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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.
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Given that the force passes through a point whose π¦-coordinate is four, find the π₯- and π§-coordinates of the point.
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We can start by drawing a sketch of the force, πΉ, on a three-dimensional set of axes.
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Weβre told in the statement that thereβs a force called πΉ which acts at a point we can call π.
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The coordinates of π are only partially known.
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We know it has a π¦-value of four but its π₯- and π§-values are to be determined.
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The force πΉ acting at point π, some distance away from the origin, creates a moment about the origin, π sub zero.
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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.
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Using this notation, we can now recall a relationship that connects π sub zero, π, and πΉ.
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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.
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And the last two rows are the components of vectors π΄ and π΅, respectively.
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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.
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When we consider the last two rows of this matrix, we see we can fill in for these components of π and πΉ.
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ππ is π₯, ππ is four, and ππ is π§.
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And πΉπ is one, πΉπ is negative two, and πΉπ is negative one.
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We want to solve for the unknown components of π, π₯, and π§.
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And remember, we know all three components.
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They complete information about π sub zero.
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We can start off solving for π₯ by calculating the π-component of π sub zero.
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That component, which we can call π sub zero π, is equal to negative two times π₯ minus four times one.
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And as we look at our expression for π sub zero, we see that its π-component is negative 34.
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So negative 34 is equal to negative two π₯ minus four.
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Or, π₯ equals negative 30 over negative two which equals 15.
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Thatβs the π₯-coordinate of π.
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Now that weβve solved for π₯, we just need to solve for the π§-component of π.
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To do that, we can look at the π-component of π sub zero.
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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.
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Now π sub zero π is equal to 20.
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And when we solve this equation for π§, we find it equals 24 over two or 12.
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Thatβs the π§-coordinate of π.
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And now we know all three coordinates of that point.