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 ^

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

A → = 2 i ^ + 3 j ^ + 4 k ^ and B → = i ^ - j ^ + k ^ ; find the value of A → × B → .

Given. A vector A → = 2 i ^ + 3 j ^ + 4 k ^ vector B → = i ^ - j ^ + k ^ The cross product of two vectors is given by A → × B → = i j k A x A y A z B x B y B z Substituting the given values in the above A → × B → = i j k 2 3 4 1 - 1 1 = i 3 × 1 - 4 - 4 + j 2 × 1 - 4 × 1 + k 2 × - 1 - 4 × 1 or A → × B → = 7 i ^ - 2 j ^ - 5 k ^

Show that the magnitude of ( A → × B → ) is equal to the area of a parallelogram whose adjoining arms are A → and B → .

Consider Vectors A → a n d B → represented by O P → a n d O Q → . Complete the parallelogram OPRQ Join P , Q . Drop Q N perpendicular on O P The cross product of the vectors is given by A → × B → = AB sin θ = OP OQ sinθ But sin θ = NQ OQ or NQ = OQ sinθ Or A → × B → = OP × NQ = area of parallelogram