Question

Find the vector equation of the plane passing through the intersection of the planes $\bar{r}$ $\cdot$ (2$\hat{i}$ - 3$\hat{j}$ + 4$\hat{k}$) = 1 and $\bar{r}$ $\cdot$ ($\hat{i}$ - $\hat{j}$) +4 = 0 and perpendicular to the plane $\bar{r}$ $\cdot$ (2$\hat{i}$ - $\hat{j}$ + $\hat{k}$) = -5.

Answer 1

Answer:

Equation of plane is

3x-2y-4z+21=0

Step-by-step explanation:

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

intersection of the planes

Take $\vec r=(x\:\hat i+y\:\hat j+z\:\hat k),the\:\:equation\:\:of\:\:the\:\:given\:\:planes\:\:are\\\\ \bar{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 1 \\\\(x\:\hat i+y\:\hat j+z\:\hat k).(2\hat{i} - 3\hat{j} + 4\hat{k}) = 1\\\\2x-3y+4k-1=0\\\\and (x\:\hat i+y\:\hat j+z\:\hat k)\cdot (\hat{i} - \hat{j}) +4 = 0 \\\\\\x-y+4=0\\\\So\\\\2x-3y+4z-1+\lambda(x-y+4)=0\\\\(2+\lambda)x-(3+\lambda)y+4z-1+4\lambda=0$

Direction ratio's of normal to the plane

(2+λ,3+λ,4)

Since the palne is perpendicular to the  $\bar{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = -5$

Hence both direction ratio's to be put in the perpendicularity condition,to find value of λ

2(2+λ)-1(3+λ)+4(1)=0

4+2λ-3-λ+4=0

λ=-5

Now put the value of λ in the equation of plane

2x-3y+4z-1+(-5)(x-y+4)=0

2x-5x-3y+5y+4z-1-20=0

-3x+2y+4z-21=0

3x-2y-4z+21=0

Is the equation of the plane.