(i) Let position vectors of points A , B and C are a → , b → and c → respectively. Point D divides line segment B C internally in the ratio 2 : 1 Find vector A D → .

Let position vector of point D be d → . d → = 2 . c → + 1 . b → 2 + 1 = 2 c → + b → 3 A D → = d → - a → A D → = 2 c → + b → 3 - a → ⇒ A D → = 2 c → + b → - 3 a → 3

(i) Let position vectors of points A , B and C are a → , b → and c → respectively. Point D divides line segment B C internally in the ratio 2 : 1 Find vector A D → .

Let position vector of point D be d → . d → = 2 . c → + 1 . b → 2 + 1 = 2 c → + b → 3 A D → = d → - a → A D → = 2 c → + b → 3 - a → ⇒ A D → = 2 c → + b → - 3 a → 3

(ii) Let A B C D is parallelogram. Position vector of points A , C and D are a → , c → and d → respectively. If E divides line segement A B internally in the ratio 3 : 2 , then find vector D E → .

Let positive vector of point B & E are b → & e → respectively. Using parallelogram law of addition, A B → + A D → = A C → ⇒ b → - a → + d → - a → = c → - a → ⇒ b → = a → + c → - d → Now, e → = 2 . a → + 3 . b → 2 + 3 = 2 a → + 3 b → 5 D E → = e → - d → ⇒ D E → = 2 a → + 3 b → 5 - d → ⇒ D E → = 2 a → 5 + 3 5 a → + c → - d → - d → ⇒ D E → = 1 5 5 a → + 3 c → - 8 d →

In a Δ A B C , A B → = 6 i ^ + 3 j ^ + 3 k ^ ; A C → = 3 i ^ - 3 j ^ + 6 k ^ , D and D ' are points trisections of side B C . Find A D → and A D ' → .

We have, B C → = A C → - A B → (from triangle law) or B C → = ( 3 i ∧ - 3 j ∧ + 6 k ∧ ) - ( 6 i ∧ + 3 j ∧ + 3 k ∧ ) = - 3 i ∧ - 6 j ∧ + 3 k ∧ Now, according to the question, B D → = D ' C → = 1 3 B C → = ( - i ∧ - 2 j ∧ + k ∧ ) ∴ A D → = B D → - B A → = B D → + A B → = ( 5 i ∧ + j ∧ + 4 k ∧ ) And A D ' → = A C → - D ' C → = ( 4 i ∧ - j ∧ + 5 k ∧ )

Show that the points A , B , C with position vectors 2 i ^ - j ^ + k ^ , i ^ - 3 j ^ - 5 k ^ and 3 i ^ - 4 j ^ - 4 k ^ respectively are the vertices of a right angled triangle. Also find the remaining angles of the triangle.

B A → = 2 - 1 i ^ + - 1 + 3 j ^ + 1 + 5 k ^ ⇒ B A → = i ^ + 2 j ^ + 6 k ^ C A → = 2 - 3 i ^ + - 1 + 4 j ^ + 1 + 4 k ^ ⇒ C A → = - i ^ + 3 j ^ + 5 k ^ C B → = 1 - 3 i ^ + - 3 + 4 j ^ + - 5 + 4 k ^ ⇒ C B → = - 2 i ^ + j ^ - k ^ ∙ Angle between a → & b → is given by cos θ = a → . b → a → b → ∙ cos C = C A → . C B → C A → C B → = - 1 - 2 + 3 1 + 5 - 1 35 6 ⇒ cos C = 0 ⇒ C = π 2 ∙ cos A = A B → . A C → A B → A C → = - 1 1 + - 2 - 3 + - 6 - 5 41 35 ⇒ A = cos - 1 35 41 ∙ cos B = B A → . B C → B A → B C → = 1 2 + 2 - 1 + 6 1 41 6 ⇒ B = cos - 1 - 6 41

If a → , b → , c → are three mutually perpendicular vectors of equal magnitude prove that a → + b → + c → is equally inclined with vectors a → , b → and c → .

If three vectors are mutually perpendicular, a → · b → = b → · c → = c → · a → = 0 Also according to the question | a → | = | b → | = | c → | = λ | a → + b → + c → | 2 = | a → | 2 + | b → | 2 + | c → | 2 + 2 ( a → · b → + b → · c → + c → · a → ) | a → + b → + c → | 2 = | a → | 2 + | b → | 2 + | c → | 2 | a → + b → + c → | 2 = 3 λ 2 | a → + b → + c → | = 3 λ Let angle between a → and a → + b → + c → is θ . ∴ cos θ = a → · a → + b → + c → a → a → + b → + c → = a → ⋅ a → λ ⋅ 3 λ 2 ∵ a → ⋅ b → = b → ⋅ c → = 0 = λ 2 3 λ 2 cos θ = 1 3 ∴ θ = cos − 1 1 3 Now find the angle between a → and a → + b → + c → which gives you cos - 1 1 3 . ( ∵ cos θ = a → . b → a → . b → ) Similarly, we can write angle between b → and a → + b → + c → is cos - 1 1 3 . Als0, we can show that c → and a → + b → + c → is cos - 1 1 3 . Hence, a → + b → + c → is equally inclined with a → , b → , c → .

Find the projection of b → + c → on a → , where a → = i ^ + 2 j ^ + k ^ , b → = i ^ + 3 j ^ + k ^ and c → = i ^ + k ^ .

The projection of a vector a → on b → is given by a → · b → b → . First we need to find the sum of vectors b → and c → . b → + c → = i ^ + 3 j ^ + k ^ + i ^ + k ^ = 2 i ^ + 3 j ^ + 2 k ^ So, the projection of b → + c → on a → = b → + c → · a → a → = 2 i ^ + 3 j ^ + 2 k ^ · i ^ + 2 j ^ + k ^ 1 2 + 2 2 + 1 2 = 2 × 1 + 3 × 2 + 2 × 1 1 + 4 + 1 = 2 + 6 + 2 6 = 10 6