### Video Transcript

A particle moved from point 𝐴
negative seven, negative one to point 𝐵 negative four, six along a straight line
under the action of the force 𝐅 equal to 𝑎𝐢 plus 𝑏𝐣. During this stage of motion, the
work done by the force was 106 units of work. The particle then moved from 𝐵 to
another point 𝐶 negative eight, negative three under the effect of the same
force. During this stage of motion, the
work done by the force was negative 138 units of work. Determine the two constants 𝑎 and
𝑏.

We begin by recalling that we can
calculate the work done by a force using the scalar or dot product. We find the scalar product of the
force vector 𝐅 and displacement vector 𝐝. In this question, we are told that
𝐅 is equal to 𝑎𝐢 plus 𝑏𝐣. And it is the value of these
constants 𝑎 and 𝑏 that we are trying to calculate.

There are two stages of motion
described in this question. Firstly, the particle moves from
point 𝐴 to point 𝐵. And during this stage, the work
done is 106 units of work. We are then told that the particle
moves from point 𝐵 to point 𝐶. And during this stage of motion,
the work done is negative 138 units of work. The displacement during the first
stage of motion is the vector 𝐀𝐁. And we can calculate this by
subtracting the position vector of point 𝐴 from the position vector of point
𝐵. We have negative four 𝐢 plus six
𝐣 minus negative seven 𝐢 minus 𝐣. Negative four minus negative seven
is equal to three, and six minus negative one is seven. Therefore, vector 𝐀𝐁 is equal to
three 𝐢 plus seven 𝐣.

We now have a force vector 𝐅 and
displacement vector 𝐝 for the first stage of motion. And we know that the scalar product
of these two vectors is equal to 106. Since the scalar product of two
vectors is the sum of the products of their individual components, we have three 𝑎
plus seven 𝑏 is equal to 106. As this is an equation with two
unknowns, we will call it equation one and move on to consider the second stage of
motion.

This time, we begin by calculating
the vector 𝐁𝐂. This is the position vector of
point 𝐶 minus the position vector of point 𝐵. Negative eight 𝐢 minus three 𝐣
minus negative four 𝐢 plus six 𝐣 is equal to negative four 𝐢 minus nine 𝐣. Once again, we will take the scalar
product of this displacement vector and the force vector 𝑎𝐢 plus 𝑏𝐣. During this stage of motion, we are
told the work done by the force was negative 138 units of work. And this gives us the equation
negative four 𝑎 minus nine 𝑏 is equal to negative 138.

As all three terms are negative, we
can multiply through by negative one, giving us four 𝑎 plus nine 𝑏 is equal to
138. We now have a pair of simultaneous
equations that we can solve to calculate the value of the constants 𝑎 and 𝑏. One way of solving a pair of
simultaneous equations is by elimination. In this question, we will eliminate
the variable 𝑎 by firstly multiplying equation one by four and equation two by
three. This gives us the two equations
12𝑎 plus 28𝑏 is equal to 424 and 12𝑎 plus 27𝑏 is equal to 414. Subtracting these two equations,
the terms containing 𝑎 cancel, and we are left with 𝑏 is equal to 10.

We can then substitute this value
of 𝑏 into anyone of our equations to calculate the value of 𝑎. Substituting into equation one, we
have three 𝑎 plus seven multiplied by 10 is equal to 106. Seven multiplied by 10 is 70. And subtracting this from both
sides, we have three 𝑎 is equal to 36. We can then divide through by
three, giving us 𝑎 is equal to 12. The value of the two constants 𝑎
and 𝑏 are 𝑎 is equal to 12 and 𝑏 is equal to 10. This means that the force acting on
the particle is equal to 12𝐢 plus 10𝐣.