Question

Find the equation of the plane passing through the line of intersection of the planes r.(i + j + k) = 1 and r.(2i + 3j - k) + 4 = 0 and parallel to X-axis.

Answer 1

Answer:

Equation of the plane $y-3z+6 =0$

Step-by-step explanation:

Equation of the plane passing through the line of intersection of the planes r.(i + j + k) = 1 and r.(2i + 3j - k) + 4 = 0

are given by P1+λP2 ,here λ is a real number ,can be calcultaed by given conditions

1) Convert the given plane in Cartesian form

let $\vec r= x \hat i+y \hat j+z \hat k$

$(x\hat i +y \hat j + z\hat k).(\hat i + \hat j + \hat k) = 1\\\\=x+y+z-1=0\\\\(x\hat i +y \hat j + z\hat k).(2\hat i + 3\hat j - \hat k) + 4 = 0 \\\\=>2x+3y-z+4=0$

Now put these equation of planes in P1+λP2

$x+y+z-1+\lambda (2x+3y-z+4)=0\:\:for\:\:any\:\:real\:\:number\:\:\lambda\\\\\\= (1+2\lambda)x+(1+3\lambda)y+(1-\lambda)z+(-1+4\lambda)=0$---eq1

since plane is parallel to x-axis,and Direction ratio of x-axis are (1,0,0)

and we also know that in equation of plane Dr's of normal to the plane are given,these are(1+2λ ,1+3λ,1-λ)

Because normal to the plane and x axis are perpendicular, so apply the condition of perpendicularity

(1+2λ)(1)+(0)(1+3λ)(0)(1-λ)=0

1+2λ=0

λ=-1/2

Now put the value of λ in the  eq1

$(1+2(\frac{-1}{2} ))x+(1+3(\frac{-1}{2} ))y+(1-(\frac{-1}{2} ))z+(-1+4(\frac{-1}{2} ))=0\\\\(1-1)x+(1-(\frac{3}{2} ))y+(\frac{3z}{2} )-3=0\\\\=\frac{-y}{2} +\frac{3z}{2}-3=0\\ \\y-3z+6 =0$

Is the required equation of the plane.