Lf A-B-C- and the magnitudes of A- B- C- are 5- 4 and 3 units respectively- Then the angle between A- and C- is-

The correct option is A cos^ 1 (3/5) Using cosine formula: #160; a^2 + c^2 2ac cosθ = b^2 therefore, cos θ = a^2 + c^2 b^22ac =5^2+3^2 4^ ...

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Standard XII

Physics

Vector Component

lf A⃗=B⃗+C⃗...

Question

lf $\to A = \to B + \to C$ and the magnitudes of $\to A, \to B, \to C$ are $5$ , $4$ and $3$ units respectively. Then the angle between $\to A$ and $\to C$ is:

c o s^{- 1} (\frac{3}{5})

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c o s^{- 1} (\frac{4}{5})

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\frac{π}{2}

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{s i n}^{- 1} (\frac{3}{4})

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Solution

The correct option is A $c o s^{- 1} (\frac{3}{5})$
Using cosine formula:
$a^{2} + c^{2} - 2 a c c o s θ = b^{2}$
therefore, $c o s θ = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{5^{2} + 3^{2} - 4^{2}}{2 \times 5 \times 3}$
$c o s θ = \frac{3}{5}$
$θ = c o s^{- 1} \frac{3}{5}$

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lf →A=→B+→C and the magnitudes of →A, →B, →C are 5, 4 and 3 units respectively. Then the angle between →A and →C is:

The correct option is A cos−1(35)Using cosine formula: a2+c2−2accosθ=b2therefore, cosθ=a2+c2−b22ac=52+32−422×5×3cosθ=35θ=cos−135

lf $\to A = \to B + \to C$ and the magnitudes of $\to A, \to B, \to C$ are $5$ , $4$ and $3$ units respectively. Then the angle between $\to A$ and $\to C$ is:

The correct option is A $c o s^{- 1} (\frac{3}{5})$
Using cosine formula:
$a^{2} + c^{2} - 2 a c c o s θ = b^{2}$
therefore, $c o s θ = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{5^{2} + 3^{2} - 4^{2}}{2 \times 5 \times 3}$
$c o s θ = \frac{3}{5}$
$θ = c o s^{- 1} \frac{3}{5}$