Exams Joint Entrance Examination Mathematics 3D Geometry Question 340 Question 340 of 521 Medium MCQ
If the point ( 2 , α , β ) (2,\alpha,\beta) ( 2 , α , β ) lies on the plane which passes through the points ( 3 , 4 , 2 ) (3,4,2) ( 3 , 4 , 2 ) and ( 7 , 0 , 6 ) (7,0,6) ( 7 , 0 , 6 ) and is perpendicular to the plane 2 x − 5 y = 15 2x-5y=15 2 x − 5 y = 15 , then 2 α − 3 β 2\alpha-3\beta 2 α − 3 β is equal to:
Solutions · 1 Post your solution Official solution
Normal vector of plane
\[ = \left| { ( i ^ j ^ k ^ 2 − 5 0 4 − 4 5 ) \begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k} \\
2 & { - 5} & 0 \\
4 & { - 4} & 5 \\
\end{pmatrix} i 2 4 j − 5 − 4 k 0 5 } \right|\]
\[ = - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)\]
equation of plane is
5(x - 7) + 2y - 3(z - 6) = 0
5x + 2y - 3z = 17
5 \[ \times \] 2 + 2\[\alpha \] - 3\[\beta \] = 17
\[ \therefore \] 2\[\alpha \] - 3\[\beta \] = 17 - 10 = 7 Answer the question to unlock