Is the triangle formed by the vectors 3 i ^ + 5 j ^ + 2 k ^ , 2 i ^ - 3 j ^ - 5 k ^ and - 5 i ^ - 2 j ^ + 3 k ^ equilateral?

Let the given vectors be A B ¯ = 3 i ^ + 5 j ^ + 2 k ^ , B C ¯ = 2 i ^ - 3 j ^ - 5 k ^ and C A ¯ = - 5 i ^ - 2 j ^ + 3 k ^ From given vectors A B ¯ + B C ¯ + C A ¯ = 0 . Therefore, given vectors are the sides of the triangle, then | A B ¯ | = 3 2 + 5 2 + 2 2 = 9 + 25 + 4 = 38 | B C ¯ | = 2 2 + 3 2 + 5 2 = 38 | C A ¯ | = 5 2 + 2 2 + 3 2 = 38 ∴ | A B ¯ | = | B C ¯ | = | C A ¯ | Therefore, given vectors forms an equilateral triangle.

Is the triangle formed by the vectors 3 i ^ + 5 j ^ + 2 k ^ , 2 i ^ - 3 j ^ - 5 k ^ and - 5 i ^ - 2 j ^ + 3 k ^ equilateral?

Let the given vectors be A B ¯ = 3 i ^ + 5 j ^ + 2 k ^ , B C ¯ = 2 i ^ - 3 j ^ - 5 k ^ and C A ¯ = - 5 i ^ - 2 j ^ + 3 k ^ From given vectors A B ¯ + B C ¯ + C A ¯ = 0 . Therefore, given vectors are the sides of the triangle, then | A B ¯ | = 3 2 + 5 2 + 2 2 = 9 + 25 + 4 = 38 | B C ¯ | = 2 2 + 3 2 + 5 2 = 38 | C A ¯ | = 5 2 + 2 2 + 3 2 = 38 ∴ | A B ¯ | = | B C ¯ | = | C A ¯ | Therefore, given vectors forms an equilateral triangle.

If α , β and γ are the angles made by the vector 3 i ^ - 6 j ^ + 2 k ^ with the positive directions of the coordinate axes then find cos α , cos β and cos γ .

Let, a → = 3 i ^ - 6 j ^ + 2 k ^ Then, cos α = a → · i ^ a → · i ^ = 3 i ^ - 6 j ^ + 2 k ^ . i ^ 3 i ^ - 6 j ^ + 2 k ^ . i ^ ⇒ cos α = 3 7 Then, cos β = a → · j ^ a → j ^ = 3 i ^ - 6 j ^ + 2 k ^ . j ^ 3 i ^ - 6 j ^ + 2 k ^ . j ^ = - 6 9 + 36 + 4 × 1 = - 6 49 ⇒ cos β = - 6 7 Then, cos α = a → · k ^ a → · k ^ = 3 i ^ - 6 j ^ + 2 k ^ . k ^ 3 i ^ - 6 j ^ + 2 k ^ . k ^ = 2 9 + 36 + 4 × 1 = 2 49 ⇒ cos γ = 2 7 ∴ cos α = 3 7 , cos β = - 6 7 , cos γ = 2 7

If a → + b → + c → = α d → , b → + c → + d → = β a → and a → , b → , c → are non-coplanar vectors, then show that a → + b → + c → + d → = 0 → .

Given, a → + b → + c → - α d → = 0 → ...(i) and b → + c → + d → - β a → = 0 → ...(ii) Now, (i)=(ii), ⇒ a → + b → + c → - α d → = - β a → + b → + c → + d → ⇒ - β = 1 and - α = 1 ⇒ β = - 1 and α = - 1 Substituting α = - 1 in (i), we get a → + b → + c → + d → = 0 . Hence, proved.

If a → , b → , c → are non-coplanar vectors, prove that the four points - a → + 4 b → - 3 c → , 3 a → + 2 b → - 5 c → , - 3 a → + 8 b → - 5 c → , - 3 a → + 2 b → + c → are coplanar.

Let, O A → = - a → + 4 b → - 3 c → , O B → = 3 a → + 2 b → - 5 c → , O C → = - 3 a → + 8 b → - 5 c → and O D → = - 3 a → + 2 b → + c → Now, A B → = O B → - O A → = a → - 4 b → + 3 c → + 3 a → + 2 b → - 5 c → = 4 a → - 2 b → - 2 c → A C → = O C → - O A → = a → - 4 b → + 3 c → - 3 a → + 8 b → - 5 c → = - 2 a → + 4 b → - 2 c → A D → = O D → - O A → = a → - 4 b → + 3 c → - 3 a → + 2 b → + c → = - 2 a ¯ - 2 b ¯ + 4 c ¯ A B → = x A C → + y A D → where x , y are scalars. ⇒ 4 a - → 2 b → - 2 c → = x [ - 2 a → + 4 b → - 2 c → ] + y [ - 2 a → - 2 b → + 4 c → ] Here, 4 = - 2 x - 2 y ...(i), - 2 = 4 x - 2 y ...(ii) - 2 = - 2 x + 4 y ...(iii) (i) - (iii) ⇒ - 6 y = 6 ⇒ y = - 1 From (iii), - 2 x = - 2 - 4 y ⇒ 2 x = - 2 + 4 ⇒ - 2 x = 2 ⇒ x = - 1 Substitute x , y in (ii), we get - 2 = - 4 + 2 ⇒ 2 = 2 Hence, given vectors are coplanar.

If a , b , c are non-coplanar vectors, prove that the four points 6 a → + 2 b → - c → , 2 a → - b → + 3 c → , - a → + 2 b → - 4 c → , - 12 a → - b → - 3 c → are coplanar.

Let, O A → = 6 a → + 2 b → - c → , O B → = 2 a → - b → + 3 c → , O C → = - a → + 2 b → - 4 c → and O D → = - 12 a → - b → - 3 c → Now, A B → = O B → - O A → = 2 a → - b → + 3 c → - 6 a → - 2 b → + c → = - 4 a → - 3 b → + 4 c → A C → = O C → - O A → = - a → + 2 b → - 4 c → - 6 a → - 2 b → + c → = - 7 a → - 3 c → A D → = O D → - O A → = - 12 a → - b → - 3 c → - 6 a → - 2 b → + c → = - 18 a → - 3 b → - 2 c → A B → = x A C → + y A D → where x , y are scalars. ⇒ - 4 a → - 3 b → + 4 c → = x [ - 7 a → - 3 c → ] + y [ - 18 a → - 3 b → - 2 c → ] Here, - 4 = - 7 x - 18 y ...(i), - 3 = - 3 y ...(ii) 4 = - 3 x - 2 y ...(iii) From (ii), y = 1 From (iii), 4 = - 3 x - 2 ⇒ - 3 x = 4 + 2 ⇒ - 3 x = 6 ⇒ x = - 2 Substitute x , y in (i), we get - 4 = 14 - 18 ⇒ - 4 = - 4 Hence, given vectors are coplanar.