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Question Video: Finding the Cross Product of Vectors Mathematics • 12th Grade

If 𝐀 = βŸ¨βˆ’2, βˆ’4, βˆ’1⟩, 𝐁 = βŸ¨βˆ’4, 1, 5⟩, and 𝐂 = ⟨0, 3, 0⟩, find 𝐀 Γ— (𝐁 + 𝐂).

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Video Transcript

If vector 𝐀 is equal to negative two, negative four, negative one; vector 𝐁 is equal to negative four, one, five; and vector 𝐂 is equal to zero, three, zero, find the cross product of vector 𝐀 and vector 𝐁 plus 𝐂.

We recall that when calculating the cross product of two vectors in three dimensions, our answer will be a vector that is perpendicular to the original two vectors. The cross product of any two vectors 𝐏 and 𝐐 is calculated by working out the determinant of the three-by-three matrix shown. The top row of our matrix will be the unit vectors 𝐒, 𝐣, and 𝐀. The second row contains the components of our first vector, in this case, 𝑃 sub π‘₯, 𝑃 sub 𝑦, and 𝑃 sub 𝑧. The bottom row of the matrix contains the components of the second vector.

In this question, we are trying to calculate the cross product of vector 𝐀 and the vector 𝐁 plus 𝐂. We will begin by calculating the vector 𝐁 plus 𝐂. This is equal to negative four, one, five plus zero, three, zero. When adding any two vectors, we simply add the corresponding components, in this case, negative four and zero, one and three, and five and zero. The vector 𝐁 plus 𝐂 is therefore equal to negative four, four, five. We can now calculate 𝐀 cross 𝐁 plus 𝐂. This is equal to the determinant of the three-by-three matrix 𝐒, 𝐣, 𝐀, negative two, negative four, negative one, negative four, four, five.

This determinant will contain three terms. Firstly, the unit vector 𝐒 multiplied by the determinant of the two-by-two matrix negative four, negative one, four, five. This is equal to the unit vector 𝐒 multiplied by negative 20 minus negative four which in turn simplifies to negative 16𝐒. The second term is the negative unit vector 𝐣 multiplied by the determinant of the two-by-two matrix negative two, negative one, negative four, five. This is equal to negative 𝐣 multiplied by negative 10 minus four which is equal to 14𝐣. Finally, we have the unit vector 𝐀 multiplied by the determinant of the two-by-two matrix negative two, negative four, negative four, four. This is equal to 𝐀 multiplied by negative eight minus 16 which is negative 24𝐀.

If vector 𝐀 is equal to negative two, negative four, negative one; vector 𝐁 is equal to negative four, one, five; and vector 𝐂 is equal to zero, three, zero, then 𝐀 cross 𝐁 plus 𝐂 is equal to negative 16𝐒 plus 14𝐣 minus 24𝐀. This could also be written as a vector in terms of its components negative 16, 14, negative 24.

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