Question

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

Answer 1

$$\begin{gathered} \text{We know that the lines} \\ \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\text{and}\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}\text{are perpendicular if} \\ {l}_1{l}_2+m_1{m}_2+n_1{n}_2=0 \\ \text{Here,} \\ {l}_1=-3;m_1=-2k;n_1=2;l_2=k;m_2=1;n_2=5 \\ \text{It is given that given lines are perpendicular.} \\ \Rightarrow l_1{l}_2+m_1{m}_2+n_1{n}_2=0 \\ \Rightarrow \left(-3\right)\left(k\right)+\left(-2k\right)\left(1\right)+\left(2\right)\left(5\right)=0 \\ \Rightarrow -3k-2k+10=0 \\ \Rightarrow -5k=-10 \\ \Rightarrow k=2 \\ \text{Substituting this value in the given equations of the lines, we get} \\ \frac{x-1}{-3}=\frac{y-2}{-4}=\frac{z-3}{2}...\left(1\right) \\ \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{5}...\left(2\right) \\ \mathbf{\text{Finding the equation of the plane}} \\ \text{Let the direction ratios of the required plane be proportional to}a,b,c. \\ \text{We know from (1) and (2) that lines (1) and (2) pass through the point (1, 2, 3) and the direction ratios of (1) and (2) are proportional to -3, -4, 2 and 2, 1, 5 respectively.} \\ \text{Since the plane contains the lines (1) and (2), the plane must pass through the point (1, 2, 3) and it must be parallel to the line.} \\ \text{So, the equation of the plane is} \\ a\left(x-1\right)+b\left(y-2\right)+c\left(z-3\right)=0...\left(3\right) \\ -3a-4b+2c=0...\left(4\right) \\ 2a+b+5c=0...\left(5\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-1& y-2& z-3\\ -3& -4& 2\\ 2& 1& 5\end{array}\right|=0 \\ \Rightarrow -22\left(x-1\right)+19\left(y-2\right)+5\left(z-3\right)=0 \\ \Rightarrow -22x+19y+5z=31 \end{gathered}$$