Question

The equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$.

& A)2x+6y+12z=13 \\\ & B)x+3y+6z=-7 \\\ & C)x+3y+6z=7 \\\ & D)2x+6y+12z=-13 \\\ \end{aligned}$$

Answer 1

We know that the equation of the plane containing the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is equal to ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda \left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \right)=0$. So, we should write the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$. We know that the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ are said to be parallel, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. It is also given that the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$. By using the concept, we can find the value of $\lambda$. By using this value of $\lambda$, we can find the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$.

Complete step-by-step answer:
We know that the equation of the plane containing the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is equal to ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda \left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \right)=0$.
Now we should find the plane containing the lines $2x-5y+z=3;x+y+4z=5$. $\begin{aligned} & \Rightarrow 2x-5y+z+\lambda \left( x+y+4z \right)=3+5\lambda \\\ & \Rightarrow 2x-5y+z+\lambda x+\lambda y+4\lambda z=3+5\lambda \\\ & \Rightarrow \left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)=3+5\lambda \\\ & \Rightarrow \left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)-3-5\lambda =0....(1) \\\ \end{aligned}$
From the question, it is clear that the equation of the plane containing the lines $2x-5y+z=3;x+y+4z=5$ and parallel to the plane $x+3y+6z=1$.
We know that the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ are said to be parallel, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$.
So, it is clear that the planes $\left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)-3-5\lambda =0$ and $x+3y+6z=1$ are parallel.
$\Rightarrow \dfrac{2+\lambda }{1}=\dfrac{\lambda -5}{3}=\dfrac{4\lambda +1}{6}.....(2)$
So, let us assume
$\Rightarrow \dfrac{2+\lambda }{1}=\dfrac{\lambda -5}{3}=\dfrac{4\lambda +1}{6}=k.....(3)$
So, from equation (3), we can write
$\Rightarrow \dfrac{2+\lambda }{1}=k$
By using cross multiplication, we get
$\Rightarrow 2+\lambda =k....(4)$
Now, from equation (3), we get
$\Rightarrow \dfrac{\lambda -5}{3}=k$
By using cross multiplication, we get
$\Rightarrow \lambda -5=3k....(5)$
Now let us substitute equation (5) in equation (4), then we get

& \Rightarrow \lambda -5=3\left( 2+\lambda \right) \\\ & \Rightarrow \lambda -5=6+3\lambda \\\ & \Rightarrow 2\lambda =-11 \\\ & \Rightarrow \lambda =\dfrac{-11}{2}.....(6) \\\ \end{aligned}$$ Now we will substitute equation (6) in equation (1), then we get $$\begin{aligned} & \Rightarrow \left( 2-\dfrac{11}{2} \right)x+\left( -\dfrac{11}{2}-5 \right)y+\left( 4\left( \dfrac{-11}{2} \right)+1 \right)z-3-5\left( \dfrac{-11}{2} \right)=0 \\\ & \Rightarrow \left( \dfrac{4-11}{2} \right)x+\left( \dfrac{-11-10}{2} \right)y+\left( 2(-11)+1 \right)z-3+\dfrac{55}{2}=0 \\\ & \Rightarrow \left( \dfrac{-7}{2} \right)x-\dfrac{21y}{2}-21z+\dfrac{49}{2}=0 \\\ & \Rightarrow \dfrac{-7x-21y-42z+49}{2}=0 \\\ & \Rightarrow -7x-21y-42z+49=0 \\\ & \Rightarrow -x-3y-6z+7=0 \\\ & \Rightarrow x+3y+6z-7=0 \\\ & \Rightarrow x+3y+6z=7.....(7) \\\ \end{aligned}$$ So, from equation (7) we will get that the equation of the plane containing the lines $$2x-5y+z=3;x+y+4z=5$$ and parallel to the plane $$x+3y+6z=1$$ is equal to $$x+3y+6z=7$$. **So, the correct answer is “Option C”.** **Note:** Students have a misconception that that the planes $${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$$ and $${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$$ are said to be parallel, then $$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{{{d}_{1}}}{{{d}_{2}}}$$. But we know that the planes $${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$$ and $${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$$ are said to be parallel, then $$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$$. So, students should avoid this misconception.