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 ^

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 ^

Forces 2 i ^ + j ^ , 2 i ^ - 3 j ^ + 6 k ^ and - i ^ + 2 j ^ - k ^ act at a point P → , with position vector 4 i ^ - 3 j ^ - k ^ . Find the moment of the resultant of these forces about the point Q → whose position vector is 6 i ^ + j ^ - 3 k ^ .

Given, forces 2 i ^ + j ^ , 2 i ^ - 3 j ^ + 6 k ^ and - i ^ + 2 j ^ - k ^ . Therefore, the resultant force F net → = 2 + 2 - 1 i ^ + 1 - 3 + 2 j ^ + 0 + 6 - 1 k ^ = 3 i ^ + 0 j ^ + 5 k ^ Position vector: P Q → = 2 i ^ + 4 j ^ - 2 k ^ To find: moment M = P Q → × F net → = 2 i ^ + 4 j ^ - 2 k ^ × 3 i ^ + 5 k ^ = 20 i ^ - 16 j ^ - 12 k ^ .

If a → and b → are two vectors such that a → × b → = 2 , then find the value of a → b → a → × b → .

We have, a → × b → = 2 And, a → b → a → × b → = a → × b → · a → × b → ⇒ a → b → a → × b → = a → × b → a → × b → cos θ where, θ is the angle between a → × b → and a → × b → . Now angle between a → × b → and a → × b → is 0 radian. ⇒ a → b → a → × b → = a → × b → 2 cos 0 ⇒ a → b → a → × b → = 2 2 × 1 ⇒ a → b → a → × b → = 4

If the volume of a parallelepiped whose adjacent edges are a → = 2 i ^ + 3 j ^ + 4 k ^ , b → = i ^ + α j ^ + 2 k ^ , c → = i ^ + 2 j ^ + α k ^ is 15 , then find the value of α , where α > 0 .

Given a → = 2 i ^ + 3 j ^ + 4 k ^ b → = i ^ + α j ^ + 2 k ^ and c → = i ^ + 2 j ^ + α k ^ Volume of the parallelepiped = [ a → b → c → ] = 15 2 3 4 1 α 2 1 2 α = 15 2 α 2 - 7 α - 9 = 0 2 α - 9 α + 1 = 0 α = 9 2 , α > 0 .

The position vectors of the four angular points of a tetrahedron are A j ^ + 2 k ^ , B 3 i ^ + k ^ , C 4 i ^ + 3 j ^ + 6 k ^ and D 2 i ^ + 3 j ^ + 2 k ^ . Find the volume of the tetrahedron A B C D .

Vertices of tetrahedron are A j ^ + 2 k ^ , B 3 i ^ + k ^ , C 4 i ^ + 3 j ^ + 6 k ^ and D 2 i ^ + 3 j ^ + 2 k ^ Therefore, A B → = Position vector of B → - Position vector of A → ⇒ A B → = 3 i ^ + k ^ - j ^ + 2 k ^ ⇒ A B → = 3 i ^ - j ^ - k ^ Similarly, A C → = 4 i ^ + 2 j ^ + 4 k ^ and A D → = 2 i ^ + 2 j ^ ∴ Volume, V = 1 6 A B → A C → A D → = 1 6 3 - 1 - 1 4 2 4 2 2 0 = 1 6 3 0 - 8 + 0 - 8 - 8 - 4 = 6 cubic unit Since, volume is not negative.

Find the altitude of a parallelepiped whose three coterminous edges are vectors A → = i ^ + j ^ + k ^ , B → = 2 i ^ + 4 j ^ - k ^ and C → = i ^ + j ^ + 3 k ^ , with A → and B → as the sides of the base of the parallelepiped.

Given edges, A → = i ^ + j ^ + k ^ B → = 2 i ^ + 4 j ^ - k ^ C → = i ^ + j ^ + 3 k ^ Volume of parallelepiped = A → B → C → = Base Area × Altitude = 4 Base area = A → × B → = 38 Altitude = A → B → C → A → × B → = 4 38 = 2 38 19 .

Examine whether the vectors a → = 2 i ^ + 3 ȷ ^ + 2 k ^ , b → = i ^ - ȷ ^ + 2 k ^ and c → = 3 i ^ + 2 ȷ ^ + 4 k ^ form a left handed or a right handed system.

Given, a → = 2 i ^ + 3 j ^ + 2 k ^ b → = i ^ - j ^ + 2 k ^ and c → = 3 i ^ + 2 j ^ + 4 k ^ To find whether they are right-handed or left-handed system. a → b → c → = 2 3 2 1 - 1 2 3 2 4 = 2 [ - 4 - 4 ] - 3 [ 4 - 6 ] + 2 [ 2 + 3 ] = - 16 + 6 + 10 = 0 Hence, it is neither a left handed or a right handed system.

Prove that the four points 4 i ^ + 5 j ^ + k ^ , - j ^ - k ^ , 3 i ^ + 9 j ^ + 4 k ^ and 4 - i ^ + j ^ + k ^ are coplanar.

Given, A ≡ 4 i ^ + 5 j ^ + k ^ B ≡ - j ^ - k ^ C ≡ 3 i ^ + 9 j ^ + 4 k ^ D ≡ 4 - i ^ + j ^ + k ^ Then, A B → = Position vector of B → - Position vector of A → ⇒ A B → = - j ^ - k ^ - 4 i ^ + 5 j ^ + k ^ ⇒ A B → = - 4 i ^ - 6 j ^ - 2 k ^ Similarly, A C → = 3 i ^ + 9 j ^ + 4 k ^ - 4 i ^ + 5 j ^ + k ^ ⇒ A C → = - i ^ + 4 j ^ + 3 k ^ And, A D → = 4 - i ^ + j ^ + k ^ - 4 i ^ + 5 j ^ + k ^ ⇒ A D → = - 8 i ^ - j ^ + 3 k ^ If three vectors are coplanar, then scalar triple product of them is zero. Now, A B → A C → A D → - 4 - 6 - 2 - 1 4 3 - 8 - 1 3 = - 4 12 + 3 - - 6 - 3 + 24 - 2 1 + 32 = - 4 15 + 6 21 - 2 33 = - 60 + 126 - 66 = - 126 + 126 = 0 Hence, proved that the given four points are coplanar.