The position vectors of three consecutive vertices of a parallelogram are i ^ + j ^ + k ^ , i ^ + 3 j ^ + 5 k ^ and 7 i ^ + 9 j ^ + 11 k ^ . Find the position vector of the fourth vertex.

Let A B C D be a parallelogram. Let a , b , c , d be the position vectors of the vertices A , B , C , D of the parallelogram. Where, a = i ^ + j ^ + k ^ , b = i ^ + 3 j ^ + 5 k ^ , c = 7 i ^ + 9 j ^ + 11 k ^ We know that the diagonals of a parallelogram bisect each other. Since the mid-point of the two position vectors x and y is given by x + y 2 . So, 1 2 a + c , 1 2 b + d ⇒ d = a + c - b = i ^ + j ^ + k ^ + 7 i ^ + 9 j ^ + 11 k ^ - i ^ - 3 j ^ - 5 k ^ = 7 i ^ + 7 j ^ + 7 k ^ = 7 i ^ + j ^ + k ^ Hence, the position vector of the fourth vertex is 7 i ^ + j ^ + k ^ .

The position vectors of three consecutive vertices of a parallelogram are i ^ + j ^ + k ^ , i ^ + 3 j ^ + 5 k ^ and 7 i ^ + 9 j ^ + 11 k ^ . Find the position vector of the fourth vertex.

Let A B C D be a parallelogram. Let a , b , c , d be the position vectors of the vertices A , B , C , D of the parallelogram. Where, a = i ^ + j ^ + k ^ , b = i ^ + 3 j ^ + 5 k ^ , c = 7 i ^ + 9 j ^ + 11 k ^ We know that the diagonals of a parallelogram bisect each other. Since the mid-point of the two position vectors x and y is given by x + y 2 . So, 1 2 a + c , 1 2 b + d ⇒ d = a + c - b = i ^ + j ^ + k ^ + 7 i ^ + 9 j ^ + 11 k ^ - i ^ - 3 j ^ - 5 k ^ = 7 i ^ + 7 j ^ + 7 k ^ = 7 i ^ + j ^ + k ^ Hence, the position vector of the fourth vertex is 7 i ^ + j ^ + k ^ .

A point P with position vector - 14 i ^ + 39 j ^ + 2 k ^ 5 divides the line joining A ( - 1 , 6 , 5 ) and B in the ratio 3 : 2 then find the point B .

Let A , B and P have position vectors a , b and p respectively. Then a = - i ^ + 6 j ^ + 5 k ^ , p = - 14 i ^ + 39 j ^ + 2 k ^ 5 Now, P divides A B internally in the ratio 3 : 2 . ∴ p = 3 b + 2 a 5 ⇒ 5 p = 3 b + 2 a ⇒ 3 b = 5 p - 2 a = 5 - 14 i ^ + 39 j ^ + 2 k ^ 5 - 2 - i ^ + 6 j ^ + 5 k ^ = - 14 i ^ + 39 j ^ + 2 k ^ + 2 i ^ - 12 j ^ - 10 k ^ = - 12 i ^ + 27 j ^ + 18 k ^ ∴ b = - 12 i ^ + 27 j ^ + 18 k ^ 3 = - 4 i ^ + 9 j ^ + 6 k ^ ∴ Coordinates of B are ( − 4 , 9 , 6 ) .

Dot-product of a vector with vectors 3 i ^ - 5 k ^ , 2 i ^ + 7 j ^ and i ^ + j ^ + k ^ are respectively - 1 , 6 and 5 . Find the vector.

Let the vector be a i ^ + b j ^ + c k ^ . Dot product with 3 i ^ - 5 k ^ = 3 a - 5 c = - 1 … … ( 1 ) Dot product with 2 i ^ + 7 j ^ = 2 a + 7 b = 6 … … ( 2 ) Dot product with i ^ + j ^ + k ^ = a + b + c = 5 . . . … ( 3 ) Equation (2) - 2 × Equation (3) ⇒ 2 a + 7 b − 2 a − 2 b - 2 c = 6 - 10 ⇒ 5 b - 2 c = - 4 … … ( 4 ) Equation (1) - 3 × equation (3) ⇒ 3 a - 5 c − 3 a − 3 b - 3 c = - 1 - 15 ⇒ 3 b + 8 c = 16 … … ( 5 ) 4 × Equation (4) + equation (5) ⇒ 20 b - 8 c + 3 b + 8 c = - 16 + 16 ⇒ b = 0 Put b = 0 in (5) we get c = 2 Put b = 0 , c = 2 in (3) a + 0 + 2 = 5 ∴ a = 3 Hence, the vector is 3 i ^ + 2 k ^ .

If a → , b → , c → are unit vectors such that a → + b → + c → = 0 → , find the value of a → ⋅ b → + b → ⋅ c → + c → ⋅ a → .

Given, a → , b → , and c → are unit vectors. So, a → = b → = c → = 1 ................ 1 Also, a → + b → + c → = 0 → So, a → + b → + c → = 0 → = 0 ............. 2 Now, a → + b → + c → 2 = a → + b → + c → ⋅ a → + b → + c → = a → · a → + a → · b → + a → · c → + b → · a → + b → · b → + b → · c → + c → · a → + c → · b → + c → · c → [Using distributivity of scalar product] Now, using the commutative property of dot product i.e. a → · b → = b → · a → = a → · a → + a → · b → + c → · a → + a → · b → + b → · b → + b → · c → + c → · a → + b → · c → + c → · c → = a → · a → + b → · b → + c → · c → + 2 a → · b → + b → · c → + c → · a → = a → 2 + b → 2 + c → 2 + 2 a → · b → + b → · c → + c → · a → [ Using a → · a → = a → 2 ] = 1 + 1 + 1 + 2 a → · b → + b → · c → + c → · a → [ Using 1 ] = 3 + 2 a → · b → + b → · c → + c → · a → ⇒ 0 = 3 + 2 a → · b → + b → · c → + c → · a → [Using 2 ] ⇒ a → ⋅ b → + b → ⋅ c → + c → ⋅ a → = - 3 2

If a line makes angles 90 ° , 135 ° , 45 ° with X , Y and Z axes respectively, then find its direction cosines.

Direction cosines of a line making angle α with X - axis, angle β with Y - axis and angle γ with Z - axis are l , m , n , where l = cos α , m = cos β , n = cos γ . Here, α = 90 ° , β = 135 ° , γ = 45 ° So, direction cosines are l = cos 90 ° = 0 m = cos 135 ° = cos 180 ° - 45 ° = - cos 45 ° = - 1 2 n = cos 45 ° = 1 2 ∴ Direction cosines are l , m , n = 0 , - 1 2 , 1 2 .