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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 3 Find the vector equation of the plane which is at a distance of 6/ā29 from the origin and its normal vector from the origin is 2š Ģ ā 3š Ģ + 4š Ģ.Vector equation of a plane at a distance ādā from the origin and unit vector to normal from origin š Ģ is š ā.š Ģ = d Unit vector of š ā = š Ģ = 1/|š ā | (š ā) Now, distance from origin = d = 6/ā29 š ā = 2š Ģ ā 3š Ģ + 4š Ģ Magnitude of š Ģ = ā(22+(ā3)^2+4^2 ) |š ā | = ā(4+9+16) = ā29 Now, š Ģ = 1/|š ā | " (" š ā")" = 1/ā29 " (2" š Ģā"3" š Ģ+"4" š Ģ")" = 2/ā29 š Ģ ā 3/ā29 š Ģ + 4/ā29 š Ģ Vector equation of plane is š ā.š Ģ = d š ā . (š/āšš š Ģāš/āšš " " š Ģ+ š/āšš " " š Ģ ) = š/āšš Cartesian equation Equation of a plane in Cartesian form which is at a distance ādā from the origin and has a normal vector š ā = šš Ģ + bš Ģ + cš Ģ is lx + my + nz = d where l, m, n are direction cosines of š ā l = š/ā(š^2 + š^2 + š^2 ) , m = š/ā(š^2 + š^2 + š^2 ) , n = š/ā(š^2 + š^2 + š^2 ) Distance form origin = d = 6/ā29 š ā = 2š Ģ ā 3š Ģ + 4š Ģ Comparing with š ā = šš Ģ + šš Ģ + šš Ģ, š = 2, b = ā3, c = 4 Also, ā(š2+š2+š2) = ā(22+(ā3)2+42) = ā(4+9+16) = ā29 So, direction cosines are l = 2/ā29 , m = ( ā3)/ā29 , n = 4/ā29 ā“ Cartesian equations of plane is lx + my + nz = d 2/ā29 x + (( ā3)/ā29)y + 4/ā29 z = 6/ā29 2x ā 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x ā 3y + 4z = 6