Question

Show that the line of intersection of the planes $x+2y+3z=8$ and $2x+3y+4z=11$ is coplanar with the line $\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$. Also find the equation of the plane containing them.

Answer 1

Given planed

$x+2y+3z=8$ ...(1)

$2x+3y+4z=11$ ...(2)

and fine

$\frac{x+1}{1}=\frac{y+1}{2}=\frac{z+1}{3}$ ...(3)

Let the plane passing through

intersection of plane $\left(1\right)$ & $\left(2\right)$

$(x+2y+3z-8)+k(2x+3y+4z-11)=0$ ....$\left(4\right)$

Now, we have to show that line of inter.

$\left(1\right)$ & $\left(2\right)$ are co-plane with k=line $\left(3\right)$

if line $\left(3\right)$ is co-planar with intersection of plane

Then, line $\left(3\right)$ is point $(-1,-1,-1)$

passing through plane $\left(4\right)$

& normal to the plane $\left(4\right)$ is 1st to 11 vector of line $\left(3\right)$

but $(-1,-1,-1)$ in eqn.

$(-1+2(-1)+3(-)-8)+k\left(2\right(-1)+3(-1)+4(-1)-11)=0$

$+14-20k=0$

$K=\frac{-7}{10}$

put $K$ is (4)

$(x+2y+3z-8)+\left(\frac{-7}{10}\right)(2x+3y+4z-11)=0$

$10x+10y+30z-80-14x-21y-28z+77=0$

$+4x-y+2z-3=0$

$4x+y-2z+3=0$

Then this is $e{q}^h$ of plane

Direction ratio of normal of plane $=4,1,-2$

So, $a_1=4,b_1=1,c_1=-2$

Direction ratios of line $=1,2,3$

so, $a_2=1,b_2=2,c_2=3$

$a_1{a}_2+b_1{b}_2+c_1{c}_2=4+2-6=0$

So, normal to the plane & line is $1^{st}$.

Hence, proved.