Three forces A-k- B-2-3k- and C- acting on a body- To keep it in the equilibrium- the value of C- is-

The correct option is A {3î 2k̂} Put A⃗ + B⃗ + C⃗ = 0 C⃗ = (A⃗+B⃗) C⃗ = ((î + ĵ + k̂) + (2î ĵ 3k̂))= {3î 2k̂} ...

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Acceleration in 2D

Three forces ...

Question

Three forces $→A={^i+^j+^k}, →B={2^i−^j−3^k}$ and $\to C$ acting on a body. To keep it in the equilibrium, the value of $\to C$ is:

−{3^i−2^k}

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−{2^i−3^k}

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3^i + 4^j

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2^i - 3^k

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Solution

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\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k

\to c = 3^i +^j + 2^k

\to a \cdot (\to b \times \to c)

\to a \cdot (\to b \times \to c)

\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k

\to c = 3^i +^j + 2^k .

\to a = 3^i - 2^j +^k,

\to b = 2^i - 4^j - 3^k, \to c = -^i + 2^j + 2^k

\to a + \to b + \to c = 4^i - 4^j

\to a =^i -^j + 2^k, \to b = 2^i + 3^j +^k, \to c =^i -^k

\to a + 2 \to b - 3 \to c

\sqrt{78}

\to a = 2^i +^j + 3^k

\to b = 3^i + 2^j +^k

\to c =^i -^j - 4^k

\to d =^i + 2^j -^k

(\to a \times \to b) \times (\to c \times \to d) =

\to r = 3^i + 2^j - 5^k, \to a = 2^i -^j +^k, \to b =^i + 3^j - 2^k

\to c = - 2^i +^j - 3^k

\to r = l \to a + m \to b + n \to c

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