Question

Find the equation of the plane passing through the intersection of the planes $\bar{r}$ $\cdot$ ($\hat{i}$ + $\hat{j}$ + $\hat{k}$) = 2 and $\bar{r}$ $\cdot$ (2$\hat{i}$ + 3$\hat{j}$ + $\hat{k}$) - 4 = 0 and parallel to X-axis.

Answer 1

Answer:

Equation of plane -y+z=0

Step-by-step explanation:

As we know that to find the equation of the plane passing through the

intersection of the planes $\bar{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 and \bar{r} \cdot (2\hat{i} + 3\hat{j} + \hat{k}) - 4 = 0$

=> $\vec r.(\hat i+\hat j+\hat k)-2=0\\=>(x\:\hat i+y\:\hat j+z\:\hat k).((\hat i+\hat j+\hat k)-2=0\\\\=> x+y+z-2=0\\.....eq1\\and \\\\\vec r.(2\hat i+3\hat j+\hat k)-4=0\\\\(x\:\hat i+y\:\hat j+z\:\hat k).((2\hat i+3\hat j+\hat k)-4=0\\\\\\2x+3y+z-4=0....eq2\\\\Equation\:\:of\:\:any\:\:plane\:\:through\:\:the\:\:intesection\:\:of\:\:these\:\:planes\:\:is \\(x+y+z-2)+\lambda(2x+3y+z-4)=0\:\:for\:\:some\:\:real\:\:number\:\:\\\\(1+2\lambda)x+(1+3\lambda)y+(1+\lambda)z-2-4\lambda=0$

Direction ratio's of the normal to the plane are (1+2λ,1+3λ,1+λ)

direction ratio's of x-axis are(1,0,0)

now plane is parallel to x-axis,that the normal to this plane is perpendicular to the x-axis.so direction ratio of x-axis.

So put the DR's in the condition of perpendicularity

1(1+2 λ)+0(1+3λ)+0(1+λ)=0

1+2λ=0

λ = -1/2

Now put the value of λ in the equation above

$(x+y+z-2)+(\frac{-1}{2}) (2x+3y+z-4)=0\\\\x+y+z-2-x-\frac{3y}{2} -\frac{z}{2}+2=0\\ \\\\y-\frac{3y}{2} +z-\frac{z}{2} =0\\ \\2y-3y+2z-z=0\\\\-y+z=0$