if f1= 3i + 5j + 6k and f2= 2i + 3j - 4k both measure the newton are displaced by d= 3i - 2j + 4k measure the metre, find the workdone

work=f1 dot d + f2 dot d

work=(3i + 5j + 6k).( 3i - 2j + 4k ) + (2i+3j-4k).(3i-2j+4k)
work= 9-10+24+6-6-16
work=add them up.

To find the work done, we can use the formula:

Work = Force ∙ Displacement ∙ cos(θ)

Where:
- Force is the magnitude of the force vector
- Displacement is the magnitude of the displacement vector
- θ is the angle between the force vector and the displacement vector

First, let's calculate the magnitudes of the force vectors:

Magnitude of Force 1 (f1):
|f1| = √(3^2 + 5^2 + 6^2) = √(9 + 25 + 36) = √70

Magnitude of Force 2 (f2):
|f2| = √(2^2 + 3^2 + (-4)^2) = √(4 + 9 + 16) = √29

Next, let's calculate the magnitude of the displacement vector:

Magnitude of Displacement (d):
|d| = √(3^2 + (-2)^2 + 4^2) = √(9 + 4 + 16) = √29

Now, let's calculate the angle θ between the force and displacement vectors:

θ = cos^(-1)((f1 • d) / (|f1| • |d|))

where • denotes the dot product.

Dot Product of f1 and d (f1 • d):
f1 • d = (3 * 3) + (5 * (-2)) + (6 * 4) = 9 - 10 + 24 = 23

Now we can substitute the values into the formula to find the work done:

Work = √70 * √29 * cos^-1(23 / (√70 * √29))

Calculating this expression will give you the work done.

To find the work done, we can use the formula:

Work = Force * Displacement * cos(theta)

Where:
- Force is the magnitude of the force vector
- Displacement is the magnitude of the displacement vector
- theta is the angle between the force and displacement vectors

First, let's calculate the magnitudes of the force and displacement vectors.

Magnitude of f1 = sqrt((3^2) + (5^2) + (6^2))
= sqrt(9 + 25 + 36)
= sqrt(70)

Magnitude of f2 = sqrt((2^2) + (3^2) + (-4^2))
= sqrt(4 + 9 + 16)
= sqrt(29)

Magnitude of displacement vector (d) = sqrt((3^2) + (-2^2) + (4^2))
= sqrt(9 + 4 + 16)
= sqrt(29)

Next, let's calculate the angle (theta) between the force and displacement vectors.

theta = cos^(-1)((f1 dot d)/(magnitude of f1 * magnitude of d))

Where "dot" represents the dot product.

f1 dot d = (3 * 3) + (5 * (-2)) + (6 * 4)
= 9 - 10 + 24
= 23

theta = cos^(-1)((23) / (sqrt(70) * sqrt(29)))

Now we can calculate the work done:

Work = sqrt(70) * sqrt(29) * cos(theta)

Plug in the values of sqrt(70), sqrt(29), and cos(theta) to get the final answer.