}{(5\mu -4\lambda +9)}}={\displaystyle \frac{y+2\lambda -2}{(2\lambda +3\mu +1)}}={\displaystyle \frac{z-3}{-3}}$$ should be perpendicular to $$L_2:{\displaystyle \frac{x-5}{5}}={\displaystyle \frac{y-3}{3}}={\displaystyle \frac{z}{0}}$$.
We get $$a_1=5\mu -4\lambda +9,b_1=(2\lambda +3\mu +1),c_1=-3$$ and $$a_2=5,b_2=3,c_2=0$$.
$$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$$
$$\Rightarrow 5(5\mu -4\lambda +9)+(3)(2\lambda +3\mu +1)+3(0)=0$$
$$\Rightarrow 25\mu -20\lambda +45+6\lambda +9\mu +3=0$$
$$\Rightarrow -14\lambda +34\mu +48=0$$
$$\Rightarrow 14\lambda -34\mu -48=0$$
$$\Rightarrow 7\lambda -17\mu -24=0.....(10)$$
We need to find the values of $$\lambda$$ and $$\mu$$, to get the equation of line $$L_3:{\displaystyle \frac{x-4\lambda +4}{(5\mu -4\lambda +9)}}={\displaystyle \frac{y+2\lambda -2}{(2\lambda +3\mu +1)}}={\displaystyle \frac{z-3}{-3}}$$.
By solving equations (9) and (10) we will get the values of $$\lambda$$ and $$\mu$$.
We should multiply equation (9) by 7.
$$70\lambda -49\mu -119=0....(11)$$
We should multiply (10) by 10.
$$70\lambda -170\mu -240=0.....(12)$$
Now we should subtract (11) and (12).
$$\begin{array}{rl}& (70\lambda -49\mu -119)-(70\lambda -170\mu -240)=0\\ & \Rightarrow (170-49)\mu +(240-119)=0\\ & \Rightarrow 121\mu +121=0\\ & \Rightarrow 121\mu =-121\\ & \Rightarrow \mu =-1....(13)\end{array}$$
From equation (13) we get the value of $$\mu$$,
Now we will substitute the value of $$\mu$$in (10).
$$\begin{array}{rl}& 7\lambda -17(-1)-24=0\\ & \Rightarrow 7\lambda +17-24=0\\ & \Rightarrow 7\lambda -7=0\\ & \Rightarrow \lambda =1.......(14)\end{array}$$
From equation (14) we get the value of $$\lambda$$.
Now we will substitute the both values of $$\lambda$$ and $$\mu$$in line equation $$L_3:{\displaystyle \frac{x-4\lambda +4}{(5\mu -4\lambda +9)}}={\displaystyle \frac{y+2\lambda -2}{(2\lambda +3\mu +1)}}={\displaystyle \frac{z-3}{-3}}$$.
$$\Rightarrow L_3:{\displaystyle \frac{x-4(1)+4}{(5(-1)-4(1)+9)}}={\displaystyle \frac{y+2(1)-2}{-(2(1)+3(-1)+1)}}={\displaystyle \frac{z-3}{-3}}$$
$$\Rightarrow L_3:{\displaystyle \frac{x-4+4}{-5-4+9}}={\displaystyle \frac{y+2-2}{(2-3+1)}}={\displaystyle \frac{z-3}{-3}}$$
$$\Rightarrow L_3:{\displaystyle \frac{x-0}{0}}={\displaystyle \frac{y-0}{0}}={\displaystyle \frac{z-3}{-3}}$$
Now we will multiply and divide the equation by (-3)
$$\Rightarrow L_3:{\displaystyle \frac{x-0}{\left({\displaystyle \frac{0}{-3}}\right)}}={\displaystyle \frac{y-0}{\left({\displaystyle \frac{0}{-3}}\right)}}={\displaystyle \frac{z-3}{\left({\displaystyle \frac{-3}{-3}}\right)}}$$
$$\Rightarrow L_3:{\displaystyle \frac{x-0}{0}}={\displaystyle \frac{y-0}{0}}={\displaystyle \frac{z-3}{1}}$$
Hence, option (C) is correct.

Note: The formula to calculate the shortest distance between the lines between $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ is equal to $${\displaystyle \frac{|\begin{array}{rl}& (x_2-x_1\text{) (}{\text{y}}_2-y_1)\text{ (}{\text{z}}_2-z_1)\\ & {\text{a}}_1{\text{b}}_1{\text{c}}_1\\ & {\text{a}}_2{\text{b}}_2{\text{c}}_2\end{array}|}{\sqrt{{(b_1{c}_2-b_2{c}_1)}^2+{(c_1{a}_2-c_2{a}_1)}^2+{(a_1{b}_2-a_2{b}_1)}^2}}}$$ where $$(a_1,b_1,c_1)$$ and $$(a_2,b_2,c_2)$$ are directional ratios of $$L_1:{\displaystyle \frac{x-x_1}{a_1}}={\displaystyle \frac{y-y_1}{b_1}}={\displaystyle \frac{z-z_1}{c_1}}$$ and $$L_2:{\displaystyle \frac{x-x_2}{a_2}}={\displaystyle \frac{y-y_2}{b_2}}={\displaystyle \frac{z-z_2}{c_2}}$$ respectively.