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forces 2i-j+k, 3i+4j+4k, 5i+3j+3k acts through points with position vector 3i+2k, 4i-5j+ak, -4i-2j+4k respectively. Find the value of a,if the system of forces is in equilibrium​

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Answer:

For the system of forces to be in equilibrium, the vector sum of the forces must be equal to zero. Therefore, we can set up the following equation:

(2i-j+k) + (3i+4j+4k) + (5i+3j+3k) = 0

We can also express the force vectors in terms of the position vectors of the points at which they act as follows

F = (3i+2k)×(2i-j+k) + (4i-5j+ak)×(3i+4j+4k) + (-4i-2j+4k)×(5i+3j+3k) = 0

Expanding and simplifying the equation we get

3i+2k - 2j + k - 4i +5aj +4k -5j +4k +5i +3j -3k = 0

-i +3j +3k +5aj = 0

Therefore, a = -1/5

The value of a is -1/5.

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