If a → , b → , c → are coplanar, prove that a → + b → , b → + c → , c → + a → are coplanar.

If a → , b → , c → are coplanar, then a → b → c → = 0 To prove: a → + b → , b → + c → , c → + a → are coplanar, which is [ a → + b → b → + c → c → + a → ] = 0 Now, [ a → + b → b → + c → c → + a → ] = ( a → + b → ) · ( b → + c → ) × ( c → + a → ) = ( b → + c → ) × ( c → + a → ) · ( a → + b → ) = b → × c → + b → × a → + c → × c → + c → × a → · ( a → + b → ) = b → × c → + b → × a → + c → × a → · ( a → + b → ) ∵ c → × c → = 0 = ( b → × c → ) · a → + ( b → × a → ) · a → + ( c → × a → ) · a → + ( b → × c → ) · b → + ( b → × a → ) · b → + ( c → × a → ) · b → = [ b → c → a → ] + 0 + 0 + 0 + 0 + [ c → a → b → ] = 2 [ a → b → c → ] {Since, [ a → b → c → ] = [ b → c → a → ] = [ c → a → b → ] } = 2 × 0 = 0 Hence, proved that a → + b → , b → + c → , c → + a → are coplanar.

If a → , b → , c → are coplanar vectors, write the vector perpendicular to a → .

Given that a → , b → , c → are coplanar vectors. Then the vector perpendicular to a → is Cross product of a → with any of the vectors b → or c → . i.e. ( a → × b → ) or ( b → × c ) → or ( a → × c → ) or ( b → × a → ) or ( c → × b → ) or ( c → × a → ) .

The position vector of three points A , B , C are given to be i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ respectively. Find a vector which is perpendicular to both A B → and A C → having magnitude 9 units.

Given the position vectors of three points A , B , C as i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ . Now, A B → = Position vector of B - Position vector of A . ⇒ A B → = 4 i ^ + 4 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A B → = 4 i ^ + 4 k ^ - i ^ - 3 j ^ - 3 k ^ ⇒ A B → = 3 i ^ - 3 j ^ + k ^ and A C → = Position vector of C - Position vector of A ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ - 3 j ^ - 3 k ^ ⇒ A C → = - 3 i ^ + j ^ - k ^ Now, A B → × A C → = i ^ j ^ k ^ 3 − 3 1 − 3 1 − 1 = 2 i ^ - 6 k ^ Also, | A B → × A C → | = 2 2 + - 6 2 = 2 10 Let r → be the vector which is perpendicular to A B → × A C → then we have, r → = ± k ⋅ A B → × A C → | A B → × A C → | , where k is a scalar. Now, A B → × A C → | A B → × A C → | = 2 i ^ − 6 k ^ 2 10 = i ^ − 3 k ^ 10 Thus, r → is given by = 9 i ^ − 3 k ^ 10

The position vector of three points A , B , C are given to be i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ respectively. Find the angle between A B → and A C → .

Given the position vectors of three points A , B , C as i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ . Now, A B → = Position vector of B - Position vector of A . ⇒ A B → = 4 i ^ + 4 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A B → = 3 i ^ - 3 j ^ + k ^ and A C → = Position vector of C - Position vector of A ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A C → = - 3 i ^ + j ^ - k ^ Now, A B → = 3 2 + - 3 2 + 1 2 = 19 and A C → = - 3 2 + 1 2 + - 1 2 = 11 and A B → · A C → = 3 i ^ - 3 j ^ + k ^ · - 3 i ^ + j ^ - k ^ ⇒ A B → · A C → = - 9 - 3 - 1 = - 13 The angle between A B → and A C → is given by: cos ⁡ θ = A B → ⋅ A C → | A B → | ⋅ | A C → | ⇒ cos ⁡ θ = - 13 19 · 11 ⇒ θ = cos - 1 - 13 19 · 11

The position vector of three points A , B , C are given to be i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ respectively. Find A B → and A C → .

Given the position vectors of three points A , B , C as i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ . Now, A B → = Position vector of B - Position vector of A . ⇒ A B → = 4 i ^ + 4 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A B → = 3 i ^ - 3 j ^ + k ^ and A C → = Position vector of C - Position vector of A ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A C → = - 3 i ^ + j ^ - k ^

Find a unit vector perpendicular to the plane A B C where A , B , C are points 1 , 1 , 2 , 2 , 3 , 5 and 1 , 5 , 5 .

Given A = ( 1 , 1 , 2 ) B = ( 2 , 3 , 5 ) C = ( 1 , 5 , 5 ) Since, A , B , C are points on a plane then A B → and A C → belongs to the same plane Therefore, we have to find a unit vector perpendicular to A B → and A C → We know that, A B → × A C → which is perpendicular to A B → and A C → A B → × A C → = i j k 1 2 3 0 4 3 = i ( 6 − 12 ) − j ( 3 − 0 ) + k ( 4 − 0 ) A B → × A C → = − 6 i − 3 j + 4 k Therefore, unit vector corresponding to A B → × A C → is given by A B → × A C → | A B → × A C → | = − 6 i − 3 j + 4 k 36 + 9 + 16 = − 6 i 61 − 3 61 j + 4 k 61