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Question: If the lines \[\dfrac{{x - 1}}{{ - 3}} = \dfrac{{y - 2}}{{ - 2k}} = \dfrac{{z - 3}}{2}\] and \[\dfra...

If the lines x−1−3=y−2−2k=z−32\dfrac{{x - 1}}{{ - 3}} = \dfrac{{y - 2}}{{ - 2k}} = \dfrac{{z - 3}}{2} and x−1k=y−21=z−35\dfrac{{x - 1}}{k} = \dfrac{{y - 2}}{1} = \dfrac{{z - 3}}{5} are perpendicular, then find the value of k and hence find the equation of the plane containing these lines.

Explanation

Solution

Hints: To solve this question, we must have a basic idea about the general equation of the line. At first, we have to find the parallel vectors corresponding to the respective lines. Then we will apply the equation of perpendicularity of the two vectors to the obtained vectors and thus the value of k can be determined. After substituting the value of k, we will get the line equations. Finally, we will apply the formula of plain containing two lines to get the plane equation.

Complete step-by-step solution:
We know the general equation of a line passing through P(x1,y1,z1)P({{x}_{1}},{{y}_{1}},{{z}_{1}}) having direction cosines or dcs l,m and n respectively is given by
x−x1l=y−y1m=z−z1n\dfrac{x-{{x}_{1}}}{l}=\dfrac{y-{{y}_{1}}}{m}=\dfrac{z-{{z}_{1}}}{n} ……………………………………… (1)
Now the equations of the two lines are given by,
L1:x−1−3=y−2−2k=z−32{{L}_{1}}:\dfrac{x-1}{-3}=\dfrac{y-2}{-2k}=\dfrac{z-3}{2} ……………………………………… (2)
L2=x−1k=y−21=z−35{{L}_{2}}=\dfrac{x-1}{k}=\dfrac{y-2}{1}=\dfrac{z-3}{5} ……………………………………… (3)
Again we know that two parallel lines have equal dcs. For line L1{{L}_{1}} dcs are −3,−2k,k-3,-2k,k and for the line L2{{L}_{2}} dcs are k,1,5k,1,5 respectively.
Then the vectors parallel to the line L1{{L}_{1}} is given by
v⃗1=−3i^−2kj^+2k^{{\vec{v}}_{1}}=-3\hat{i}-2k\hat{j}+2\hat{k} ……………………………………… (4)
And the vector parallel to the line L2{{L}_{2}} is given by
v⃗2=ki^+j^+5k^{{\vec{v}}_{2}}=k\hat{i}+\hat{j}+5\hat{k} ……………………………………… (5)
But according to the question the two lines L1{{L}_{1}} and L2{{L}_{2}} are perpendicular to each other. Then the parallel vectors corresponding to the lines must be perpendicular to each other. Hence v⃗1⊥v⃗2{{\vec{v}}_{1}}\bot {{\vec{v}}_{2}}.
We know the dot product of two perpendicular vectors is zero. Therefore
v⃗1⋅v⃗2=0{{\vec{v}}_{1}}\cdot {{\vec{v}}_{2}}=0 ……………………………………… (6)
Substituting the respective values of v⃗1{{\vec{v}}_{1}} and v⃗2{{\vec{v}}_{2}} from eq. (4) and eq. (5) in eq. (6), we will get
⇒(−3i^−2kj^+2k^)⋅(ki^+j^+5k^)=0\Rightarrow \left( -3\hat{i}-2k\hat{j}+2\hat{k} \right)\cdot \left( k\hat{i}+\hat{j}+5\hat{k} \right)=0
⇒(−3)(k)i^2+(−2k)(1)j^2+(2)(5)k^2=0\Rightarrow (-3)(k){{\hat{i}}^{2}}+(-2k)(1){{\hat{j}}^{2}}+(2)(5){{\hat{k}}^{2}}=0
We know that, i^2=j^2=k^2=1{{\hat{i}}^{2}}={{\hat{j}}^{2}}={{\hat{k}}^{2}}=1
⇒−3k−2k+10=0\Rightarrow -3k-2k+10=0
⇒5k=10\Rightarrow 5k=10
⇒k=2\Rightarrow k=2……………………………………… (7)
Now equation of the lines is obtained by substituting the value of k obtained in eq. (7) in eq. (2) and eq. (3)
L1:x−1−3=y−2−2×2=z−32{{L}_{1}}:\dfrac{x-1}{-3}=\dfrac{y-2}{-2\times 2}=\dfrac{z-3}{2}
⇒x−1−3=y−2−4=z−32\Rightarrow \dfrac{x-1}{-3}=\dfrac{y-2}{-4}=\dfrac{z-3}{2}……………………………………… (8)
And
⇒L2 :x−12=y−21=z−35\Rightarrow {{L}_{2}}\text{ }:\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{5} ……………………………………… (9)
We know the formula that the equation of plane containing the lines x−x1l1=y−y1m1=z−z1n1\dfrac{x-{{x}_{1}}}{{{l}_{1}}}=\dfrac{y-{{y}_{1}}}{{{m}_{1}}}=\dfrac{z-{{z}_{1}}}{{{n}_{1}}} and x−x1l2=y−y1m2=z−z1n2\dfrac{x-{{x}_{1}}}{{{l}_{2}}}=\dfrac{y-{{y}_{1}}}{{{m}_{2}}}=\dfrac{z-{{z}_{1}}}{{{n}_{2}}} is given by

