Question

If the distance between the plane, $23x-10y-2z+48=0$ and the plane containing the lines $\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}$​ and $\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }$$(\lambda \in \mathrm{ℝ})$ is equal to $\frac{k}{\sqrt{633}}$ then $k$ is equal to

Answer 1

Step 1. Find the the point of intersection of the two line

$\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}=t$ and $\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }=s$

We know that

The equation of line $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ can be represented as $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}=\lambda$

Then any point on the line is $\left(\lambda a+x_0,\lambda b+y_0,\lambda c+z_0\right)$

So point on the line are

$(2t-1,4t+3,3t-1)$ and $(2s-3,6s-2,\lambda s+1)$

Therefore the intersection of the two line is

$$\begin{gathered} 2t-1=2s-3.....\left(i\right) \\ 4t+3=6s-2...\left(ii\right) \\ 3t-1=\lambda s+1....\left(iii\right) \end{gathered}$$

Solving $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} 4t=4s-4 \\ 4t=6s-5 \\ --+ \\ \_\_\_\_\_\_\_\_ \\ -2s+1=0 \\ \therefore s=\frac{1}{2} \\ \therefore 4t=6s-5 \\ \Rightarrow t=\frac{6s-5}{4} \\ \Rightarrow t=\frac{6\left({\displaystyle \frac{1}{2}}\right)-5}{4} \\ \Rightarrow t=\frac{-2}{4} \\ \Rightarrow t=-\frac{1}{2} \end{gathered}$$

Putting in $\left(iii\right)$ we have

$$\begin{gathered} 3t-1=\lambda s+1 \\ \Rightarrow 3\left(-\frac{1}{2}\right)-1=\lambda \left(\frac{1}{2}\right)+1 \\ \Rightarrow -\frac{5}{2}-1=\lambda \left(\frac{1}{2}\right) \\ \Rightarrow \lambda =-7 \end{gathered}$$

So the point is $(2s-3,6s-2,\lambda s+1)=\left(2\left(\frac{1}{2}\right)-3,6\left(\frac{1}{2}\right)-2,-7\left(\frac{1}{2}\right)+1\right)=\left(1-3,3-2,\frac{-7+2}{2}\right)=$$\left(-2,1,-\frac{5}{2}\right)$

Step 2. Find the distance of the plane from the point

We know that the distance of the plane $ax+by+cz+d=0$ from point $\left(x_1,y_1,z_1\right)$ is $d=\frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}}$

Therefore the distance of the plane $23x-10y-2z+48=0$ from point $\left(-2,1,-\frac{5}{2}\right)$ is $d=\frac{\left|23\left(-2\right)-10\left(1\right)-2\left(-{\displaystyle \frac{5}{2}}\right)+48\right|}{\sqrt{{23}^2+{10}^2+{\left(-2\right)}^2}}=\frac{3}{\sqrt{633}}$

Now comparing $d$ with $\frac{k}{\sqrt{633}}$ the value of $k$ is $3$

Hence, the value of $k$ is $3$