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 \(\vec r.(5\hat i+3\hat j-6\hat k)+8=0.\)
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 \(\vec r.(5\hat i+3\hat j-6\hat k)+8=0.\)
Solution and Explanation
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).
\((x\hat i+y\hat j+z \hat k).(33\hat i+45\hat j+50\hat k)-41=0\)
\(⇒33x+45y+50z-41=0\).
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