Question

If A= 4i + 6j - 3 k and B = - 2i - 5j + 7k, find the angle between A and B.​

Answer 1

$\huge\bf\red{Answer}$

Answer 2

Given :-

• A = 4 i + 6 j - 3 k
• B = - 2 i - 5 j + 7 k

To find :-

• the angle between A and B

Knowledge required :-

• dot product of two vectors

      A.B = | A | | B | cos θ    

( where A and B are two vectors , | A | and | B | are their magnitudes and  is the angle θ )

• Dot Product of two vectors with orthogonal notations Let , vector A = xi + yj + zk and vector B = pi + qj + rk is given by  

      A.B = x.p + y.q + z.r

(where i , j , k are three orthogonal vectors)

     

• Magnitude of a vector V in the form  m i + n j + o k  is given by

     |V| = √( m² + n² + o² )

Solution :-

Let, angle b/w vectors A and B is θ

Calculating dot product of vectors A and B

→  A.B = (4i + 6j - 3k) (-2i - 5j + 7k)

→  A.B = (4)(-2) + (6)(-5) + (-3)(7)

→  A.B = -8 - 30 - 21

→  A.B = -59

Calculating Magnitude of vector A

→  | A | = √((4)² + (6)² + (-3)²)

→ | A | = √(16 + 36 + 9)

→ | A | = √61

Calculating magnitude of vector B

→  | B | = √((-2)² + (-5)² + (7)²)

→  | B | = √(4 + 25 + 49)

→ | B | = √78

Calculating the angle b/w vectors A and B

Putting values in

A.B = | A | | B | cos θ

→  - 59 = ( √61 ) ( √78 ) cos θ

→ - 59 = 68.98  cos θ

→  cos θ = - 59 / 68.98

→ θ = cos⁻¹ ( -59 / 68.98)

or,

θ = 148.795°  (approx.)