Two forces F 1-2 -2 - N and F 2-3 -4 k- N are acting on a particle- The component of force F 1 along force F 2 isA- 5-2B- 6-5D- 5-6

The correct option is B 65Magnitude of component of F_1 along F_2 can be given by nbsp; |F_1| cosθ = F_1 . F⃗_⃗2⃗|F_2|=65 Magnitude of component of F_1 along ...

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Standard XII

Physics

Multiplication with Vectors

Two forces F ...

Question

Two forces $_{1} = (2^i + 2^j) N$ and $_{2} = (3^j + 4^k) N$ are acting on a particle. The component of force $_{1}$ along force $_{2}$ is

\frac{5}{6}

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\frac{5}{3}

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\frac{6}{5}

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\frac{5}{2}

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Solution

The correct option is C $\frac{6}{5}$
Magnitude of component of $_{1}$ along $_{2}$ can be given by
$|_{1} | cos θ = \frac{_{1} ._{2}}{|_{2} |} = \frac{6}{5}$
Magnitude of component of $_{1}$ along $_{2} = \frac{6}{5}$

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Two forces −→F1=(2^i+2^j) N and −→F2=(3^j+4^k) N are acting on a particle. The component of force −→F1 along force −→F2 is

The correct option is C 65Magnitude of component of −→F1 along −→F2 can be given by |−→F1| cosθ=−→F1.→F2|−→F2|=65 Magnitude of component of −→F1 along −→F2=65

Two forces $_{1} = (2^i + 2^j) N$ and $_{2} = (3^j + 4^k) N$ are acting on a particle. The component of force $_{1}$ along force $_{2}$ is

The correct option is C $\frac{6}{5}$
Magnitude of component of $_{1}$ along $_{2}$ can be given by
$|_{1} | cos θ = \frac{_{1} ._{2}}{|_{2} |} = \frac{6}{5}$
Magnitude of component of $_{1}$ along $_{2} = \frac{6}{5}$