Question

The vector equation of the plane passing through the intersection of the planes $\vec{ r } .(\hat{ i }+\hat{ j }+\hat{ k })=1$ and $\vec{ r } .(\hat{ i }-2 \hat{ j })=-2,$ and the point (1,0,2) is :

• A. $\vec{ r } .(\hat{ i }+7 \hat{ j }+3 \hat{ k })=\frac{7}{3}$
• B. $\vec{ r } \cdot(3 \hat{ i }+7 \hat{ j }+3 \hat{ k })=7$
• C. $\vec{ r } \cdot(\hat{ i }+7 \hat{ j }+3 \hat{ k })=7$
• D. $\vec{ r } .(\hat{ i }-7 \hat{ j }+3 \hat{ k })=\frac{7}{3}$

Answer 1

The Correct Option is C

 To find the vector equation of the plane passing through the intersection of two given planes and a specific point, we can follow these steps:

The given planes can be represented by their vector equations: 
\(\vec{ r } .(\hat{ i }+\hat{ j }+\hat{ k })=1\) 
\(\vec{ r } .(\hat{ i }-2 \hat{ j })=-2\)

Let the equation of the plane passing through the intersection of these planes be: 
\(\vec{ r } \cdot (\hat{ A}) = D\), 
where \(\hat{ A}\) is a vector normal to the plane.

For a plane passing through the intersection of two planes, the vector \(\hat{ A}\) can be written as a linear combination of the normals of the given planes: 
\(\hat{ A} = \hat{ i }+\hat{ j }+\hat{ k } + \lambda (\hat{ i }-2 \hat{ j })\)

Simplifying, we obtain: 
\(\hat{ A} = (1+\lambda)\hat{ i } + (1-2\lambda)\hat{ j } + \hat{ k }\)

Since the plane passes through the point (1, 0, 2), we substitute this point into the plane equation to find \(\lambda\): 
\(1(1+\lambda) + 0(1-2\lambda) + 2 = D\)

Hence, the plane equation becomes: 
\((1+\lambda) + 2 = D\) 
\(D = 3 + \lambda\)

Now, assume the form of the plane equation as: 
\(\vec{ r } \cdot (\hat{ i }+7 \hat{ j }+3 \hat{ k }),\) 
where: 
Multiplying through the conditions to form: 
\((1+\lambda, 1-2\lambda, 1) = (1, 7, 3)\\) 
Solving these: 
\(\lambda = 2, \, D = 7\)\)

Thus, the equation of the plane becomes: 
\(\vec{ r } \cdot (\hat{ i }+7 \hat{ j }+3 \hat{ k })=7\)

The correct answer is: 
\(\vec{ r } \cdot (\hat{ i }+7 \hat{ j }+3 \hat{ k })=7\)