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Question Video: Finding the Dot and the Cross Product of Vectors Mathematics

Given 𝐀 = (1, 5, βˆ’5), 𝐁 = (2, 4, 3), and 𝐂 = (0, 5, βˆ’4), find 𝐀 β‹… (𝐁 Γ— 𝐂).

02:13

Video Transcript

Given vectors 𝐀, 𝐁, and 𝐂, where 𝐀 has components one, five, negative five; 𝐁 has components two, four, three; and 𝐂 has components zero, five, negative four, find the scalar triple product of 𝐀, 𝐁, and 𝐂.

We’re asked to calculate the scalar triple product of vectors 𝐀, 𝐁, and 𝐂. And we know that this is equivalent to calculating the determinant of the matrix consisting of the components of the three vectors as shown. So with our vectors 𝐀, 𝐁, and 𝐂, we want to find the determinant of the matrix whose first line consists of the components of vector 𝐀, whose second line has components of vector 𝐁, and whose third line has the components of vector 𝐂.

And remember to calculate the determinant of a three-by-three matrix using the first row as our pivot, we take the first top-left element 𝐀 π‘₯ and multiply this by the determinant of the two-by-two matrix in the bottom-right corner. We then take the negative of the second element in the top row and multiply this by the determinant shown. And finally, we add the third element in the top row, multiply it by the determinant formed by the four elements in the bottom-left-hand corner.

In our case, this translates into one times the two-by-two matrix with elements four, three, five, negative four minus five times the determinant of the matrix with elements two, three, zero, and negative four plus a negative five times the determinant of the two-by-two matrix with elements two, four, zero, five. And now remembering that the determinant of a two-by-two matrix with elements π‘Ž, 𝑏, 𝑐, 𝑑 is π‘Žπ‘‘ minus 𝑏𝑐, we have one times four times negative four minus three times five minus five times two times negative four minus three times zero minus five again times two times five minus four times zero.

That is one times negative 16 minus 15 minus five times negative eight minus zero minus five times 10 minus zero. Evaluating this gives us negative 31 plus 40 minus 50, which is negative 41. The scalar triple product of vectors 𝐀, 𝐁, and 𝐂 is therefore negative 41.

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