Solveeit Logo
Login

Question

Question: The equation of the line of shortest distance between the lines \[\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\df...

The equation of the line of shortest distance between the lines x+44=y−2−2=z−30\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\dfrac{z-3}{0} and x−55=y−33=z0\dfrac{x-5}{5}=\dfrac{y-3}{3}=\dfrac{z}{0}, is

& (A)\text{ }\dfrac{x+4}{0}=\dfrac{y-2}{0}=\dfrac{z-3}{1} \\\ & (B)\text{ }\dfrac{x-5}{0}=\dfrac{y-3}{0}=\dfrac{z}{1} \\\ & (C)\text{ }\dfrac{x}{0}=\dfrac{y}{0}=\dfrac{z-3}{1} \\\ & (D)\text{ None of these} \\\ \end{aligned}$$
Explanation

Solution

We needed to remember that the line of shortest distance between lines L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} and L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} is L3:x−x3a3=y−y3b3=z−z3c3{{L}_{3}}:\dfrac{x-{{x}_{3}}}{{{a}_{3}}}=\dfrac{y-{{y}_{3}}}{{{b}_{3}}}=\dfrac{z-{{z}_{3}}}{{{c}_{3}}} which must be perpendicular and also passes through the both L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} and L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}. By this point, we can solve the problem. Now we should equate L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} to a constant λ\lambda . In the similar way, we should equate L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} to a constant μ\mu . Now we should find the line passing through these two points. Let us assume this line as L3:x−x3a3=y−y3b3=z−z3c3{{L}_{3}}:\dfrac{x-{{x}_{3}}}{{{a}_{3}}}=\dfrac{y-{{y}_{3}}}{{{b}_{3}}}=\dfrac{z-{{z}_{3}}}{{{c}_{3}}}. This line should be perpendicular to both L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} and L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} This will give us the L3:x−x3a3=y−y3b3=z−z3c3{{L}_{3}}:\dfrac{x-{{x}_{3}}}{{{a}_{3}}}=\dfrac{y-{{y}_{3}}}{{{b}_{3}}}=\dfrac{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 λ\lambda and μ\mu .

Complete step-by-step solution:
From the given information, let us assume L1:x+44=y−2−2=z−30{{L}_{1}}:\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\dfrac{z-3}{0} and L2:x−55=y−33=z0{{L}_{2}}:\dfrac{x-5}{5}=\dfrac{y-3}{3}=\dfrac{z}{0}.
Let L1:x+44=y−2−2=z−30=λ.....(1){{L}_{1}}:\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\dfrac{z-3}{0}=\lambda .....(1)
From the equation (1) we get
x+44=λ\dfrac{x+4}{4}=\lambda
By using cross multiplication, we get
⇒x=4λ−4......(2)\Rightarrow x=4\lambda -4......(2)
In the same way, from equation (1) we get
y−2−2=λ\dfrac{y-2}{-2}=\lambda
By using cross multiplication, we get
⇒y=−2λ+2......(3)\Rightarrow y=-2\lambda +2......(3)
In the same way, from equation (3) we get
z−30=λ\dfrac{z-3}{0}=\lambda
By using cross multiplication, we get
⇒z=3......(4)\Rightarrow z=3......(4)
From equation (2), (3) and (4) let us assume a point on L1:x+44=y−2−2=z−30{{L}_{1}}:\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\dfrac{z-3}{0}is A(4λ−4,−2λ+2,3)A\left( 4\lambda -4,-2\lambda +2,3 \right).
Let L2:x−55=y−33=z0=μ....(5){{L}_{2}}:\dfrac{x-5}{5}=\dfrac{y-3}{3}=\dfrac{z}{0}=\mu ....(5)
From the equation (5) we get
x−55=μ\dfrac{x-5}{5}=\mu
By using cross multiplication, we get
⇒x=5μ+5.....(6)\Rightarrow x=5\mu +5.....(6)
In the same way, from equation (5) we get
y−33=μ\dfrac{y-3}{3}=\mu
By using cross multiplication, we get
⇒y=3μ+3.....(7)\Rightarrow y=3\mu +3.....(7)
In the same way, from equation (5) we get
z0=μ\dfrac{z}{0}=\mu
By using cross multiplication, we get
⇒z=0......(8)\Rightarrow z=0......(8)
From equation (5), (6) and (7) let us assume a point on L2:x−55=y−33=z0=μ....(5){{L}_{2}}:\dfrac{x-5}{5}=\dfrac{y-3}{3}=\dfrac{z}{0}=\mu ....(5) is B(5μ+5,3μ+3,0)B\left( 5\mu +5,3\mu +3,0 \right).
We know that the equation of line passing through A(x1,y1,z1)A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and B(x2,y2,z2)B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}
Now we can find the equation of line passing through A(4λ−4,−2λ+2,3)A\left( 4\lambda -4,-2\lambda +2,3 \right) and B(5μ+5,3μ+3,0)B\left( 5\mu +5,3\mu +3,0 \right) is L3:x−(4λ−4)(5μ+5)−(4λ−4)=y−(−2λ+2)(3μ+3)−(−2λ+2)=z−30−3{{L}_{3}}:\dfrac{x-(4\lambda -4)}{\left( 5\mu +5 \right)-\left( 4\lambda -4 \right)}=\dfrac{y-\left( -2\lambda +2 \right)}{\left( 3\mu +3 \right)-\left( -2\lambda +2 \right)}=\dfrac{z-3}{0-3}
⇒L3:x−4λ+4(5μ−4λ+9)=y+2λ−2(2λ+3μ+1)=z−3−3\Rightarrow {{L}_{3}}:\dfrac{x-4\lambda +4}{\left( 5\mu -4\lambda +9 \right)}=\dfrac{y+2\lambda -2}{\left( 2\lambda +3\mu +1 \right)}=\dfrac{z-3}{-3}
We know that the line of shortest distance between lines L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} and L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} is L3:x−x3a3=y−y3b3=z−z3c3{{L}_{3}}:\dfrac{x-{{x}_{3}}}{{{a}_{3}}}=\dfrac{y-{{y}_{3}}}{{{b}_{3}}}=\dfrac{z-{{z}_{3}}}{{{c}_{3}}}which must be perpendicular to both }L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} and L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}.

