Question

The vector equation of the plane passing through the origin and the line of intersection of the plane \[r.a = \lambda \] and \[r.b = \mu \;\] is
A) \[r.(\lambda a - \mu b) = 0\]
B) \[\;\;\;\;r.(\lambda b - \mu a) = 0\]
C) \[\;\;r.(\lambda a + \mu b) = 0\]
D) \[r.(\lambda b + \mu a) = 0\]

Answer 1

Hint: in this question we have to find a vector equation of plane passing through origin and line of intersection of two given planes. In order to get this we have to write the equation of required plane as a combination of two given planes. After that we have to the value of constant by using the condition given in the question then put this value of constant in equation.



Formula Used:Equation of required plane is given by
\[{P_1} + k{P_2} = 0\]
Where
\[{P_1}\]is a equation of first given plane
\[{P_2}\]is a equation of second given plane
K is arbitrary constant



Complete step by step solution:Equation of required plane which is passing through the line of intersection of a plane \[r.a = \lambda \]and plane \[r.b = \mu \;\]is given by
\[{P_1} + k{P_2} = 0\]
Where
\[{P_1}\]is a equation of first given plane
\[{P_2}\]is a equation of second given plane
k is arbitrary constant
\[(\;r.a - \lambda ) + k(r.b - \mu \;) = 0\]
After rearranging the above equation we get
\[r.(a + kb) = \lambda + k\mu \]
It is given in the question that this plane is passing through the origin
\[0.(a + kb) = \lambda + k\mu \]
\[\lambda + k\mu = 0\]
\[k = \dfrac{{ - \lambda }}{\mu }\]
After putting the value of k in equation \[r.(a + kb) = \lambda + k\mu \]we get
\[r.(a - \dfrac{\lambda }{\mu }b) = \lambda - \dfrac{\lambda }{\mu }\mu \]
\[r.(\mu a - \lambda b) = 0\]
Now the equation of required plane is
\[r\;.\;(\lambda b - \mu a) = 0\]



Option ‘B’ is correct



Note: Note :< /b>
Here we need to remember that, the required equation passing through origin. And then applying this condition we get unknown which is present in the equation of required plane.