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Question: The work done by the force \[\overset{\to }{\mathop{F}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to ...

The work done by the force F=i+j+k\overset{\to }{\mathop{F}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, acting on a particle is displaced from A (3, 3, 3) to the point B (4, 4, 4) is,
(a) 2 units
(b) 3 units
(c) 4 units
(d) 7 units

Explanation

Solution

Hint: First calculate the distance by using the formula AB=ba\overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\, and then use the formula W=F.ABW=\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{AB}}\, to get the value of work done. Use the formula p.q=(ai+bj+ck).(di+ej+fk)=ad+be+cf\overset{\to }{\mathop{p}}\,.\overset{\to }{\mathop{q}}\,=\left( \overset{\to }{\mathop{ai}}\,+\overset{\to }{\mathop{bj}}\,+\overset{\to }{\mathop{ck}}\, \right).\left( \overset{\to }{\mathop{di}}\,+\overset{\to }{\mathop{ej}}\,+\overset{\to }{\mathop{fk}}\, \right)=ad+be+cf to calculate the dot product.

Complete step-by-step answer:
To find out the work done by the given force we should write the given equation first, therefore,
F=i+j+k\overset{\to }{\mathop{F}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, ……………………………………….. (1)
Also, A (3, 3, 3) and B (4, 4, 4) ……………………………………… (2)
As we have given the two points A and B therefore it’s position vector will become,
a=3i+3j+3k\overset{\to }{\mathop{a}}\,=3\overset{\to }{\mathop{i}}\,+3\overset{\to }{\mathop{j}}\,+3\overset{\to }{\mathop{k}}\, and b=4i+4j+4k\overset{\to }{\mathop{b}}\,=4\overset{\to }{\mathop{i}}\,+4\overset{\to }{\mathop{j}}\,+4\overset{\to }{\mathop{k}}\, ……………………………………… (3)
The distance between two points is given by the formula, AB=ba\overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\, therefore,
AB=ba\overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\,
If we put the values of equation (3) in the above formula we will get,
AB=(4i+4j+4k)(3i+3j+3k)\therefore \overset{\to }{\mathop{AB}}\,=\left( 4\overset{\to }{\mathop{i}}\,+4\overset{\to }{\mathop{j}}\,+4\overset{\to }{\mathop{k}}\, \right)-\left( 3\overset{\to }{\mathop{i}}\,+3\overset{\to }{\mathop{j}}\,+3\overset{\to }{\mathop{k}}\, \right)
Further simplification in the above equation will give,
AB=4i+4j+4k3i3j3k\therefore \overset{\to }{\mathop{AB}}\,=4\overset{\to }{\mathop{i}}\,+4\overset{\to }{\mathop{j}}\,+4\overset{\to }{\mathop{k}}\,-3\overset{\to }{\mathop{i}}\,-3\overset{\to }{\mathop{j}}\,-3\overset{\to }{\mathop{k}}\,
By rearranging the above equation we will get,
AB=(4i3i)+(4j3j)+(4k3k)\therefore \overset{\to }{\mathop{AB}}\,=\left( 4\overset{\to }{\mathop{i}}\,-3\overset{\to }{\mathop{i}}\, \right)+\left( 4\overset{\to }{\mathop{j}}\,-3\overset{\to }{\mathop{j}}\, \right)+\left( 4\overset{\to }{\mathop{k}}\,-3\overset{\to }{\mathop{k}}\, \right)
AB=i+j+k\therefore \overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, ………………………………………………………. (4)
To proceed further in the solution we should know the formula of work done given below,
Formula:
W=F.dW=\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{d}}\, Where d\overset{\to }{\mathop{d}}\, is the distance covered which is AB\overset{\to }{\mathop{AB}}\, in this case,
Therefore the work by the given force is given by,
W=F.ABW=\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{AB}}\,
If we put the values of equation (1) and equation (4) in the above equation we will get,
W=(i+j+k).(i+j+k)W=\left( \overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, \right).\left( \overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, \right)
Now to proceed further in the solution we should know the the formula given below,
Formula:
If p=ai+bj+ck\overset{\to }{\mathop{p}}\,=\overset{\to }{\mathop{ai}}\,+\overset{\to }{\mathop{bj}}\,+\overset{\to }{\mathop{ck}}\, and q=di+ej+fk\overset{\to }{\mathop{q}}\,=\overset{\to }{\mathop{di}}\,+\overset{\to }{\mathop{ej}}\,+\overset{\to }{\mathop{fk}}\, then their dot product is given by, p.q=(ai+bj+ck).(di+ej+fk)=ad+be+cf\overset{\to }{\mathop{p}}\,.\overset{\to }{\mathop{q}}\,=\left( \overset{\to }{\mathop{ai}}\,+\overset{\to }{\mathop{bj}}\,+\overset{\to }{\mathop{ck}}\, \right).\left( \overset{\to }{\mathop{di}}\,+\overset{\to }{\mathop{ej}}\,+\overset{\to }{\mathop{fk}}\, \right)=ad+be+cf.
By using the above formula in ‘W’ we will get,
W=1×1+1×1+1×1\therefore W=1\times 1+1\times 1+1\times 1
W=1+1+1\therefore W=1+1+1
Therefore, W = 3 units.
Therefore the work done by a given force from point A to point B is equal to 3 units.
Therefore the correct answer is option (b).

Note: You can also solve this problem by calculating F\left| \overset{\to }{\mathop{F}}\, \right| and distance AB by using distance formula and then using the formula W=F.ABW=\left| \overset{\to }{\mathop{F}}\, \right|.AB to get the quick answer.