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Example 14 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at May 29, 2018 by

Example 14 - Find direction cosines of unit vector perpendicular - Examples

Example 14 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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Example 14 Find the direction cosines of the unit vector perpendicular to the plane .(6 3 2 ) + 1 = 0 passing through the origin. Vector equation of a plane at a distance d from the origin and unit vector to normal from origin is . = d Unit vector of = = 1 ( ) Given, equation of plane is .(6 3 2 ) + 1 = 0 .(6 3 2 ) = 1 Multiplying with 1 on both sides, .(6 3 2 ) = 1 1 . ( 6 + 3 + 2 ) = 1 So; = 6 + 3 + 2 Magnitude of = 6 2+32+22 = 36+9+4 = 49 = 7 Now, = 1 ( ) = 1 7 ( 6 + 3 + 2 ) = + + Direction cosines of unit vector perpendicular to the given plane i.e. in are , , .

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