Question

Three vectors A,B,C are such that Vector A = B+ C and their magnitude are in the ratio of 5 ratio 4 ratio 3 respectively find the angle between A and C​

Answer 1

Answer:

53°

Explanation:

Given

$\vec{A} = \vec{B} + \vec{C}$

'Or'

$\vec{A} - \vec{C} = \vec{B}$

Ratio of magnitude = 5:4:3

To Find

Angle between A and C

Now

On taking self dot product on both sides

$=>(\vec{A } - \vec{C}). (\vec{A} - \vec{C})= \vec{B}.\vec{B} \\ = > {A}^{2} + {C}^{2} - 2\vec{A }.\vec{C} = {B}^{2}$

Now

Let the angle between A and C be Θ

then

${A}^{2} + {C}^{2} - 2\vec{A }.\vec{C} \cos \theta = {B}^{2}$

On solving and putting Values

$= > \cos \theta = \frac{{A}^{2} + {C}^{2} - {B}^{2} }{2AC} \\ = > \cos \theta = \frac{ {(5)}^{2} + {(3)}^{2} - {(4)}^{2} }{2(5)(3)} \\ = > \cos \theta = \frac{25 + 9 - 16}{30} \\ = > \cos \theta = \frac{ \cancel18}{ \cancel30} \\ = > \cos \theta = \frac{3}{5} \\ = > \theta = { \cos }^{ - 1} ( \frac{3}{5} ) \\ = > \huge \boxed{\theta = 53 \degree}$

Therefore

Angle between A and C will be 53°.