Question

Find the equation of the plane which contain the line of intersection of the planes $\hat r.(\hat i+2\hat j+3\hat k)-4=0$, $\vec r.(2\hat i+\hat j-\hat k)+5=0$ and which is perpendicular to the plane $\vec r.(5\hat i+3\hat j-6\hat k)+8=0.$

Answer 1

The equations of the given planes are

$\hat r.(\hat i+2\hat j+3\hat k)-4=0$ ...(1)

$\vec r.(2\hat i+\hat j-\hat k)+5=0$ ...(2)

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

[$\hat r.(\hat i+2\hat j+3\hat k)-4$] + λ[$\vec r.(2\hat i+\hat j-\hat k)+5$] $= 0$

$\vec r.[(2λ+1)\hat i+(λ+2)\hat j+(3-λ)\hat k]+(5λ-4)=0$ ...(3)

The plane in equation (3) is perpendicular to the plane,

$\vec r.(5\hat i+3\hat j-6\hat k)+8=0$

$∴5(2λ+1)+3(λ+2)-6(3-λ)=0$

$⇒19λ-7-0 ⇒λ=\frac {7}{19}$

Substituting λ=7/19 in equation(3), we obtain

$⇒\vec r.[\frac {33}{19}\hat i+\frac {45}{19}\hat j+\frac {50}{19}\hat k]-\frac {41}{19}=0$

⇒$\vec r.(33\hat i+45\hat j+50 \hat k)-41=0$ ...(4)

This is the vector equation of the required plane.

The cartesian equation of this plane can be obtained by substituting $\vec r=x\hat i+y\hat j+z\hat k$ in equation (3).

$(x\hat i+y\hat j+z \hat k).(33\hat i+45\hat j+50\hat k)-41=0$

$⇒33x+45y+50z-41=0$.