Manipal MET2016MathematicsThree Dimensional Geometry
The vector equation of the plane through the point i +2 j - k and perpendicular to the line of intersection of the plane r (3 i - j + k )=1 and r ( i -4 j +2 k )=2 is
Options
- Ar (2 i +7 j -13 k )=1
- Br (2 i -7 j -13 k )=1
- Cr (2 i +7 j +13 k )=0
- DNone of the above
Correct answer
B. r (2 i -7 j -13 k )=1
Step-by-step solution
The line of intersection of the planes r (3 i - j + k )=1 and r ( i +4 j -2 k )=2 is common to both the planes. Therefore, it is perpendicular to normals to the two planes i.e. n ₁=3 i - j + k and n ₂= i +4 j -2 k . Hence, it is parallel to the vector n ₁ n ₂=-2 i +7 j +13 k Thus, we have to find the equation of the plane passing through a = i +2 j - k and normal to the vector n = n ₁ n ₂ . The equation of the required plane is ( r - a ) n or r n = a n or r (-2 i +7 j +13 k )=( i +2 i + k ) (-2 i +7 j +13 k ) or r