AP EAMCET201921 Apr 2019Evening ShiftMathematicsThree Dimensional GeometryActual
The vector equation of the plane passing through the points ( 1 , - 2 , 5 ) , ( 0 , - 5 , - 1 ) and ( - 3 , 5 , 0 ) is
Options
- Ar ¯ = ( 1 - λ - 4 μ ) i ¯ - ( 2 + 3 λ - 7 μ ) j ¯ + ( 5 - 6 λ - 5 _
- Br ¯ = ( 1 + λ + 4 μ ) i ¯ - ( 2 - 3 λ + 7 μ ) j ¯ + ( 5 - 6 λ - 5 _
- Cr ¯ = ( 1 - λ + 4 μ ) i ¯ - ( 2 + 3 λ + 7 μ ) j ¯ + ( 5 - 6 λ + 5 _
- Dr ¯ = ( 1 + λ - 4 μ ) i ¯ + ( 2 + 3 λ - 7 μ ) j ¯ + ( 5 + 6 λ - 5 _
Correct answer
A. r ¯ = ( 1 - λ - 4 μ ) i ¯ - ( 2 + 3 λ - 7 μ ) j ¯ + ( 5 - 6 λ - 5 _
Step-by-step solution
The vectors of the points are, a → = i ^ - 2 j ^ + 5 k ^ b → = - 5 j ^ - k ^ c → = - 3 i ^ + 5 j ^ The vector equation of the plane, r → = a → + λ ( b → - a → ) + μ ( c → - a → ) ⇒ r → = ( i ^ - 2 j ^ + 5 k ^ ) + λ [ ( - 5 j ^ - k ^ ) - ( i ^ - 2 j ^ + 5 k ^ ) ] + μ [ ( - 3 i ^ + 5 j ^ ) - ( i ^ - 2 j ^ + 5 k ^ ) ] ⇒ r → = i ^ - 2 j ^ + 5 k ^ + λ ( - i ^ - 3 j ^ - 6 k ^ ) + μ ( - 4 i ^ + 7 j ^ - 5 k ^ ) ͡