Vectors A → and B → can be expressed as A → = 10 i ^ - 12 j ^ + 5 k ^ and B = 7 i ^ + 8 j ^ - 12 k ^ , where i ^ , j ^ and k ^ are unit vectors along x , y , z axes respectively. Find the resultant of the two vectors and its magnitude.

Given A → = 10 i ^ - 12 j ^ + 5 k ^ and B = 7 i ^ + 8 j ^ - 12 k ^ The sum of two vectors is given by A → + B → = 10 i ^ - 12 j ^ + 5 k ^ + 7 i → + 8 j → - 12 j → R = A → + B → = x 1 + x 2 i ^ + y 1 + y 2 j ^ + z 1 + z 2 k ^ Putting the values, we have, R = 10 + 7 i ^ + - 12 + 8 j ^ + 5 + - 12 k ^ R = 17 i ^ - 4 j ^ - 7 k ^ The magnitude of the resultant vector is given by R = A x 2 + A y 2 + A z 2 = 17 2 + - 4 2 + - 7 2 = 354 = 18 . 81

Vectors A → and B → can be expressed as A → = 10 i ^ - 12 j ^ + 5 k ^ and B = 7 i ^ + 8 j ^ - 12 k ^ , where i ^ , j ^ and k ^ are unit vectors along x , y , z axes respectively. Find the resultant of the two vectors and its magnitude.

Given A → = 10 i ^ - 12 j ^ + 5 k ^ and B = 7 i ^ + 8 j ^ - 12 k ^ The sum of two vectors is given by A → + B → = 10 i ^ - 12 j ^ + 5 k ^ + 7 i → + 8 j → - 12 j → R = A → + B → = x 1 + x 2 i ^ + y 1 + y 2 j ^ + z 1 + z 2 k ^ Putting the values, we have, R = 10 + 7 i ^ + - 12 + 8 j ^ + 5 + - 12 k ^ R = 17 i ^ - 4 j ^ - 7 k ^ The magnitude of the resultant vector is given by R = A x 2 + A y 2 + A z 2 = 17 2 + - 4 2 + - 7 2 = 354 = 18 . 81

Two vectors A → and B → are A → = 5 i ^ + 3 j ^ - 4 k ^ and B → = 5 i ^ + 2 j ^ + 4 k ^ . Find the unit vectors along A → + B → and A → - B → .

Given A → = 5 i ^ + 3 j ^ - 4 k ^ and B → = 5 i ^ + 2 j ^ + 4 k ^ . We have A → + B → = 5 i ^ + 3 j ^ - 4 k ^ + 5 i ^ + 2 j ^ + 4 k ^ = 10 i ^ + 5 j ^ . and A → + B → = 10 2 + 5 2 = 125 = 5 5 ,The unit vector along this direction is, A → + B → A → + B → = 10 i ^ + 5 j ^ 5 5 = 2 5 i ^ + 1 5 j ^ . Now, we have A → - B → = 5 i ^ + 3 j ^ - 4 k ^ - 5 i ^ + 2 j ^ + 4 k ^ = j ^ - 8 k ^ . and A → - B → = 1 2 + 8 2 = 65 . The unit vector along this direction is, A → + B → A + B = j ^ - 8 k ^ 65 = 1 65 i ^ - 8 65 k ^ .

Find the angle between the vectors A → = 2 i ^ + 3 j ^ and B → = - 3 i ^ + 2 j ^ .

Given two vectors A → = 2 i ^ + 3 j ^ and B → = - 3 i ^ + 2 j ^ The cross product of two vectors having an angle θ between them is given by A → . B → = AB sinθ , therefore we have for their cross product, A → . B → = AB sinθ = 2 i ^ + 3 j ^ ^ - 3 i ^ + 2 j ^ = - 6 + 6 = 0 Since, their cross product is zero.we get sin θ = 0 o r θ = 90 ° The vectors are mutually perpendicular.

A → = 2 i ^ + 3 j ^ + 4 k ^ and B → = i ^ - j ^ + k ^ are two vectors. Find A → × B → .

Given A → = 2 i ^ + 3 j ^ + 4 k ^ and B → = i ^ - j ^ + k ^ The cross product of two vectors is given by, A → × B → = A B sin θ , where θ = angle between the two vectors It's direction is defined by the right-hand screw rule A → × B → = i ^ j ^ k ^ A x A y A z B x B y B z = i ^ j ^ k ^ 2 3 4 1 - 1 1 = i ^ 3 × 1 - 4 × - 1 + j ^ 2 × 1 - 4 × 1 + k ^ 2 × - 1 - 3 × 1 = 7 i ^ + 2 j ^ - 5 k ^

. A ( 1 , 3 , 4 ) , B ( 0 , 2 , 1 ) and C ( 1 , - 2 , 3 ) are three points. Find the vectors A B → and B C → in terms of i ^ , j ^ and k ^ . Find the unit vector perpendicular to the plane containing A B → and B C → .

Given point A ( 1 , 3 , 4 ) , B ( 0 , 2 , 1 ) and C ( 1 , - 2 , 3 ) AB → = x 2 - x 1 i ^ , y 2 - y 1 j ^ , z 2 - z 1 k ^ = ( ( 0 - 1 ) i ^ , ( 2 - 3 ) j ^ , ( 1 - 4 ) k ^ - i , ^ - j ^ + 3 k ^ BC → = x 2 - x 1 i ^ , y 2 - y 1 j ^ , z 2 - z 1 k ^ = ( ( 1 - 0 ) i ^ , ( 2 - 2 ) j ^ , ( 3 - 1 ) k ^ = i ^ , - 4 j ^ , 2 k ^ The vector perpendicular to the plane containing the vector A B → and B C → will be the cross product on these vectors P → = AB → × BC → = - i ^ , j ^ , - 3 k ^ × i ^ , - 4 j ^ , 2 k ^ = P → = - 2 - 12 i ^ + 2 - 3 j ^ + 4 - 3 k ^ = 14 i ^ - j ^ - 5 k ^ P → = P x 2 + P y 2 + P z 2 = 14 2 + 1 2 + 5 2 = 222 = 14 . 9 The unit vector of this vector is given by n ^ = AB → × BC → P = 14 14 . 9 i ^ , - 1 14 . 9 j ^ + 5 14 . 9 k ^