From the above condition, we get
L3:x−4λ+4(5μ−4λ+9)=y+2λ−2(2λ+3μ+1)=z−3−3{{L}_{3}}:\dfrac{x-4\lambda +4}{\left( 5\mu -4\lambda +9 \right)}=\dfrac{y+2\lambda -2}{\left( 2\lambda +3\mu +1 \right)}=\dfrac{z-3}{-3}is perpendicular to both L1:x+44=y−2−2=z−30{{L}_{1}}:\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\dfrac{z-3}{0} and L2:x−55=y−33=z0{{L}_{2}}:\dfrac{x-5}{5}=\dfrac{y-3}{3}=\dfrac{z}{0}
We know that a line L1:x−x1a1=y−y1b1=z−z1c1{{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} is said to be perpendicular to L2:x−x2a2=y−y2b2=z−z2c2{{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} , if a1a2+b1b2+c1c2=0{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0.
So, LineL3:x−4λ+4(5μ−4λ+9)=y+2λ−2(2λ+3μ+1)=z−3−3{{L}_{3}}:\dfrac{x-4\lambda +4}{\left( 5\mu -4\lambda +9 \right)}=\dfrac{y+2\lambda -2}{\left( 2\lambda +3\mu +1 \right)}=\dfrac{z-3}{-3} should be perpendicular to L1:x+44=y−2−2=z−30{{L}_{1}}:\dfrac{x+4}{4}=\dfrac{y-2}{-2}=\dfrac{z-3}{0}.
We get a1=5μ−4λ+9,b1=(2λ+3μ+1),c1=−3{{a}_{1}}=5\mu -4\lambda +9,{{b}_{1}}=\left( 2\lambda +3\mu +1 \right),{{c}_{1}}=-3 and a2=4,b2=−2,c2=0{{a}_{2}}=4,{{b}_{2}}=-2,{{c}_{2}}=0.
a1a2+b1b2+c1c2=0{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0
⇒4(5μ−4λ+9)+(−2)(2λ+3μ+1)+3(0)=0\Rightarrow 4\left( 5\mu -4\lambda +9 \right)+(-2)\left( 2\lambda +3\mu +1 \right)+3(0)=0
⇒20μ−16λ+36−4λ−6μ−2=0\Rightarrow 20\mu -16\lambda +36-4\lambda -6\mu -2=0
⇒−20λ+14μ+34=0\Rightarrow -20\lambda +14\mu +34=0
⇒20λ−14μ−34=0\Rightarrow 20\lambda -14\mu -34=0
⇒10λ−7μ−17=0.....(9)\Rightarrow 10\lambda -7\mu -17=0.....(9)
In the similar manner, we know that lineL3:x−4λ+4(5μ−4λ+9)=y+2λ−2(2λ+3μ+1)=z−3−3{{L}_{3}}:\dfrac{x-4\lambda +4}{\left( 5\mu -4\lambda +9 \right)}=\dfrac{y+2\lambda -2}{\left( 2\lambda +3\mu +1 \right)}=\dfrac{z-3}{-3} should be perpendicular to L2:x−55=y−33=z0{{L}_{2}}:\dfrac{x-5}{5}=\dfrac{y-3}{3}=\dfrac{z}{0}.
We get a1=5μ−4λ+9,b1=(2λ+3μ+1),c1=−3{{a}_{1}}=5\mu -4\lambda +9,{{b}_{1}}=\left( 2\lambda +3\mu +1 \right),{{c}_{1}}=-3 and a2=5,b2=3,c2=0{{a}_{2}}=5,{{b}_{2}}=3,{{c}_{2}}=0.
a1a2+b1b2+c1c2=0{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0
⇒5(5μ−4λ+9)+(3)(2λ+3μ+1)+3(0)=0\Rightarrow 5\left( 5\mu -4\lambda +9 \right)+(3)\left( 2\lambda +3\mu +1 \right)+3(0)=0
⇒25μ−20λ+45+6λ+9μ+3=0\Rightarrow 25\mu -20\lambda +45+6\lambda +9\mu +3=0
⇒−14λ+34μ+48=0\Rightarrow -14\lambda +34\mu +48=0
⇒14λ−34μ−48=0\Rightarrow 14\lambda -34\mu -48=0
⇒7λ−17μ−24=0.....(10)\Rightarrow 7\lambda -17\mu -24=0.....(10)
We need to find the values of λ\lambda and μ\mu , to get the equation of line L3:x−4λ+4(5μ−4λ+9)=y+2λ−2(2λ+3μ+1)=z−3−3{{L}_{3}}:\dfrac{x-4\lambda +4}{\left( 5\mu -4\lambda +9 \right)}=\dfrac{y+2\lambda -2}{\left( 2\lambda +3\mu +1 \right)}=\dfrac{z-3}{-3}.
By solving equations (9) and (10) we will get the values of λ\lambda and μ\mu .
We should multiply equation (9) by 7.
70λ−49μ−119=0....(11)70\lambda -49\mu -119=0....(11)
We should multiply (10) by 10.
70λ−170μ−240=0.....(12)70\lambda -170\mu -240=0.....(12)
Now we should subtract (11) and (12).

