Question

Three vectors A,B & C are such that A=B+C and their magnitudes are 5,4 and 3 respectively. Find the angle between A and C

Answer 1

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Answer 2

Answer:

$Cos^{-1}\frac{3}{5}$ is the angle between  vector A and C

Explanation:

Given in the question that, there are three vectors$\vec{A}$  , $\vec{B}$ and $\vec{C}$ in which

$\vec{A} = \vec{B} + \vec{C}$  and their magnitudes are 5 , 4 and 3.

From the question we have,

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

⇒$|\bar{B}| = |\vec{A} | - |\vec{C}|$

Now squaring both side we get,

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

Put the value of magnitude of vector A, B and C in (i),

$4^2 = 5 ^2 + 3^2 - 2 . (5)(3)cos\theta$

⇒16 = 25 + 9 - 30$cos\theta$

⇒30 $Cos\theta$ =  = 25 + 9 - 16

⇒30 $Cos\theta$ = 18

⇒$Cos\theta = \frac{18}{30}$ = $\frac{3}{5}$

$\theta =$ $cos^{-1}\frac{3}{5}$

Hence, the angle between A and C is $Cos^{-1}\frac{3}{5}$.

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