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Question: Find the equation of the plane which contains two parallel lines \(\dfrac{x-4}{1}=\dfrac{y-3}{-4}=\d...

Find the equation of the plane which contains two parallel lines x−41=y−3−4=z−25\dfrac{x-4}{1}=\dfrac{y-3}{-4}=\dfrac{z-2}{5} and x−31=y+2−4=z5\dfrac{x-3}{1}=\dfrac{y+2}{-4}=\dfrac{z}{5}?

Explanation

Solution

We know that if we have given two straight line equations which are parallel to each other and are in the following form:
x−x1a=y−y1b=z−z1c; x−x2a=y−y2b=z−z2c \begin{aligned} & \dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}; \\\ & \dfrac{x-{{x}_{2}}}{a}=\dfrac{y-{{y}_{2}}}{b}=\dfrac{z-{{z}_{2}}}{c} \\\ \end{aligned}
Then the equation of the plane containing those two parallel lines is as follows: ∣x−x1y−y1z−z1 x2−x1y2−y1z2−z1 abc ∣=0\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ a & b & c \\\ \end{matrix} \right|=0
So, we are going to use the above equations in the given problem by substituting the values of (x1,y1,z1,x2,y2,z2,a,b,c)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}},{{x}_{2}},{{y}_{2}},{{z}_{2}},a,b,c \right) from the following relations:(x1,y1,z1)=(4,3,2);(x2,y2,z2)=(3,−2,0)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 4,3,2 \right);\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 3,-2,0 \right). Also, we are going to substitute the value of (a,b,c)=(1,−4,5)\left( a,b,c \right)=\left( 1,-4,5 \right)

Complete step by step answer:
The equation of the parallel lines given in the above problem is as follows:
x−41=y−3−4=z−25\dfrac{x-4}{1}=\dfrac{y-3}{-4}=\dfrac{z-2}{5}
x−31=y+2−4=z5\dfrac{x-3}{1}=\dfrac{y+2}{-4}=\dfrac{z}{5}
In the above problem, we are asked to find the equation of a plane which contains the above two lines.
We know that if we have given two parallel lines then we can find the equation of the plane containing those lines. Let us consider the two parallel lines in 3-dimension as follows:
x−x1a=y−y1b=z−z1c; x−x2a=y−y2b=z−z2c \begin{aligned} & \dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}; \\\ & \dfrac{x-{{x}_{2}}}{a}=\dfrac{y-{{y}_{2}}}{b}=\dfrac{z-{{z}_{2}}}{c} \\\ \end{aligned}
Now, the equation of plane containing the above two lines are calculated as follows:
∣x−x1y−y1z−z1 x2−x1y2−y1z2−z1 abc ∣=0\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ a & b & c \\\ \end{matrix} \right|=0…….. (1)
Comparing the given equation of parallel lines with the parallel lines which we have shown above we got the values of (x1,y1,z1,x2,y2,z2,a,b,c)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}},{{x}_{2}},{{y}_{2}},{{z}_{2}},a,b,c \right) and are equal to:
(x1,y1,z1)=(4,3,2);(x2,y2,z2)=(3,−2,0);(a,b,c)=(1,−4,5)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 4,3,2 \right);\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 3,-2,0 \right);\left( a,b,c \right)=\left( 1,-4,5 \right)
Substituting the above values in eq. (1) we get,
∣x−4y−3z−2 3−4−2−30−2 1−45 ∣=0 ⇒∣x−4y−3z−2 −1−5−2 1−45 ∣=0 \begin{aligned} & \left| \begin{matrix} x-4 & y-3 & z-2 \\\ 3-4 & -2-3 & 0-2 \\\ 1 & -4 & 5 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow \left| \begin{matrix} x-4 & y-3 & z-2 \\\ -1 & -5 & -2 \\\ 1 & -4 & 5 \\\ \end{matrix} \right|=0 \\\ \end{aligned}
Expanding the above determinant along first row we get,
(x−4)(−5(5)−(−4)(−2))−(y−3)(−1(5)−1(−2))+(z−2)(−1(−4)−(−5))=0 ⇒(x−4)(−25−8)−(y−3)(−5+2)+(z−2)(4+5)=0 ⇒(x−4)(−33)−(y−3)(−3)+(z−2)(9)=0 \begin{aligned} & \left( x-4 \right)\left( -5\left( 5 \right)-\left( -4 \right)\left( -2 \right) \right)-\left( y-3 \right)\left( -1\left( 5 \right)-1\left( -2 \right) \right)+\left( z-2 \right)\left( -1\left( -4 \right)-\left( -5 \right) \right)=0 \\\ & \Rightarrow \left( x-4 \right)\left( -25-8 \right)-\left( y-3 \right)\left( -5+2 \right)+\left( z-2 \right)\left( 4+5 \right)=0 \\\ & \Rightarrow \left( x-4 \right)\left( -33 \right)-\left( y-3 \right)\left( -3 \right)+\left( z-2 \right)\left( 9 \right)=0 \\\ \end{aligned}
Eliminating brackets by multiplying the numbers with the variable expressions we get,
−33x−4(−33)+3y−9+9z−18=0 ⇒−33x+3y+9z+132−27=0 ⇒−33x+3y+9z+105=0 \begin{aligned} & -33x-4\left( -33 \right)+3y-9+9z-18=0 \\\ & \Rightarrow -33x+3y+9z+132-27=0 \\\ & \Rightarrow -33x+3y+9z+105=0 \\\ \end{aligned}
Dividing -3 on both the sides of the above equation we get,
−33x−3+3y−3+9z−3+105−3=0 ⇒11x−y−3z−35=0 \begin{aligned} & \dfrac{-33x}{-3}+\dfrac{3y}{-3}+\dfrac{9z}{-3}+\dfrac{105}{-3}=0 \\\ & \Rightarrow 11x-y-3z-35=0 \\\ \end{aligned}
Adding 35 on both the sides of the above equation we get,
11x−y−3z=3511x-y-3z=35
From the above solution, we have found the equation of the plane containing the two parallel lines as 11x−y−3z=3511x-y-3z=35.

Note: To solve the above problem, you must know the relation of the two parallel lines equation lying in the plane and the equation of plane derived from that. Also, you should know how to solve the determinant. Failing to know any of these concepts will cripple you in this problem.