If a → and b → are non-collinear vectors and p → = x + 4 y a → + 2 x + y + 1 b → and q → = y - 2 x + 2 a → + 2 x - 3 y - 1 b → , find x and y such that 3 p → = 2 q → .

We have p → = x + 4 y a → + 2 x + y + 1 b → and q → = y - 2 x + 2 a → + 2 x - 3 y - 1 b → , such that 3 p → = 2 q → where a → and b → are non-collinear vectors. 3 p → = 2 q → ⇒ 3 ( x + 4 y a → + 2 x + y + 1 b → ) = 2 ( y - 2 x + 2 a → + 2 x - 3 y - 1 b → ) ⇒ ( 3 x + 12 y ) a → + 6 x + 3 y + 3 b → = ( 2 y - 4 x + 4 ) a → + 4 x - 6 y - 2 b → ⇒ 3 x + 12 y = 2 y - 4 x + 4 and 6 x + 3 y + 3 = 4 x - 6 y - 2 ( ∵ a → , b → are non-collinear vectors, a → is not a scalar multiple of b → ) ⇒ 10 y + 7 x = 4 . . . ( 1 ) and 2 x + 9 y = - 5 . . . ( 2 ) Solving above two equations, we get x = 2 and y = - 1 which are the required values.

If a → and b → are non-collinear vectors and p → = x + 4 y a → + 2 x + y + 1 b → and q → = y - 2 x + 2 a → + 2 x - 3 y - 1 b → , find x and y such that 3 p → = 2 q → .

We have p → = x + 4 y a → + 2 x + y + 1 b → and q → = y - 2 x + 2 a → + 2 x - 3 y - 1 b → , such that 3 p → = 2 q → where a → and b → are non-collinear vectors. 3 p → = 2 q → ⇒ 3 ( x + 4 y a → + 2 x + y + 1 b → ) = 2 ( y - 2 x + 2 a → + 2 x - 3 y - 1 b → ) ⇒ ( 3 x + 12 y ) a → + 6 x + 3 y + 3 b → = ( 2 y - 4 x + 4 ) a → + 4 x - 6 y - 2 b → ⇒ 3 x + 12 y = 2 y - 4 x + 4 and 6 x + 3 y + 3 = 4 x - 6 y - 2 ( ∵ a → , b → are non-collinear vectors, a → is not a scalar multiple of b → ) ⇒ 10 y + 7 x = 4 . . . ( 1 ) and 2 x + 9 y = - 5 . . . ( 2 ) Solving above two equations, we get x = 2 and y = - 1 which are the required values.

The position vectors of four points A , B , C , D are a → + b → + 2 c → , 2 a → + 5 b → + c → , 2 a → - 6 b → - c → and 4 a → + 2 b → - 3 c → respectively. Show that C D is parallel to A B and A B = 1 2 C D .

The position vectors of four points A , B , C , D are a → + b → + 2 c → , 2 a → + 5 b → + c → , 2 a → - 6 b → - c → and 4 a → + 2 b → - 3 c → respectively. We have to show that C D is parallel to A B and A B = 1 2 C D . Two vectors are parallel if one vector is a scalar multiple of other. C D → = O D → - O C → = 4 a → + 2 b → - 3 c → - ( 2 a → - 6 b → - c → ) = 2 a → + 8 b → - 2 c → A B → = O B → - O A → = 2 a → + 5 b → + c → - ( a → + b → + 2 c → ) = a → + 4 b → - c → Clearly, C D → = 2 ( A B → ) i.e., A B → = 1 2 C D → Hence, C D is parallel to A B .

Vectors a → and b → are non-collinear. For what value of x are the vectors c → = x - 2 a → + b → and d → = 3 + 2 x a → - 2 b → collinear?

Vectors a → and b → are non-collinear ⇒ a → is not a scalar multiple of b → . . . ( 1 ) Now, vectors c → = x - 2 a → + b → and d → = 3 + 2 x a → - 2 b → are collinear if c → = k d → ⇒ x - 2 a → + b → = k [ 3 + 2 x a → - 2 b → ] ⇒ x - 2 a → + b → = 3 + 2 x k a → - 2 k b → ⇒ x - 2 = k ( 2 x + 3 ) and - 2 k = 1 ⇒ x - 2 = - 1 2 ( 2 x + 3 ) (using ( 1 ) ) ⇒ 2 x - 4 = - 2 x - 3 ⇒ 4 x = 1 ⇒ x = 1 4

Prove by vector method that the vectors A → = i ^ + j ^ - 2 k ^ , B → = - 3 i ^ + 3 j ^ + 4 k ^ , C → = 4 i ^ - 2 j ^ - 6 k ^ can form the sides of a triangle.

