### Video Transcript

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

So in this question, weβre told that we have some point whose position vector π« relative to the origin is 14π’ hat minus π£ hat plus 12π€ hat. Weβre told that some force given by this vector π
is acting at this point. As we can see, the π₯- and π¦-components of this vector π
are represented by the constants π and π. And itβs our job to find the values of these constants. To enable us to do this, weβve been given some information about the moment of the force π
, in other words, the moment thatβs produced by π
. Specifically, weβre told that the component of this moment about the π₯-axis is 73 and the component of this moment about the π¦-axis is 242.

Letβs recall that a force acting at a distance from a point produces a moment about that point. So 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, weβre told some of the components of π, specifically the components about the π₯-axis and the π¦-axis.

Now, whenever we think about components of a moment about our coordinate axes, generally, we want to consider the overall moment about the origin. The reason for this is that if we calculate a moment vector about the origin, then the components of this moment vector are the components that cause rotation about the coordinate axes. So the π₯-component of π, for example, tells us the component of that moment about the π₯-axis. Likewise, the π¦-component of a moment vector calculated about the origin is the component about the π¦-axis. And the π§-component is the component about the π§-axis.

So when the question tells us that the components of the moment about the π₯-axis and the π¦-axis are 73 and 242, respectively, this means that if we were to calculate the moment vector about the origin, it would have an π₯-component of 73, which weβll write 73 times π’ hat, and a π¦-component of 242, which weβll write 242π£ hat. And since we donβt know the π§-component, weβll just call this π sub π§ times π€ hat. Remember that this is the moment vector about the origin. This means itβs equal to the cross product of the position vector, where the force acts relative to the origin and the force vector itself.

This position vector is the one denoted by a lowercase π« in the 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
. Itβs important to remember that this is a vector cross product, not just a multiplication. So itβs not possible to make π
the subject of the equation just by dividing both sides by π«. Instead, we actually need to compute this cross product. And the way that we do this is by evaluating this three-by-three determinant. Here, the elements in the top row are the unit vectors π’ hat, π£ hat, and π€ hat. The elements in the middle row are the π₯-, π¦-, and π§-components, respectively, of the position vector π«. And the elements in the bottom row are the π₯-, π¦-, and π§-components of the force vector π
.

Note that the order in which these two vectors are written down determines the order that we write them into this determinant. The vector π« is written down first in this cross product, which means we write it into the middle row of this determinant. And the vector π
, which is written second, goes into the bottom row. Writing these two vectors the other way around results in a different answer.

The next step is to populate the determinant with the different components of the vectors π« and π
. The π₯-component of π« is 14π’. So in its place, we write 14 in the determinant. Note that we donβt include the unit vector. The π¦-component of π« is negative π£ which without the unit vector is negative one. And the π§-component of π« is 12π€, so we write 12 into our determinant. Then in the bottom row of the determinant, we write the components of the force vector π
but, again, without their unit vectors, giving us π, π, and negative one.

Next, weβre ready to calculate this determinant. This is effectively done in three parts. First, we have the unit vector π’ hat multiplied by negative one times negative one minus 12 times π. Next, we subtract the unit vector π£ hat multiplied by 14 times negative one minus 12π. And finally, we have π€ hat multiplied by 14 times π minus negative one times π. We can then simplify each of these terms. Looking at the first term, negative one times negative one can be simplified to just one. So this term can be written as π’ hat times one minus 12π.

Looking at the second term, 14 times negative one can be simplified to negative 14. And since π£ hat is negative and everything inside the parentheses is negative, we can simplify this further by taking out a factor of negative one. In other words, we change negative π£ hat to positive π£ hat, and we change the sign of everything inside the parentheses, which leaves us with π£ hat times 14 plus 12π. Looking at the last term inside the parentheses, we start with 14π and weβre subtracting negative one times π. Negative one times π is just negative π and subtracting negative π is the same as adding positive π.

So altogether, we found that the cross product of π« and π
is equal to π’ hat times one minus 12π plus π£ hat times 14 plus 12π plus π€ hat times 14π plus π. This is an algebraic expression for the moment vector in terms of π and π. In order to find the values of π and π, we can equate the π’-components and the π£-components of the moment as it was given in the question and the expression for that moment that we calculated using π« and π
.

Letβs start by equating the π₯-components of these expressions. This gives us 73π’ hat equals one minus 12π times π’ hat. We can start by removing π’ hat from this equation because it appears on both sides, subtracting one from both sides of the equation and then dividing both sides by negative 12. 72 divided by negative 12 is negative six. And this is half of our answer.

Now, we just need to find π. And to do this, we equate the π¦-components of both of our expressions for π. This gives us 242 equals 14 plus 12π. Subtracting 14 from both sides of the equation gives us 228 equals 12π, and then we can divide both sides of the equation by 12 to give us 228 over 12 equals π. 228 over 12 is 19 which means π is 19. And this is the second half to our answer. The two previously unknown components of our force vector π
are given by π equals 19 and π equals negative six.