# Question Video: Finding Two Unknown Components in a Force Vector given the Vector of the Moment of the Force about the Origin in Three Dimensions Mathematics

If π = β19π’ + πΏπ£ + 2π€ acts at a point π΄(β3, 5, β3) and the moment of π about the origin point is equal to 4π’ + 63π£ + 101π€, find the value of πΏ.

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### Video Transcript

If π equal to negative 19π’ plus πΏπ£ 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 πΏ.

So here weβre thinking about force π which acts at a point π΄. And weβre 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 πΏ. And itβs our job to find the value of πΏ. Because this question talks about the moment vector thatβs produced by a three-dimensional force acting at a certain point in space, letβs 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 weβve been given in the question. π is the displacement vector of the point at which the force acts relative to the point about which weβre calculating moments.

In this question, weβre told that the force acts at point π΄ with coordinates negative three, five, negative three. And the question talks about a moment thatβs calculated about the origin point. This means that the vector π weβll use in our calculation is equal to the displacement vector that goes from the origin to point π΄. This is the same as saying itβs the displacement vector of π΄ relative to the origin. This vector is easy to find. Weβre 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. Itβs equal to four π’ plus 63π£ plus 101π€. So looking at this equation, it seems like we know a lot about the vectors used. Weβre 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, itβs 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 isnβt correct. Firstly, it isnβt possible to divide a vector by another vector. And secondly, this equation uses a cross product. This isnβt the same as general multiplication. So even if we could divide both sides of the equation by the vector π, this wouldnβt actually help us rearrange the equation. In fact, the cross product operation doesnβt actually have an inverse, which means thereβs no way of rearranging this equation to make π the subject.

So, how do we find this unknown component of π? Well, hereβs the idea. First, we obtain an expression for the moment vector π by taking the vector π as itβs given in this question and substituting it into this equation along with π. Since π is expressed in terms of this unknown quantity πΏ, the expression we end up with for π will also be in terms of πΏ. Next, we compare this expression for π, which is in terms of πΏ, 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, weβll be able to form equations that we can solve to find the value of πΏ. So letβs 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. Letβs 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 πΏπ£ plus two π€. So the elements in the bottom row of this determinant are negative 19, πΏ, 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 πΏ. 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 πΏ 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 weβre subtracting negative three πΏ, which is the same as adding three πΏ. Looking at the next term, negative three times two is negative six. And weβre then subtracting negative three times negative 19. Well, negative three times negative 19 is positive 57. So weβre 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 πΏ minus five times negative 19. Five times negative 19 is negative 95. And since weβre subtracting this, this is equivalent to adding 95. And just to make things a bit more consistent, letβs swap around the negative three πΏ and positive 95 to give us 95 minus three πΏ. And finally, weβll move the unit vectors π’ and π€ to the end of those terms.

Okay, so now weβve completed the first thing on our list. We obtained an expression for the moment vector π in terms of this unknown quantity πΏ.

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 weβve 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 weβre 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 πΏ 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 πΏ is equal to 101. Both of these equations are linear equations with a single unknown quantity, πΏ. This means that solving either one of these equations should tell us the value of πΏ. If we start with the equation on the left, we can see that subtracting 10 from both sides gives us three πΏ equals negative six. And then dividing both sides by three tells us that πΏ is equal to negative two.

Just to be safe, letβs 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 πΏ equals six. And then dividing both sides of the equation by negative three also gives us πΏ equals negative two. So thereβs 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 weβre 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 πΏπ£ 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 πΏ must be negative two.