The position vectors of two points A and B are OA ¯ = 2 i ^ - j ^ - k ^ and OB ¯ = 2 i ^ - j ^ + 2 k ^ , respectively. The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is

The position vectors of two points A and B are OA ¯ = 2 i ^ - j ^ - k ^ and OB ¯ = 2 i ^ - j ^ + 2 k ^ , respectively. Let The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is O P ¯ When a point whose position vector is O P ¯ divides the line segment joining the position vectors O A ¯ and O B ¯ in the ratio of m : n , then O P ¯ = n O A ¯ + m O B ¯ m + n ⇒ O P ¯ = O A ¯ + 2 O B ¯ 2 + 1 ⇒ O P ¯ = 2 i ^ - j ^ - k ^ + 2 2 i ^ - j ^ + 2 k ^ 3 ⇒ O P ¯ = 2 i ^ - j ^ + k ^ Thus the required position vector is 2 i ^ - j ^ + k ^ .

Find a unit vector perpendicular to each of the vectors a → and b → where a → = 5 i ^ + 6 j ^ - 2 k ^ and b → = 7 i ^ + 6 j ^ + 2 k ^ .

Given vectors: a → = 5 i ^ + 6 j ^ - 2 k ^ and b → = 7 i ^ + 6 j ^ + 2 k ^ We know that, the unit vector perpendicular to the vectors a → and b → is n ^ = ± a → × b → | a → × b → | Now, a → × b → = i ^ j ^ k ^ 5 6 - 2 7 6 2 = 12 + 12 i ^ - 10 + 14 j ^ + 30 - 42 k ^ = 24 i ^ - 24 j ^ - 12 k ^ And also, a → × b → = 24 i ^ - 24 j ^ - 12 k ^ = 24 2 + - 24 2 + - 12 2 = 576 + 576 + 144 = 1296 = 36 Therefore, n ^ = ± ( a → × b → ) | a → × b → | = ± 24 i ^ - 24 j ^ - 12 k ^ 36 = ± 2 3 i ^ - 2 3 j ^ - 1 3 k ^ Hence, the unit vector perpendicular to both a → and b → is ± 2 3 i ^ - 2 3 j ^ - 1 3 k ^ .

Find the volume of the parallelepiped whose adjacent edges are represented by 2 a → , - b → and 3 c → , where a → = i ^ - j ^ + 2 k ^ , b → = 3 i ^ + 4 j ^ - 5 k ^ , and c → = 2 i ^ - j ^ + 3 k ^ .

Given vectors: a → = i ^ - j ^ + 2 k ^ , b → = 3 i ^ + 4 j ^ - 5 k ^ , and c → = 2 i ^ - j ^ + 3 k ^ Now, 2 a → = 2 i ^ - j ^ + 2 k ^ = 2 i ^ - 2 j ^ + 4 k ^ - b → = - 3 i ^ - 4 j ^ + 5 k ^ 3 c → = 3 2 i ^ - j ^ + 3 k ^ = 6 i ^ - 3 j ^ + 9 k ^ The volume of the parallelepiped whose adjacent edges are represented by 2 a → , - b → and 3 c → is 2 a → · ( - b → × 3 c → ) Now, - b → × 3 c → = i ^ j ^ k ^ - 3 - 4 5 6 - 3 9 = - 36 + 15 i ^ - - 27 - 30 j ^ + 9 + 24 k ^ = - 21 i ^ + 57 j ^ + 33 k ^ And also, 2 a → · ( - b → × 3 c → ) = 2 i ^ - 2 j ^ + 4 k ^ · - 21 i ^ + 57 j ^ + 33 k ^ = 2 - 21 - 2 57 + 4 33 = - 42 - 114 + 132 = - 24 Therefore, the volume of parallelepiped is - 24 or 24 cubic units.

Find the volume of the parallelepiped whose adjacent edges are represented by 2 a → , - b → and 3 c → , where a → = i ^ - j ^ + 2 k ^ , b → = 3 i ^ + 4 j ^ - 5 k ^ , and c → = 2 i ^ - j ^ + 3 k ^ .

