Question

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

Answer 1

Hint: First, before proceeding with this, we must know that the parallel vectors of the given lines can be found using the coefficients in the denominator of the equation of the line. Then, we are given with the conditions that both the lines are perpendicular and can be used dot product of the vectors as $\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}}\centerdot \overrightarrow{{{b}_{2}}}}{\left| \overrightarrow{{{b}_{1}}} \right|\left| \overrightarrow{{{b}_{2}}} \right|}$ and by solving, we get the value of k. Then, by substituting the value of k in two lines and then by using them, we get the equation of the plane by using determinants.

Complete step-by-step solution
In this question, we are supposed to find the value of k and also find the equation of plane containing the lines with equation as $\dfrac{x-1}{-3}=\dfrac{y-2}{-2k}=\dfrac{z-3}{2}$and $\dfrac{x-1}{k}=\dfrac{y-2}{1}=\dfrac{z-3}{5}$which are perpendicular.
So, before proceeding with this, we must know that the parallel vectors of the given lines can be found using the coefficients in the denominator of the equation of a line.
So, by using this condition, we get the vector $\overrightarrow{{{b}_{1}}}$parallel to equation of line $\dfrac{x-1}{-3}=\dfrac{y-2}{-2k}=\dfrac{z-3}{2}$is given by:
$\overrightarrow{{{b}_{1}}}=-3\widehat{i}-2k\widehat{j}+2\widehat{k}$
Similarly, by using this condition, we get the vector $\overrightarrow{{{b}_{2}}}$parallel to equation of line $\dfrac{x-1}{k}=\dfrac{y-2}{1}=\dfrac{z-3}{5}$is given by:
$\overrightarrow{{{b}_{2}}}=k\widehat{i}+\widehat{j}+5\widehat{k}$
Now, we are given with the conditions that both the lines are perpendicular and can be used dot product of the vectors as $\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}}\centerdot \overrightarrow{{{b}_{2}}}}{\left| \overrightarrow{{{b}_{1}}} \right|\left| \overrightarrow{{{b}_{2}}} \right|}$.
So, by using the formula when $\theta ={{90}^{\circ }}$, we get the expression as:
$\overrightarrow{{{b}_{1}}}\centerdot \overrightarrow{{{b}_{2}}}=0$
Now, by substituting the value of both the vectors, we get:
$\begin{align}
  & \left( -3\widehat{i}-2k\widehat{j}+2\widehat{k} \right)\left( k\widehat{i}+\widehat{j}+5\widehat{k} \right)=0 \\
 & \Rightarrow -3k-2k+10=0 \\
 & \Rightarrow -5k+10=0 \\
 & \Rightarrow 5k=10 \\
 & \Rightarrow k=\dfrac{10}{5} \\
 & \Rightarrow k=2 \\
\end{align}$
So, we get the value of k as 2.
Then, by substituting the value of k in two lines equation, we get:
$\dfrac{x-1}{-3}=\dfrac{y-2}{-4}=\dfrac{z-3}{2}$ and $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{5}$
Now, to get the equation of plane by using the above two lines conditions, we get:
$\left| \begin{matrix}
   x-1 & y-2 & z-3 \\
   -3 & -4 & 2 \\
   2 & 1 & 5 \\
\end{matrix} \right|=0$
Then, by expanding the above determinant along first row, we get:
$\left( x-1 \right)\left( -4\times 5-2\times 1 \right)-\left( y-2 \right)\left( -3\times 5-2\times 2 \right)+\left( z-3 \right)\left( -3\times 1+4\times 2 \right)=0$
Now, by solving the above expression, we get the equation of plane as:
$\begin{align}
  & \left( x-1 \right)\left( -20-2 \right)-\left( y-2 \right)\left( -15-4 \right)+\left( z-3 \right)\left( -3+8 \right)=0 \\
 & \Rightarrow \left( x-1 \right)\left( -22 \right)-\left( y-2 \right)\left( -19 \right)+\left( z-3 \right)\left( 5 \right)=0 \\
 & \Rightarrow -22x+22+19y-38+5z-15=0 \\
 & \Rightarrow 22x-19y-5z+31=0 \\
\end{align}$
So, we get the equation of the plane as $22x-19y-5z+31=0$containing lines $\dfrac{x-1}{-3}=\dfrac{y-2}{-2k}=\dfrac{z-3}{2}$and $\dfrac{x-1}{k}=\dfrac{y-2}{1}=\dfrac{z-3}{5}$.
Hence, the value of k is 2 and equation of plane is $22x-19y-5z+31=0$.

Note: Now, to solve these types of questions we need to know some of the basic formulas for finding the equation of the plane when both lines are given. So, when the equations of the lines re of the form $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$, then equation of plane is given by:
$\left| \begin{matrix}
   x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|=0$.