x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\\ \end{matrix} \right|=0$$ ……………………………………… (10) Therefore applying this formula equation of the plane containing the lines$${{L}_{1}}:\dfrac{x-1}{-3}=\dfrac{y-2}{-4}=\dfrac{z-3}{2}$$and $${{L}_{2}}:\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{5}$$is given by $$\begin{aligned} & =\left| \begin{matrix} x-1 & y-2 & z-3 \\\ -3 & -4 & 2 \\\ 2 & 1 & 5 \\\ \end{matrix} \right|=0 \\\ & =(x-1)\left| \begin{matrix} -4 & 2 \\\ 1 & 5 \\\ \end{matrix} \right|-(y-2)\left| \begin{matrix} -3 & 2 \\\ 2 & 5 \\\ \end{matrix} \right|+(z-3)\left| \begin{matrix} -3 & -4 \\\ 2 & 1 \\\ \end{matrix} \right|=0 \end{aligned}$$ $$\begin{aligned} & =(x-1)\left| \begin{matrix} -4 & 2 \\\ 1 & 5 \\\ \end{matrix} \right|-(y-2)\left| \begin{matrix} -3 & 2 \\\ 2 & 5 \\\ \end{matrix} \right|+(z-3)\left| \begin{matrix} -3 & -4 \\\ 2 & 1 \\\ \end{matrix} \right|=0 \\\ & =(x-1)(-4\times 5-2\times 1)-(y-2)(-3\times 5-2\times 2)+(z-3)(-3\times 1-(-4)\times 2)=0 \\\ \end{aligned}$$ $$=-22(x-1)-(-19)(y-2)+5(z-3)=0$$ $$=-22x+22+19y-38+5z-15=0$$ On solving, we get $$=-22x+19y+5z-31=0$$…………………………… **This is the required equation of plane.** **Note:** The direction cosines of a line are the cosines of the angles made by the line with coordinate axes. $$\hat{i},\hat{j},\hat{k}$$ are unit vectors along X, Y and Z axes respectively. The dot product of the vectors $$\left( {{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k} \right)$$ and $$\left( {{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k} \right)$$ is $$\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{z}_{1}}{{z}_{2}} \right)$$. The determinant $$\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|$$ can be solved as follows, $$\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|={{a}_{1}}\left| \begin{matrix} {{b}_{2}} & {{b}_{3}} \\\ {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|-{{a}_{2}}\left| \begin{matrix} {{b}_{1}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{3}} \\\ \end{matrix} \right|+{{a}_{3}}\left| \begin{matrix} {{b}_{1}} & {{b}_{2}} \\\ {{c}_{1}} & {{c}_{2}} \\\ \end{matrix} \right|={{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{a}_{2}}\left( {{b}_{1}}{{c}_{3}}-{{c}_{1}}{{b}_{3}} \right)+{{a}_{3}}\left( {{b}_{1}}{{c}_{2}}-{{c}_{1}}{{b}_{2}} \right)$$.