Given vectors: a → = i ^ - j ^ + 2 k ^ , b → = 3 i ^ + 4 j ^ - 5 k ^ , and c → = 2 i ^ - j ^ + 3 k ^ Now, 2 a → = 2 i ^ - j ^ + 2 k ^ = 2 i ^ - 2 j ^ + 4 k ^ - b → = - 3 i ^ - 4 j ^ + 5 k ^ 3 c → = 3 2 i ^ - j ^ + 3 k ^ = 6 i ^ - 3 j ^ + 9 k ^ The volume of the parallelepiped whose adjacent edges are represented by 2 a → , - b → and 3 c → is 2 a → · ( - b → × 3 c → ) Now, - b → × 3 c → = i ^ j ^ k ^ - 3 - 4 5 6 - 3 9 = - 36 + 15 i ^ - - 27 - 30 j ^ + 9 + 24 k ^ = - 21 i ^ + 57 j ^ + 33 k ^ And also, 2 a → · ( - b → × 3 c → ) = 2 i ^ - 2 j ^ + 4 k ^ · - 21 i ^ + 57 j ^ + 33 k ^ = 2 - 21 - 2 57 + 4 33 = - 42 - 114 + 132 = - 24 Therefore, the volume of parallelepiped is - 24 or 24 cubic units.

Find a unit vector perpendicular to each of the vectors a → and b → where a → = 5 i ^ + 6 j ^ - 2 k ^ and b → = 7 i ^ + 6 j ^ + 2 k ^ .

Given vectors: a → = 5 i ^ + 6 j ^ - 2 k ^ and b → = 7 i ^ + 6 j ^ + 2 k ^ We know that, the unit vector perpendicular to the vectors a → and b → is n ^ = ± a → × b → | a → × b → | Now, a → × b → = i ^ j ^ k ^ 5 6 - 2 7 6 2 = 12 + 12 i ^ - 10 + 14 j ^ + 30 - 42 k ^ = 24 i ^ - 24 j ^ - 12 k ^ And also, a → × b → = 24 i ^ - 24 j ^ - 12 k ^ = 24 2 + - 24 2 + - 12 2 = 576 + 576 + 144 = 1296 = 36 Therefore, n ^ = ± ( a → × b → ) | a → × b → | = ± 24 i ^ - 24 j ^ - 12 k ^ 36 = ± 2 3 i ^ - 2 3 j ^ - 1 3 k ^ Hence, the unit vector perpendicular to both a → and b → is ± 2 3 i ^ - 2 3 j ^ - 1 3 k ^ .

The position vectors of two points A and B are OA ¯ = 2 i ^ - j ^ - k ^ and OB ¯ = 2 i ^ - j ^ + 2 k ^ , respectively. The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is

The position vectors of two points A and B are OA ¯ = 2 i ^ - j ^ - k ^ and OB ¯ = 2 i ^ - j ^ + 2 k ^ , respectively. Let The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is O P ¯ When a point whose position vector is O P ¯ divides the line segment joining the position vectors O A ¯ and O B ¯ in the ratio of m : n , then O P ¯ = n O A ¯ + m O B ¯ m + n ⇒ O P ¯ = O A ¯ + 2 O B ¯ 2 + 1 ⇒ O P ¯ = 2 i ^ - j ^ - k ^ + 2 2 i ^ - j ^ + 2 k ^ 3 ⇒ O P ¯ = 2 i ^ - j ^ + k ^ Thus the required position vector is 2 i ^ - j ^ + k ^ .

If θ is the angle between two vectors i ^ - 2 j ^ + 3 k ^ and 3 i ^ - 2 j ^ + k ^ , find sin θ .