& \left( 70\lambda -49\mu -119 \right)-\left( 70\lambda -170\mu -240 \right)=0 \\\ & \Rightarrow (170-49)\mu +(240-119)=0 \\\ & \Rightarrow 121\mu +121=0 \\\ & \Rightarrow 121\mu =-121 \\\ & \Rightarrow \mu =-1....(13) \\\ \end{aligned}$$ From equation (13) we get the value of $$\mu $$, Now we will substitute the value of $$\mu $$in (10). $$\begin{aligned} & 7\lambda -17(-1)-24=0 \\\ & \Rightarrow 7\lambda +17-24=0 \\\ & \Rightarrow 7\lambda -7=0 \\\ & \Rightarrow \lambda =1.......(14) \\\ \end{aligned}$$ From equation (14) we get the value of $$\lambda $$. Now we will substitute the both values of $$\lambda $$ and $$\mu $$in line equation $${{L}_{3}}:\dfrac{x-4\lambda +4}{\left( 5\mu -4\lambda +9 \right)}=\dfrac{y+2\lambda -2}{\left( 2\lambda +3\mu +1 \right)}=\dfrac{z-3}{-3}$$. $$\Rightarrow {{L}_{3}}:\dfrac{x-4(1)+4}{\left( 5(-1)-4(1)+9 \right)}=\dfrac{y+2(1)-2}{-\left( 2(1)+3(-1)+1 \right)}=\dfrac{z-3}{-3}$$ $$\Rightarrow {{L}_{3}}:\dfrac{x-4+4}{-5-4+9}=\dfrac{y+2-2}{(2-3+1)}=\dfrac{z-3}{-3}$$ $$\Rightarrow {{L}_{3}}:\dfrac{x-0}{0}=\dfrac{y-0}{0}=\dfrac{z-3}{-3}$$ Now we will multiply and divide the equation by (-3) $$\Rightarrow {{L}_{3}}:\dfrac{x-0}{\left( \dfrac{0}{-3} \right)}=\dfrac{y-0}{\left( \dfrac{0}{-3} \right)}=\dfrac{z-3}{\left( \dfrac{-3}{-3} \right)}$$ $$\Rightarrow {{L}_{3}}:\dfrac{x-0}{0}=\dfrac{y-0}{0}=\dfrac{z-3}{1}$$ **Hence, option (C) is correct.** **Note:** The formula to calculate the shortest distance between the lines between $${{L}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$$ and $${{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$$ is equal to $$\dfrac{\left| \left. \begin{aligned} & ({{x}_{2}}-{{x}_{1}}\text{) (}{{\text{y}}_{2}}-{{y}_{1}})\text{ (}{{\text{z}}_{2}}-{{z}_{1}}) \\\ & \text{ }{{\text{a}}_{1}}\;\;\;\;\;\;\;\;\;\;\;\;\; {{\text{b}}_{1}}\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;{{\text{c}}_{1}} \\\ & \text{ }{{\text{a}}_{2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\text{b}}_{2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\; {{\text{c}}_{2}} \\\ \end{aligned} \right| \right.}{\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}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$$ and $${{L}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$$ respectively.