       1. Chapter 11 Class 12 Three Dimensional Geometry
2. Serial order wise
3. Miscellaneous

Transcript

Misc 19 (Method 1) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes . ( + 2 ) = 5 and . (3 + + ) = 6 . The vector equation of a line passing through a point with position vector and parallel to a vector is = + Given, the line passes through (1, 2, 3) So, = 1 + 2 + 3 Given, line is parallel to both planes Line is perpendicular to normal of both planes. i.e. is perpendicular to normal of both planes. We know that is perpendicular to both & So, is cross product of normals of planes . ( + 2 ) = 5 and . (3 + + ) = 6 Required normal = 1 1 2 3 1 1 = ( 1(1) 1(2)) (1(1) 3(2)) + (1(1) 3( 1)) = ( 1 2) (1 6) + (1 + 3) = 3 + 5 + 4 Thus, = 3 + 5 + 4 Now, Putting value of & in formula = + = ( + 2 + 3 ) + ( 3 + 5 + 4 ) Therefore, the equation of the line is ( + 2 + 3 ) + ( 3 + 5 + 4 ). Misc 19 (Method 2) Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes . ( + 2 ) = 5 and . (3 + + ) = 6 . The vector equation of a line passing through a point with position vector and parallel to a vector is = + Given, the line passes through (1, 2, 3) So, = 1 + 2 + 3 Let = 1 + 2 + 3 A line parallel to a plane is perpendicular to the normal of the plane. And two lines and are perpendicular if . = 0 Given, the line is parallel to planes So, our equations are 1 2 + 2 3 = 0 3 1 + 2 + 3 = 0 Thus, = 1 + 2 + 3 = 3k + 5k + 4k Now, Putting value of & in formula = + = (1 + 2 + 3 ) + ( 3k + 5k + 4k ) = ( + 2 + 3 ) + k ( 3 + 5 + 4 ) = ( + 2 + 3 ) + ( 3 + 5 + 4 ) Therefore, the equation of the line is ( + 2 + 3 ) + ( 3 + 5 + 4 ).

Miscellaneous 