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If vector 𝐀 is equal to negative five, zero, one and vector 𝐁 is equal to three, one, negative three, find 𝐀 cross 𝐀 minus two 𝐁.
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In this question, we need to find the cross product of two vectors.
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We recall that when finding the cross product, the answer is a vector that is perpendicular to the original two vectors.
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Given two three-dimensional vectors 𝐂 and 𝐃, their cross product is equal to the determinant of the three-by-three matrix shown, where the top row is made up of the unit vectors 𝐢, 𝐣, and 𝐤.
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The second row is made up of the components of vector 𝐂, denoted 𝐂 sub 𝑥, 𝐂 sub 𝑦, and 𝐂 sub 𝑧.
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The bottom row of the matrix is made up of the components of vector 𝐃.
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In this question, we need to find the cross product of vector 𝐀 and the vector 𝐀 minus two 𝐁.
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We can begin by working out the vector two 𝐁.
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We do this by multiplying vector 𝐁 by the scalar or constant two.
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Multiplying each of the components by the scalar two gives us the vector six, two, negative six.
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We need to subtract this vector from vector 𝐀.
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When subtracting two vectors, we simply subtract the corresponding components, in this case negative five minus six, zero minus two, and one minus negative six.
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This gives us the vector negative 11, negative two, seven.
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We can now find the cross product of vector 𝐀 and the vector 𝐀 minus two 𝐁.
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This will be equal to the determinant of the three-by-three matrix 𝐢, 𝐣, 𝐤, negative five, zero, one, negative 11, negative two, seven.
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We calculate the determinant of a three-by-three matrix in three parts.
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Firstly, we multiply the unit vector 𝐢 by the determinant of the two-by-two matrix zero, one, negative two, seven.
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This gives us the unit vector 𝐢 multiplied by zero minus negative two, which in turn is equal to two 𝐢.
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Next, we multiply the negative of the unit vector 𝐣 by the determinant of the two-by-two matrix negative five, one, negative 11, seven.
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This is equal to negative 𝐣 multiplied by negative 35 minus negative 11, which is equal to 24𝐣.
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Finally, we multiply the unit vector 𝐤 by the determinant of the two-by-two matrix negative five, zero, negative 11, negative two.
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This is equal to 10𝐤.
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The determinant of our three-by-three matrix is two 𝐢 plus 24𝐣 plus 10𝐤.
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We can therefore conclude that this is the cross product of vector 𝐀 and the vector 𝐀 minus two 𝐁.
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Our final answer is two 𝐢 plus 24𝐣 plus 10𝐤.
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This could also be written in component form: two, 24, 10.