Let the given vectors be a → = i ^ - 2 j ^ + 3 k ^ and b → = 3 i ^ - 2 j ^ + k ^ If θ is the angle between two vectors a → and b → , then cos θ = a → · b → | a → | | b → | Therefore, cos θ = ( i ^ - 2 j ^ + 3 k ^ ) · ( 3 i ^ - 2 j ^ + k ^ ) | i ^ - 2 j ^ + 3 k ^ | | 3 i ^ - 2 j ^ + k ^ | ⇒ cos θ = 1 ( 3 ) + ( - 2 ) ( - 2 ) + 3 ( 1 ) 1 + 4 + 9 9 + 4 + 1 ⇒ cos θ = 3 + 4 + 3 14 ⇒ cos θ = 10 14 ⇒ cos θ = 5 7 Now, sin θ = 1 - cos 2 θ = 1 - 5 7 2 = 1 - 25 49 = 49 - 25 49 = 24 49 = 2 6 7 Hence, the value of sin θ = 2 6 7 .

Let a → = 4 i ^ + 5 j ^ - k ^ , b → = i ^ - 4 j ^ + 5 k ^ , and c → = 3 i ^ + j ^ - k ^ . Find a d → which is perpendicular to both c → , and b → and d → · a → = 21

Given a → = 4 i ^ + 5 j ^ - k ^ , b → = i ^ - 4 j ^ + 5 k ^ and c → = 3 i ^ + j ^ - k ^ since d → is perpendicular to both c → and b → , it is parallel to c → × b → . Suppose d → = λ ( c → × b → ) for some scalar λ d → = λ i ^ j ^ k ^ 3 1 - 1 1 - 4 5 = λ [ ( 5 - 4 ) i ^ - ( 15 + 1 ) j ^ + ( - 12 - 1 ) k ^ ] = λ [ i ^ - 16 j ^ - 13 k ^ ] Now, d → · a → = 21 (Given) ⇒ ( 4 i ^ + 5 j ^ - k ^ ) · λ ( i ^ - 16 j ^ - 13 k ^ ) = 21 λ ( 4 - 80 + 13 ) = 21 λ = - 1 3 ∴ d → = - 1 3 ( i ^ - 16 j ^ - 13 k ^ ) ⇒ d → = - 1 3 i ^ + 16 3 j ^ + 13 3 k ^

If i ^ + j ^ + k ^ , 2 i ^ + 5 j ^ , 3 i ^ + 2 j ^ - 3 k ^ and i ^ - 6 j ^ - k ^ respectively are the position vectors of points A , B , C and D , then find the angle between the straight lines A B and C D . Find whether A B → and C D → are collinear or not.

Given position vectors are A ( i ^ + j ^ + k ^ ) , B ( 2 i ^ + 5 j ^ ) , C ( 3 i ^ + 2 j ^ - 3 k ^ ) and D ( i ^ - 6 j ^ - k ^ ) Now, A B → = ( 2 - 1 ) i ^ + ( 5 - 1 ) j ^ - k ^ = i ^ + 4 j ^ - k ^ and C D → = ( 1 - 3 ) i ^ + ( - 6 - 2 ) j ^ + ( - 1 + 3 ) k ^ = - 2 i ^ - 8 j ^ + 2 k ^ Now, A B → = ( 1 ) 2 + ( 4 ) 2 + ( - 1 ) 2 = 1 + 16 + 1 = 18 ⇒ A B → = 3 2 and C D → = ( - 2 ) 2 + ( - 8 ) 2 + ( 2 ) 2 = 4 + 64 + 4 = 72 ⇒ C D → = 6 2 Let θ is the angle between A B → and C D → ∴ cos θ = A B → · C D → | A B → | | C D → | ∴ cos θ = ( i ^ + 4 j ^ - k ^ ) · ( - 2 i ^ - 8 j ^ + 2 k ^ ) 3 2 6 2 ⇒ cos θ = - 2 - 32 - 2 36 ∵ i ^ · i ^ = j ^ · j ^ = k ^ · k ^ = 1 and i ^ · j ^ = j ^ · k ^ = k ^ · i ^ = 0 ⇒ cos θ = - 36 36 = - 1 ⇒ cos θ = - 1 ⇒ cos θ = cos π ⇒ θ = π Hence, the angle between A B → and C D → is π As the angle between A B → and C D → is π ∴ A B → and C D → are collinear.