# list out some questions based on vectors , dot and cross product

Priyanshu Gujjar
94 Points
2 years ago
The cross product or vector product is a binary operation on two vectors in three-dimensional space (R3) and is denoted by the symbol x. Two linearly independent vectors a and b, the cross product, a x b, is a vector that is perpendicular to both a and b and therefore normal to the plane containing them.
hmhm
16 Points
2 years ago
Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg  Three identical rigid circular cylinders A, B and C each of mass m are arranged on smooth inclined surfaces as shown in figure. The value of  is least such that it prevents the arrangement from collapsing. Then A) 1 1 tan 2 3  −   =     B) 1 1 tan 3 3  −   =     C) Normal reaction between A & B is zero D) Normal reaction on C is 3 mg

Heer Shah
19 Points
2 years ago
If you want to do questions for dot and cross product then there is special chapter in amit agrawal book .(Ch2).
Its a good book you can try from it.