Find a unit vector perpendicular to the plane of two vectors a → = i ^ - j ^ + 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ .

Given, that a → = i ^ - j ^ + 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ . Now, a → × b → = i ^ j ^ k ^ 1 - 1 2 2 3 - 1 a → × b → = i ^ 1 - 6 - j ^ - 1 - 4 + k ^ 3 + 2 a → × b → = - 5 i ^ + 5 j ^ + 5 k ^ = 5 - i ^ + j ^ + k ^ So, a → × b → = 5 2 + 5 2 + 5 2 = 5 3 Therefore, unit vector perpendicular to a → & b → = ± a → × b → a → × b → = ± 1 3 - i ^ + j ^ + k ^ .

Find a unit vector perpendicular to the plane of two vectors a → = i ^ - j ^ + 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ .

Given, that a → = i ^ - j ^ + 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ . Now, a → × b → = i ^ j ^ k ^ 1 - 1 2 2 3 - 1 a → × b → = i ^ 1 - 6 - j ^ - 1 - 4 + k ^ 3 + 2 a → × b → = - 5 i ^ + 5 j ^ + 5 k ^ = 5 - i ^ + j ^ + k ^ So, a → × b → = 5 2 + 5 2 + 5 2 = 5 3 Therefore, unit vector perpendicular to a → & b → = ± a → × b → a → × b → = ± 1 3 - i ^ + j ^ + k ^ .

Let a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ , c → = 2 i ^ - j ^ + 4 k ^ . Find a vector d → which is perpendicular to a → and b → and c → · d → = 15 .

Given, a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ c → = 2 i ^ - j ^ + 4 k ^ . c → · d → = 15 Now, d → = λ a → × b → = λ i ^ j ^ k ^ 1 4 2 3 - 2 7 ⇒ d → = λ 32 i ^ - j ^ - 14 k ^ Now, using c → · d → = 15 λ 64 + 1 - 4 × 14 = 15 ⇒ λ = 5 3 ⇒ d → = 5 3 32 i ^ - j ^ - 14 k ^ .

Let a → , b → and c → be unit vectors such that a → · b → = 0 = a → · c → . If the angle between b → and c → is π 6 , then find a → .

Given, a → · b → = 0 = a → · c → Angle between b → and c → is π 6 To find: a → Now we can see that a → is parallel to b → × c → as a → ⊥ b → & a → ⊥ c → , So, it can be expressed as a → = λ b → × c → ⇒ a → = λ b → × c → ⇒ a → = λ b → c → sin π 6 ⇒ 1 = λ · 1 · 1 · sin π 6 ⇒ λ = ± 2 Hence, a → = ± 2 b → × c → .

Show that the area of the parallelogram having diagonals 3 i ^ + j ^ - 2 k ^ and i ^ - 3 j ^ + 4 k ^ is 5 3 .

The diagonals of the parallelogram are given as a → = 3 i ^ + j ^ - 2 k ^ b → = i ^ - 3 j ^ + 4 k ^ , respectively. To find: area of parallelogram. a → × b → = i ^ j ^ k ^ 3 1 - 2 1 - 3 4 a → × b → = i ^ 4 - 6 - j ^ 12 + 2 + k ^ - 9 - 1 a → × b → = - 2 i ^ - 14 j ^ - 10 k ^ a → × b → = - 2 2 + - 14 2 + - 10 2 a → × b → = 300 = 10 3 Therefore, Area = 1 2 × a ⇀ × b ⇀ = 1 2 × 10 3 = 5 3 .

A force F → = 2 i ^ + j ^ - k ^ acts at point A whose position vector is 2 i ^ - j ^ . Find the moment of force F → about the origin.

Given, F → = 2 i ^ + j ^ - k ^ r → = 2 i ^ - j ^ + 0 k ^ F → >about the origin. r → × F → = i ^ j ^ k ^ 2 - 1 0 2 1 - 1 = i ^ + 2 j ^ + 4 k ^

Find the moment of F → about point 1 , - 1 , 2 when force F → = 3 i ^ + 2 j ^ - 4 k ^ is acting on point 2 , - 1 , 3 .

Given, A → 1 , - 1 , 2 B → 2 , - 1 , 3 To find the moment we have to find out the moment arm - B A → = 1 - 2 i ^ + 1 - 1 j ^ + 2 - 3 k ^ = - i ^ + 0 j ^ - k ^ As we know, moment is the product of force applied to the distance between the axis and the line of action (moment arm). Therefore, Required moment is B A → × F ⇀ = - i ^ + 0 j ^ - k ^ × 3 i ^ + 2 j ^ - 4 k ^ = 2 i ^ - 7 j ^ - 2 k ^