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If vector π is equal to negative π’ minus two π£, vector π is equal to negative four π’ minus four π£, the cross product of vector π and π is negative three π€, and the cross product of vector π and vector π is four π€, find vector π.
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In this question, weβre given vectors π and π which lie in the coordinate plane with π’ and π£ as unit vectors.
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We are also given the cross product of vector π and π together with the cross product of vector π and π.
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And we are asked to find vector π.
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We will let this vector be π₯π’ plus π¦π£, where π₯ and π¦ are constants.
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We recall that if we have two vectors π¦ and π§ such that π¦ is equal to ππ’ plus ππ£ and π§ is equal to ππ’ plus βπ£, then the cross product of vectors π¦ and π§ is equal to the determinant of the two-by-two matrix π, π, π, β multiplied by the unit vector π€, where the unit vector π€ is perpendicular to the π₯π¦-plane.
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Letβs begin by considering the cross product of vectors π and π.
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Using our general rule, this must be equal to the determinant of the two-by-two matrix negative one, negative two, π₯, π¦ multiplied by the unit vector π€.
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The determinant of a matrix is equal to negative one multiplied by π¦ minus π₯ multiplied by negative two.
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This is equal to negative π¦ plus two π₯.
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So the cross product of vector π and vector π is negative π¦ plus two π₯ multiplied by the unit vector π€.
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We already know that this is equal to negative three π€.
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We can therefore conclude that negative three is equal to negative π¦ plus two π₯.
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Adding π¦ and three to both sides of this equation, we have π¦ is equal to two π₯ plus three.
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We will call this equation one and now repeat the process for the cross product of vector π and π.
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This is equal to the determinant of the two-by-two matrix π₯, π¦, negative four, negative four multiplied by π€.
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The determinant of the matrix is π₯ multiplied by negative four minus negative four multiplied by π¦.
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This simplifies to negative four π₯ plus four π¦ such that the cross product of π and π is negative four π₯ plus four π¦ multiplied by the unit vector π€.
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We are told that this is equal to four π€.
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And negative four π₯ plus four π¦ is therefore equal to four.
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We can divide both sides of this equation by four such that one is equal to negative π₯ plus π¦.
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And adding π₯ to both sides, we have π¦ is equal to π₯ plus one.
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We now have a pair of simultaneous equations we can solve to calculate the values of π₯ and π¦.
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Since π¦ is equal to two π₯ plus three and π₯ plus one, then these two expressions must be equal to one another.
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We can then subtract π₯ and three from both sides of the equation.
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Two π₯ minus π₯ is π₯, and one minus three is negative two, giving us π₯ is equal to negative two.
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We can then substitute this value of π₯ into equation one or equation two.
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In Equation two, we have π¦ is equal to negative two plus one, which gives us π¦ is equal to negative one.
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Vector π is therefore equal to negative two π’ minus π£.
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If π is equal to negative π’ minus two π£, π is equal to negative four π£ minus four π£, the cross product of π and π is negative three π€, and the cross product of π and π is four π€, then vector π is equal to negative two π’ minus π£.