### Video Transcript

If the force π
, which is equal to ππ’ plus ππ£ minus π€, is acting at a point whose position vector is π« equal to 14π’ minus π£ plus 12π€ and the π₯- and π¦-components of the moment of the force π
about the origin are 73 and 242 units of moment, respectively, find the values of π and π.

In this question, we are told that we have a point with position vector π« equal to 14π’ minus π£ plus 12π€. There is a force vector π
equal to ππ’ plus ππ£ minus π€ acting at this point. The π₯- and π¦-components of this vector π
are represented by the constants π and π. And it is our job in this question to find these values.

In order to do this, we have been given some information about the moment of the force π
, that is, the moment produced by π
. Specifically, we are told that the π₯- and π¦-components of this moment are 73 and 242 units, respectively. We recall that a force acting at a distance from a point produces a moment about that point. In this question, we have a force which is acting at some distance from the origin, and this is producing a moment.

A moment can be represented with a vector, which we generally call π¦. And we can calculate π¦ by finding the cross product of the vectors π and π
, where π
is the force vector which produces a moment about some point and π is the position vector at which π
acts relative to that point. Now, in this question, we are told some of the components of π¦, specifically the π₯- and π¦-components. The π₯-component is 73, and the π¦-component, 242. We can therefore write the moment vector π¦ as 73π’ plus 242π£ plus π sub π§ π€, where π sub π§ is unknown as we are not given the π§-component.

This moment vector will be equal to the cross product of the position vector where the force acts and the force vector itself. The position vector is denoted by lowercase π« in this question. We can use this vector equation to find the solution to the question using the known components of the moment vector π¦ and the position vector π« to find the unknown π₯- and π¦-components of the force vector π
.

It is important to note at this stage that weβ²re using a vector cross product and not just a multiplication. This means that it is not possible to make π
the subject of the equation. Instead, we need to compute the cross product. And we do this by evaluating the three-by-three determinant shown. The elements in the first row are the unit vectors π’, π£, and π€. In the second row, we have the π₯-, π¦-, and π§-components, respectively, of the position vector π«. And in the third row, we have the π₯-, π¦-, and π§-components of the force vector π
. Note that the order here is important. We must write them in the same order in the determinant.

We can now populate the determinant with the components of vector π«. They are 14, negative one, and 12. Likewise, the components of vector π
are π, π, and negative one. We are now in a position to calculate the determinant. This is done in three parts. Firstly, we have the unit vector π’ multiplied by negative one multiplied by negative one minus π multiplied by 12. This simplifies to π’ multiplied by one minus 12π. Next, we subtract the unit vector π£ multiplied by 14 multiplied by negative one minus π multiplied by 12. This simplifies to negative π£ multiplied by negative 14 minus 12π, which in turn can be rewritten as π£ multiplied by 14 plus 12π. Finally, we add the unit vector π€ multiplied by 14 multiplied by π minus π multiplied by negative one. This is equal to π€ multiplied by 14π minus negative π, which in turn equals π€ multiplied by 14π plus π.

We now have an expression for the determinant of the three-by-three matrix π’, π£, π€, 14, negative one, 12, π, π, negative one. As this is the cross product of vector π« and vector π
, this is also equal to vector π¦.

We are now in a position to find the values of π and π by equating the π’- and π£-components in our two equations. Equating the π’-components, we have 73 is equal to one minus 12π. We can subtract 73 and add 12π to both sides of this equation, giving us 12π is equal to negative 72. Dividing through by 12 gives us π is equal to negative six. We can now repeat this process for the π£-components. We have 242 is equal to 14 plus 12π. We solve this equation by firstly subtracting 14 from both sides. We can then divide through by 12 such that π is equal to 19.

We can therefore conclude that the two unknown components of force π
are π equals 19 and π equals negative six. And this means that the vector force π
is equal to 19π’ minus six π£ minus π€.