In a regular hexagon A B C D E F , show that A E → is equal to A C → + A F → - A B → .

Given: A regular hexagon A B C D E F . To prove : A E → = A C → + A F → - A B → Proof Using triangle law of vector addition, A B → + B C → = A C → . . . i A F → + F E → = A E → . . . i i Now, from the figure, B C → = F E → . . . i i i [Same sense and all the sides of hexagon are of same length] Using equation i in equation i i , A B → + F E → = A C → ⇒ F E → = A C → - A B → . . . i v Using equation ( i v ) in equation i i , A E → = A F → + A C → - A B → ⇒ A E → = A C → + A F → - A B → Hence proved.

In a regular hexagon A B C D E F , show that A E → is equal to A C → + A F → - A B → .

Given: A regular hexagon A B C D E F . To prove : A E → = A C → + A F → - A B → Proof Using triangle law of vector addition, A B → + B C → = A C → . . . i A F → + F E → = A E → . . . i i Now, from the figure, B C → = F E → . . . i i i [Same sense and all the sides of hexagon are of same length] Using equation i in equation i i , A B → + F E → = A C → ⇒ F E → = A C → - A B → . . . i v Using equation ( i v ) in equation i i , A E → = A F → + A C → - A B → ⇒ A E → = A C → + A F → - A B → Hence proved.

For the pentagon O A B C D , prove that 3 O D → + D A → + D B → + D C → = O A → + O B → + O C → . Note: Question Corrected.

Given: A pentagon O A B C D 3 O D → + D A → + D B → + D C → = O D → + D A → + O D → + D B → + O D → + D C → = O A → + O B → + O C → [Using Triangle law of vector addition] Hence, proved.

In a regular hexagon A B C D E F , prove that A B → + A C → + A D → + A E → + A F → = 3 A D → .

Given a regular hexagon A B C D E F To prove: A B → + A C → + A D → + A E → + A F → = 3 A D → From the figure, B D → = A E → … i [Vectors of same magnitude and same sense] C D → = A F → … ii [Vectors of same magnitude and same sense] Now, L . H . S . = A B → + A C → + A D → + A E → + A F → = A B → + A C → + A D → + B D → + C D → = ( A B → + B D → ) + A D → + ( A C → + C D → ) = A D → + A D → + A D → [Using Triangle law of Vector Addition] = 3 A D → = R . H . S . Hence, proved.

For the pentagon A B C D E , prove that A B → + B C → + C D → + D E → + E A → = 0 → .

Given: A pentagon A B C D E To prove: A B → + B C → + C D → + D E → + E A → = 0 → From the given figure, B C → = - D E → … i [Same magnitude but opposite in direction] E B → = - C D → … ii [Same magnitude but opposite in direction] E A → + A B → = E B → … iii [Triangle law of Vector Addition] Now, L . H . S . = A B → + B C → + C D → + D E → + E A → = A B → + B C → + C D → + D E → + E A → = A B → + E A → + C D → + D E → + B C → = E B → + C D → + D E → + B C → [Using equation iii ] = - C D → + C D → + D E → - D E → [Using equation i and equation ii ] = 0 → = R . H . S . Hence, proved.

The position vectors of A , B , C , D are a → , b → , 2 a → + 3 b → and a → – 2 b → , respectively. Show that D B → = 3 b → – a → and A C → = a → + 3 b → .

Given, Position vector of A = a → Position vector of B = b → Position vector of C = 2 a → + 3 b → Position vector of D = a → - 2 b → To prove: D B → = 3 b → - a → and A C → = a → + 3 b → . D B → = ( Position vector of B ) - ( Position vector of D ) = b → - a → - 2 b → = b → - a → + 2 b → = 3 b → - a → A C → = ( Position vector of C ) - ( Position vector of A ) = 2 a → + 3 b → - a → = a → + 3 b → Hence, proved.

If P ( – 1 , 2 ) and Q ( 3 , – 7 ) are two points, express the vector P Q → in terms of unit vectors i ^ and j ^ . Also, find distance between point P and Q . What is the unit vector in the direction of P Q → ?

Given, points P ( – 1 , 2 ) and Q ( 3 , – 7 ) Then, O P → = - i ^ + 2 j ^ and O Q → = 3 i ^ - 7 j ^ To find P Q and find distance between P and Q and unit vector along P Q → . P Q → = O Q → - O P → ⇒ P Q → = 3 i ^ - 7 j ^ - - i ^ + 2 j ^ ⇒ P Q → = 4 i ^ - 9 j ^ Distance, P Q → = 16 + 81 = 97 unit : Unit vector, P Q ^ = P Q → P Q → = 4 97 i ^ - 9 97 j ^ .

If O P → = 2 i ^ + 3 j ^ - k ^ and O Q → = 3 i ^ - 4 j ^ + 2 k ^ , find the modulus and direction cosines of P Q → .

Given O P → = 2 i ^ + 3 j ^ - k ^ O Q → = 3 i ^ - 4 j ^ + 2 k ^ ∴ P Q → = O Q → - O P → ⇒ P Q → = i ^ - 7 j ^ + 3 k ^ To find: P Q → and direction cosines of P Q → ⇒ | P Q → | = 1 + 49 + 9 ⇒ P Q → = 59 Direction cosines of any vector A B → = a i ^ + b j ^ + c k ^ is given by a A B → , b A B → and c A B → . ∴ Direction cosine of P Q → is 1 59 , - 7 59 , 3 59

Show that the points A , B and C with position vectors a → = 3 i ^ - 4 j ^ - 4 k ^ , b → = 2 i ^ - j ^ + k ^ and c → = i ^ - 3 j ^ - 5 k ^ respectively, form the vertices of a right angled triangle.

Given vertices of a triangle are A 3 i ^ - 4 j ^ - 4 k ^ , B 2 i ^ - j ^ + k ^ and C i ^ - 3 j ^ - 5 k ^ . To prove: The given points are the vertices of a right-angled triangle. Let the origin be O . Then the position vectors of the vertices will be, O A → = a → = 3 i ^ - 4 j ^ - 4 k ^ O B → = b → = 2 i ^ - j ^ + k ^ O C → = c → = i ^ - 3 j ^ - 5 k ^ Here, vector of sides of the triangle are, Vector of side A B A B → = Position vector of B - Position vector of A ⇒ A B → = O B → - O A → ⇒ A B → = - i ^ + 3 j ^ + 5 k ^ Vector of side B C , B C → = O C → - O B → ⇒ B C → = - i ^ - 2 j ^ - 6 k ^ Vector of side A C , A C → = O C → - O A → ⇒ A C → = - 2 i ^ + j - k ^ Therefore, A B → = - 1 2 + 3 2 + 5 2 ⇒ A B → = 35 Magnitude of B C → , B C → = - 1 2 + - 2 2 - 6 2 ⇒ B C → = 41 Magnitude of A C → , A C → = - 2 2 + 1 2 + - 1 2 ⇒ A C → = 6 Applying Pythagoras theorem, B C 2 = A B 2 + A C 2 ⇒ 41 2 = 35 2 + 6 2 ⇒ 41 = 41 Pythagoras theorem is verified. Hence, the points A , B and C are the vertices of a right-angled triangle.