Question

Show that each of the given three vectors is a unit vector: Also, show that they are mutually perpendicular to each other.

Answer 1

The given vectors are,

a → = 1 7 ( 2 i ^ +3 j ^ +6 k ^ )

b → = 1 7 ( 3 i ^ −6 j ^ +2 k ^ )

c → = 1 7 ( 6 i ^ +2 j ^ −3 k ^ )

Magnitude of vector a → is,

| a → |= ( 2 7 ) 2 + ( 3 7 ) 2 + ( 6 7 ) 2 = ( 4 49 )+( 9 49 )+( 36 49 ) =1

So, a → is a unit vector.

Magnitude of vector b → is,

| b → |= ( 3 7 ) 2 + ( −6 7 ) 2 + ( 2 7 ) 2 = ( 9 49 )+( 36 49 )+( 4 49 ) =1

So, b → is a unit vector.

Magnitude of vector c → is,

| c → |= ( 6 7 ) 2 + ( 2 7 ) 2 + ( −3 7 ) 2 = ( 36 49 )+( 4 49 )+( 9 49 ) =1

So, c → is a unit vector.

Dot product of a → and b → is,

a → ⋅ b → = 1 7 ( 2 i ∧ +3 j ∧ +6 k ∧ )⋅ 1 7 ( 3 i ∧ −6 j ∧ +2 k ∧ ) = 2 7 × 3 7 − 3 7 × 6 7 + 6 7 × 2 7 = 6 49 − 18 49 + 12 49 =0

Dot product of b → and c → is,

b → ⋅ c → = 1 7 ( 3 i ∧ −6 j ∧ +2 k ∧ )⋅ 1 7 ( 6 i ∧ +2 j ∧ −3 k ∧ ) = 3 7 × 6 7 − 6 7 × 2 7 − 2 7 × 3 7 = 18 49 − 12 49 − 6 49 =0

Dot product of c → and a → is,

c → ⋅ a → = 1 7 ( 6 i ∧ +2 j ∧ −3 k ∧ )⋅ 1 7 ( 2 i ∧ +3 j ∧ +6 k ∧ ) = 6 7 × 2 7 + 2 7 × 3 7 − 3 7 × 6 7 = 12 49 + 6 49 − 18 49 =0

If the dot product between two vectors is zero, then they are perpendicular to each other.

Thus, given vectors a → , b → and c → are mutually perpendicular to each other.