### Video Transcript

If π
equal to negative 19π’ plus πΏπ£ plus two π€ acts at a point π΄ with coordinates negative three, five, negative three and the moment of π
about the origin point is equal to four π’ plus 63π£ plus 101π€, find the value of πΏ.

So here weβre thinking about force π
which acts at a point π΄. And weβre told that this force produces a moment about the origin that can be represented by this vector. We can notice that we have numerical values for the π₯- and π§-components of π
, but the π¦-component is represented by this constant πΏ. And itβs our job to find the value of πΏ. Because this question talks about the moment vector thatβs produced by a three-dimensional force acting at a certain point in space, letβs start by recalling the equation π equals π cross π
.

This equation tells us that we can calculate a moment vector by taking the cross product of a displacement vector called π and a force vector π
. In this equation, π
is simply the force that produces the moment. So this corresponds to the vector called π
that weβve been given in the question. π is the displacement vector of the point at which the force acts relative to the point about which weβre calculating moments.

In this question, weβre told that the force acts at point π΄ with coordinates negative three, five, negative three. And the question talks about a moment thatβs calculated about the origin point. This means that the vector π weβll use in our calculation is equal to the displacement vector that goes from the origin to point π΄. This is the same as saying itβs the displacement vector of π΄ relative to the origin. This vector is easy to find. Weβre told in the question that the coordinates of π΄ are negative three, five, negative three. In other words, the point π΄ lies negative three units in the π₯-direction, five units in the π¦-direction, and negative three units in the π§-direction from the origin. This means that the displacement vector that takes us from the origin to π΄ is given by negative three π’ plus five π£ minus three π€.

We can also note that the moment vector π is given in the question. Itβs equal to four π’ plus 63π£ plus 101π€. So looking at this equation, it seems like we know a lot about the vectors used. Weβre told π in the question, we can easily figure out the components of π, and we know two of the three components of π
. The only unknown quantity in this equation is the π¦-component of π
.

Now, given that we want to find this unknown component of π
, itβs tempting to try to rearrange this equation to make π
the subject. For example, we might try to divide both sides of the equation by the vector π, as this looks like it might tell us the vector π
, including its unknown component. Unfortunately, this idea isnβt correct. Firstly, it isnβt possible to divide a vector by another vector. And secondly, this equation uses a cross product. This isnβt the same as general multiplication. So even if we could divide both sides of the equation by the vector π, this wouldnβt actually help us rearrange the equation. In fact, the cross product operation doesnβt actually have an inverse, which means thereβs no way of rearranging this equation to make π
the subject.

So, how do we find this unknown component of π
? Well, hereβs the idea. First, we obtain an expression for the moment vector π by taking the vector π
as itβs given in this question and substituting it into this equation along with π. Since π
is expressed in terms of this unknown quantity πΏ, the expression we end up with for π will also be in terms of πΏ. Next, we compare this expression for π, which is in terms of πΏ, to the actual components of π, which are given in the question. By comparing the components from our expression to the components given in the question, weβll be able to form equations that we can solve to find the value of πΏ. So letβs start by calculating an expression for π.

We know that π is equal to the cross product of π, which we found, and π
, which is given in the question. This cross product is given by this three-by-three determinant, where the elements in the top row are the unit vectors π’, π£, and π€. The elements in the middle row are the π₯-, π¦-, and π§-components of the displacement vector π, written without their unit vectors. And the elements in the bottom row are the components of π
, also written without unit vectors. Letβs start by substituting in the components of π. The π₯-component of π is negative three π’. So we write negative three here in our determinant. The π¦-component of π is five π£. So we write five into this spot. And the π§-component is negative three π€. So we write negative three here.

Next, we do the same thing for the components of the vector π
. We know that π
is equal to negative 19π’ plus πΏπ£ plus two π€. So the elements in the bottom row of this determinant are negative 19, πΏ, and two. Next, we need to compute this determinant. This is done in three parts. First, we have the unit vector π’ multiplied by five times two minus negative three times πΏ. Next, we subtract the unit vector π£ hat multiplied by negative three times two minus negative three times negative 19. And finally, we add the unit vector π€ multiplied by negative three times πΏ minus five times negative 19.

We can now simplify each of these terms. Looking at the first term, we have five times two, which is 10. And weβre subtracting negative three πΏ, which is the same as adding three πΏ. Looking at the next term, negative three times two is negative six. And weβre then subtracting negative three times negative 19. Well, negative three times negative 19 is positive 57. So weβre left with negative six minus 57, which is negative 63. Because we have negative 63 times negative π£, in total, this term becomes positive 63π£.

Now, looking at the final term, we can see that inside the parentheses we have negative three πΏ minus five times negative 19. Five times negative 19 is negative 95. And since weβre subtracting this, this is equivalent to adding 95. And just to make things a bit more consistent, letβs swap around the negative three πΏ and positive 95 to give us 95 minus three πΏ. And finally, weβll move the unit vectors π’ and π€ to the end of those terms.

Okay, so now weβve completed the first thing on our list. We obtained an expression for the moment vector π in terms of this unknown quantity πΏ.

We can now move on to the second part of our solution, comparing our expression for π to the actual components of π that are given in the question. The idea behind doing this is that the expression weβve found describes exactly the same moment as this vector given in the question. In fact, we can see that the π¦-component of our expression is the same as the π¦-component given in the question, which is a good sign that weβre on the right track. The fact that these two vectors express the same quantity means that the π₯-components and π§-components of these vectors must also be the same.

This enables us to come up with two equations. Firstly, because the π₯-component of our expression must be equal to the π₯-component of this vector, we can say that 10 plus three πΏ must be equal to four. And because the π§-component of our expression must be equal to the π§-component given in the question, we can also say that 95 minus three πΏ is equal to 101. Both of these equations are linear equations with a single unknown quantity, πΏ. This means that solving either one of these equations should tell us the value of πΏ. If we start with the equation on the left, we can see that subtracting 10 from both sides gives us three πΏ equals negative six. And then dividing both sides by three tells us that πΏ is equal to negative two.

Just to be safe, letβs solve this equation too and see if it gives us the same answer. Subtracting 95 from both sides of the equation gives us negative three πΏ equals six. And then dividing both sides of the equation by negative three also gives us πΏ equals negative two. So thereβs our final answer. By coming up with our own expression for π using the information given in the question and comparing this to the numerical values of the components of π that weβre given in the question as well, we were able to find the unknown component of the force.

So, when a force equal to negative 19π’ plus πΏπ£ plus two π€ acts at the point π΄ with coordinates negative three, five, negative three and we know that the moment produced about the origin point is equal to four π’ plus 63π£ plus 101π€, the value of πΏ must be negative two.