### Video Transcript

If the force πΉ, where πΉ equals negative two π plus πΏ π minus nine π, is acting on the point π΄ β four, five, negative two β and the moment π sub π΅ of the force about the point π΅ β negative four, negative four, three β is negative 91π plus 82π plus two π, determine the value of πΏ.

Given the force πΉ and also given the coordinates of points π΄ and π΅ and, finally, given the moment π sub π΅ of the force about the point π΅, we want to solve for the value of πΏ given in the equation for πΉ. Weβll start off by drawing in the points π΄ and π΅ on a set of coordinate axes.

With our π-, π-, and π-directions written in, we can locate points π΄ and π΅ on this graph. Point π΄ has coordinates π equals four π, equals five, and π equals negative two. And point π΅ has coordinates π equals negative four, π equals negative four, and π equals three. The force πΉ acts from point π΄. Weβre not entirely sure what direction because we donβt know itβs π-component.

If we connect points π΄ and π΅ by a vector, we can call that vector π
. Itβs the distance vector between the point π΄ at which the force acts and the point π΅ at which our moment is measured. Based on our diagram, we can say that π
is equal to π΄ minus π΅. And when we enter in the coordinates of those two points and calculate π
, we find itβs a vector with components eight π, nine π, and negative five π.

Now weβll record π
, πΉ, and π sub π΅ off to the side so we can begin to solve for πΏ. Now that we know the displacement vector π
, the force πΉ, and the moment about point π΅, π sub π΅, we can recall that this moment is equal to the cross product of π
and πΉ. This cross product is equal to the determinant of this matrix, with our unit vectors written in followed by the components of our two vectors.

When we apply this relationship to our scenario, we see itβs the πΉ sub π-component we want to solve for. If we know that, then weβll know πΏ. We can figure out πΉ sub π or πΏ by focusing either on the π- or the π-component of π sub π΅. Just to choose one of those two ways, letβs focus on the π-component and let that lead us to the value of πΏ.

The πth component of the moment, which we can call π sub π΅π, is equal to π
sub π sub times πΉ sub π minus π
sub π times πΉ sub π. And we can replace πΉ sub π with πΏ because thatβs what it is. Now the πth component of π sub π΅ is negative 91. Thatβs equal to the πth component of π
, which is positive nine, times the πth component of πΉ, which is negative nine, minus the πth component of π
, negative five, multiplied by πΏ.

Negative nine times nine is negative 81. And adding 81 to both sides gives us the expression negative 10 is equal to five πΏ, or πΏ is equal to negative two. Thatβs the value of the πth component of πΉ.