Question

If lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then find the value of $k$ and hence find the equation of the plane containing these lines.

Answer 1

Any point on $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ is $(1+2\lambda ,-1+3\lambda ,1+4\lambda )$

Any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$ is

$(3+\mu ,k+2\mu ,\mu )$.

For the two lines to intersect, the following three equations must be satisfied simultaneously:

$1+2\lambda =3+\mu \Rightarrow 2-2\lambda +\mu =0-1+3\lambda =k+2\mu \Rightarrow k+1-3\lambda +2\mu =01+4\lambda =\mu \Rightarrow -1-4\lambda +\mu =0$

Solving the above equations, we get, $\lambda =-1.5,\mu =-5,k=4.5$

the point of intersection is therefore, $(-2,-5.5,-5)$.

The normal to the plane containing the lines is obtained by the cross product of the vectors:

$(2\hat{i}+3\hat{j}+4\hat{k})\times (\hat{i}+2\hat{j}+\hat{k})=-5\hat{i}+2\hat{j}+\hat{k}$.

Since the plane contains both the lines it must contain the point of intersection as well, so in order to find the constant $d$ in the equation of the plane $\overrightarrow{r}.\overrightarrow{n}=d,$ where $\overrightarrow{r}$ is any point on the plane and $\overrightarrow{n}$ is the normal vector to the plane, we substitute the point of intersection for $\overrightarrow{r}$ in the equation:

$d=(-2\hat{i}-5.5\hat{j}-5\hat{k}).(-5\hat{i}+2\hat{j}+\hat{k})=-6$.

Therefore, the equation of the plane is $5x-2y-z-6=0.$