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

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

Which vector is to be added with the vectors 3 i ^ + 4 j ^ - 5 k ^ and i ^ - 2 j ^ + 7 k ^ to give a resultant i ^ ?

Given the vector A → = 3 i ^ + 4 j ^ - 5 k ^ B → = i ^ - 2 j ^ + 7 k ^ To have the resultants be unit so that their sum is zero. To have this one should subtract a vector which is the sum of these vectors. Therefore, the vector required C → = ( 3 i ^ + 4 j ^ - 5 k ^ ) + ( i ^ - 2 j ^ + 7 k ^ ) = - 3 i ^ - 2 j ^ - 2 k ^

A → = 2 i ^ - 3 j ^ - k ^ and B → = 6 i ^ - 9 j ^ - 3 k ^ ; prove that A → and B → are parallel.

Given. A vector A → = 2 i ^ - 3 j ^ - k ^ vector B → = 6 i ^ - 9 j ^ - 3 k ^ We know that if the cross product of two vectors is zero, then those vectors will be parallel i.e A → × B = i j k A x A y A z B x B y B z → = 0 A → × B → = i j k 2 3 - 1 6 - 9 - 3 → = i 3 × - 3 - - 9 × - 1 + j 2 × - 3 - 6 × - 1 + k 2 × - 3 - 6 × - 1 or A → × B = 0 → Hence, these vectors are parallel.