### Video Transcript

The moment of the force πΉ about the origin is π sub zero, where πΉ equals π minus two π minus π and π sub zero equals 20π plus 27π minus 34π. Given that the force passes through a point whose π¦-coordinate is four, find the π₯- and π§-coordinates of the point.

We can start by drawing a sketch of the force, πΉ, on a three-dimensional set of axes. Weβre told in the statement that thereβs a force called πΉ which acts at a point we can call π. The coordinates of π are only partially known. We know it has a π¦-value of four but its π₯- and π§-values are to be determined. The force πΉ acting at point π, some distance away from the origin, creates a moment about the origin, π sub zero. Since π sub zero is acting around the origin, we can define π, not just as a single point in space, but as a vector that connects πΉ with π sub zero.

Using this notation, we can now recall a relationship that connects π sub zero, π, and πΉ. The cross product of two vectors π΄ and π΅ is equal to the determinant of this matrix, where our columns are the π-, π-, and π-directions. And the last two rows are the components of vectors π΄ and π΅, respectively. If we apply this relationship to our scenario, we can say that π sub zero equals π cross πΉ, which can also be written as the determinant of this matrix. When we consider the last two rows of this matrix, we see we can fill in for these components of π and πΉ. ππ is π₯, ππ is four, and ππ is π§. And πΉπ is one, πΉπ is negative two, and πΉπ is negative one.

We want to solve for the unknown components of π, π₯, and π§. And remember, we know all three components. They complete information about π sub zero. We can start off solving for π₯ by calculating the π-component of π sub zero. That component, which we can call π sub zero π, is equal to negative two times π₯ minus four times one. And as we look at our expression for π sub zero, we see that its π-component is negative 34. So negative 34 is equal to negative two π₯ minus four. Or, π₯ equals negative 30 over negative two which equals 15. Thatβs the π₯-coordinate of π.

Now that weβve solved for π₯, we just need to solve for the π§-component of π. To do that, we can look at the π-component of π sub zero. π sub zero π is equal to four times negative one minus negative two times π§, which we can simplify to two π§ minus four. Now π sub zero π is equal to 20. And when we solve this equation for π§, we find it equals 24 over two or 12. Thatβs the π§-coordinate of π. And now we know all three coordinates of that point.