Question

Find the equation of the plane which contains the line of intersection of the planes x$+$2y$+$3$z-$4$=$0 and 2$x+y-z$ $+$ 5$=$0 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3, $-$1) and parallel to the plane obtained above.

Answer 1

The equation of the family of planes passing through the intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 is

(x + 2y + 3z − 4) + k(2x + y − z + 5) = 0, where k is some constant

$$\begin{gathered} \Rightarrow \left(2k+1\right)x+\left(k+2\right)y+\left(3-k\right)z=4-5k \\ \Rightarrow \frac{x}{\left({\displaystyle \frac{4-5k}{2k+1}}\right)}+\frac{y}{\left({\displaystyle \frac{4-5k}{k+2}}\right)}+\frac{z}{\left({\displaystyle \frac{4-5k}{3-k}}\right)}=1 \end{gathered}$$
It is given that x-intercept of the required plane is twice its z-intercept.

$$\begin{gathered} \left(\frac{4-5k}{2k+1}\right)=2\left(\frac{4-5k}{3-k}\right) \\ \Rightarrow \left(4-5k\right)\left(3-k\right)=\left(4k+2\right)\left(4-5k\right) \\ \Rightarrow \left(4-5k\right)\left(3-k-4k-2\right)=0 \\ \Rightarrow \left(4-5k\right)\left(1-5k\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 4-5k=0\mathrm{or}1-5k=0 \\ \Rightarrow k=\frac{4}{5}\mathrm{or}k=\frac{1}{5} \end{gathered}$$
When $k=\frac{4}{5}$, the equation of the plane is $\left(2\times \frac{4}{5}+1\right)x+\left(\frac{4}{5}+2\right)y+\left(3-\frac{4}{5}\right)z=4-5\times \frac{4}{5}\Rightarrow 13x+14y+11z=0$.

This plane does not satisfies the given condition, so this is rejected.

When $k=\frac{1}{5}$, the equation of the plane is $\left(2\times \frac{1}{5}+1\right)x+\left(\frac{1}{5}+2\right)y+\left(3-\frac{1}{5}\right)z=4-5\times \frac{1}{5}\Rightarrow 7x+11y+14z=15$.

Thus, the equation of the required plane is 7x + 11y + 14z = 15.

Also, the equation of the plane passing through the point (2, 3, −1) and parallel to the plane 7x + 11y + 14z = 15 is

$$\begin{gathered} 7\left(x-2\right)+11\left(y-3\right)+14\left(z+1\right)=0 \\ \Rightarrow 7x+11y+14z=33 \end{gathered}$$