Examples

Example 1

Example, 2 Important

Example, 3

Example, 4 Important

Example, 5 Important

Example, 6 Important

Example, 7

Example 8 Important

Example 9

Example 10 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams You are here

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Question 7 Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Question 9 Important Deleted for CBSE Board 2025 Exams

Question 10 Important Deleted for CBSE Board 2025 Exams

Question 11 Important Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams

Question 13 Important Deleted for CBSE Board 2025 Exams

Question 14 Deleted for CBSE Board 2025 Exams

Question 15 Important Deleted for CBSE Board 2025 Exams

Question 16 Deleted for CBSE Board 2025 Exams

Question 17 Important Deleted for CBSE Board 2025 Exams

Question 18 Important Deleted for CBSE Board 2025 Exams

Question 19 Important Deleted for CBSE Board 2025 Exams

Question 20 Important Deleted for CBSE Board 2025 Exams

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