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).

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

\(⇒33x+45y+50z-41=0\).