Given vectors are A → = i ^ + j ^ - 2 k ^ , B → = - 3 i ^ + 3 j ^ + 4 k ^ , C → = 4 i ^ - 2 j ^ - 6 k ^ . By triangle law of vector addition, if two vectors are represented by both magnitude and direction of two sides of a triangle then their sum is represented by third sides of the triangle taken in reverse order By triangle law of vector addition, if two vectors B → + C → = ( - 3 i ^ + 3 j ^ + 4 k ^ ) + ( 4 i ^ - 2 j ^ - 6 k ^ ) B → + C → = ( - 3 + 4 ) i ^ + ( 3 - 2 ) j ^ + ( 4 - 6 ) k ^ B → + C → = i ^ + j ^ - 2 k ^ B → + C → = A → Thus, by triangle of vector addition, A → = i ^ + j ^ - 2 k ^ , B → = - 3 i ^ + 3 j ^ + 4 k ^ , C → = 4 i ^ - 2 j ^ - 6 k ^ can form the sides of a triangle.

If the four points with position vectors - 2 i ^ + j ^ + k ^ , i ^ + j ^ + k ^ , j ^ - k ^ and λ j ^ + k ^ are coplanar, then find the value of λ .

Let a → = - 2 i ^ + j ^ + k ^ , b → = i ^ + j ^ + k ^ , c → = j ^ - k ^ and d → = λ j ^ + k ^ be the coplanar vectors. Then ∃ x , y , z ∈ ℝ such that x a → + y b → + z c → + w d → = 0 → and x + y + z + w = 0 ⇒ x ( - 2 i ^ + j ^ + k ^ ) + y ( i ^ + j ^ + k ^ ) + z ( j ^ - k ^ ) + w ( λ j ^ - k ^ ) = ( 0 i ^ + 0 j ^ + 0 k ^ ) ⇒ - 2 x i ^ + x j ^ + x k ^ + y i ^ + y j ^ + y k ^ + z j ^ - z k ^ + w λ j ^ - w k ^ = ( 0 i ^ + 0 j ^ + 0 k ^ ) ⇒ ( - 2 x + y ) i ^ + ( x + y + z + λ w ) j ^ + ( x + y - z + w ) k ^ = ( 0 i ^ + 0 j ^ + 0 k ^ ) ⇒ ( - 2 x + y ) = 0 . . . ( 1 ) , ( x + y + z + λ w ) = 0 . . . ( 2 ) , ( x + y - z + w ) = 0 . . . ( 3 ) and x + y + z + w = 0 . . . ( 4 ) Subtracting ( 3 ) from ( 4 ) , we get z = 0 Substituting z in ( 2 ) and ( 3 ) , we get x + y + λ w = 0 and x + y + w = 0 ⇒ λ w = w ⇒ λ = 1

Show that the points - 2 i ^ + 3 j ^ + 5 k ^ , i ^ + 2 j ^ + 3 k ^ and 7 i ^ - k ^ are collinear.

If possible, let x ( - 2 i ^ + 3 j ^ + 5 k ^ ) + y ( i ^ + 2 j ^ + 3 k ^ ) + ( 7 i ^ - k ^ ) = 0 . . . ( 1 ) ⇒ - 2 x i ^ + 3 x j ^ + 5 x k ^ + y i ^ + 2 y j ^ + 3 y k ^ + 7 i ^ - k ^ = 0 ⇒ ( - 2 x + y + 7 ) i ^ + ( 3 x + 2 y ) j ^ + ( 5 x + 3 y - 1 ) k ^ = 0 ⇒ - 2 x + y + 7 = 0 . . . ( 1 ) , 3 x + 2 y = 0 . . . ( 2 ) and 5 x + 3 y - 1 = 0 . . . ( 3 ) Solving ( 1 ) and ( 2 ) , we get x = 2 and y = - 3 which also satisfies equation ( 3 ) . Also, x + y + 1 = 2 - 3 + 1 = 0 . Hence, the given points with position vectors - 2 i ^ + 3 j ^ + 5 k ^ , i ^ + 2 j ^ + 3 k ^ and 7 i ^ - k ^ are collinear.