How can you represent a vector by co-ordinates? Represent the vector A B → when A and B have co-ordinates ( - 1 , 3 , 5 ) and ( 2 , - 1 , 3 ) respectively.

Given coordinate of A = - 1 , 3 , 5 and B = 2 , - 1 , 3 We can represent a vector in three dimensions as A → = A x i ^ + A y j ^ + A z k ^ Where i ^ , j ^ , k ^ are the unit vectors in three directions. So we have A = - 1 i ^ + 3 j ^ + 5 k ^ and B = 2 i ^ - j ^ + 3 k ^ As we know AB → = A x - B x i ^ + A y - B y j ^ + A z - B z k ^ Putting the given value, we have A B → = - 1 + 2 i ^ + 3 - - 1 j ^ + 5 - 3 ) k ^ o r = 3 i ^ - 4 j ^ - 2 k ^

A → = 5 i ^ + 6 j ^ + 8 k ^ and B → = 3 i ^ + 7 j ^ - 10 k ^ represent two vectors. Find their sum.

Given A = 5 i ^ + 6 j ^ + 8 k ^ and B = 3 i ^ + 7 j ^ - 10 k ^ These vectors are in three coordinates, therefore their sum is represented in three coordinate as the sum of two vector quantities A → + B → = A x + B x i ^ + A y + B y j ^ + A z + B z k ^ Putting the given value, we have A → + B → = 5 + 3 i ^ + 6 + 7 j ^ + ( 8 - 10 ) k ^ or = 8 i ^ + 13 j ^ - 2 k ^

The vector sum of two vectors P → and Q → is R → . If Q → is reversed, the resultant is S → . Show that R 2 + S 2 = 2 P 2 + Q 2 .

Given the sum of two vectors P → + Q → = R → . If Q is reverse then the sum of two vectors P → + - Q → = S → In the first case, we have R 2 = P 2 + Q 2 + 2 PQ cos θ .....................(1) In the second case : S 2 = P 2 + Q 2 + ( - 2 PQ cos θ ) ...................................(2) BY adding the two equations, we have, R 2 + S 2 = 2 P 2 + 2 Q 2 = 2 P 2 + Q 2

Show that the vectors a → = 2 i ^ + 3 j ^ and b → = 6 i ^ - 4 j ^ are at right angles to each other.

Given a → = 2 i ^ + 3 j ^ and b → = 6 i ^ + 4 j ^ if two vectors A → , B → are at right angle to each other. Then their dot product becomes zero. A → . B → = AB cos θ = AB × 0 = 0 We have a → . b → = 2 × 6 + 3 × ( - 4 ) = 0 Therefore, these vectors are at right angle to each other.

Find the direction cosines of the vector r → = 2 i ^ + j ^ + 3 k ^ .

Given a vector r → = 2 i ^ + j ^ + 3 k ^ To find the direction cosines of the vector one has to divide the corresponding coordinate of the vector by the length of the vector. The coordinates of the unit vector is equal to its direction cosines. The magnitude of the vector r = r 2 x + r 2 y + r 2 z = 2 2 + 1 2 + 3 2 = 14 Therefore, the direction cosines are 2 14 , 1 14 , 3 14