Question Video: Rules of the Scalar Triple Product | Nagwa Question Video: Rules of the Scalar Triple Product | Nagwa

Question Video: Rules of the Scalar Triple Product Mathematics • Third Year of Secondary School

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Consider 𝐀 = <1, 2, 3>, 𝐁 = <βˆ’4, βˆ’5, 3>, and 𝐂 = <1, 2, βˆ’2>. Find 𝐀 β‹… (𝐁 Γ— 𝐂). Find 𝐁 β‹… (𝐀 Γ— 𝐂).

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Video Transcript

Consider the vector 𝐀 equals one, two, three; 𝐁 equals negative four, negative five, three; and 𝐂 equals one, two, negative two. Find the scalar triple product of 𝐀, 𝐁, and 𝐂. Find the scalar triple product of 𝐁, 𝐀, and 𝐂.

When we read these questions, we defined this expression to be the scalar triple product of our three vectors. To find the scalar triple product as represented, we find the dot product of the vector 𝐀 with the cross product of 𝐁 and 𝐂. So, of course, we could begin by finding the cross product of 𝐁 and 𝐂 and then finding the dot product of that with 𝐀, or we can use a formula. The scalar triple product of 𝐀, 𝐁, and 𝐂, which doesn’t actually need these parentheses, where the vector 𝐀 has elements 𝐴 sub π‘₯, 𝐴 sub 𝑦, and 𝐴 sub 𝑧 and the vector 𝐁 has elements 𝐡 sub π‘₯, 𝐡 sub 𝑦, 𝐡 sub 𝑧 and 𝐂 is 𝐢 sub π‘₯, 𝐢 sub 𝑦, 𝐢 sub 𝑧, is given by the determinant of the three-by-three matrix containing these elements.

This means the scalar triple product of 𝐀, 𝐁, and 𝐂 is going to be equal to the determinant of the three-by-three matrix that contains the elements one, two, three; negative four, negative five, three; and one, two, negative two. So, next, we’ll remind ourselves how to find this determinant. We take the elements in the top row and multiply them by the determinant of the two-by-two matrix that remains when we eliminate the elements in that row and that column. So, in this case, we multiply one by the determinant of the matrix negative five, three, two, negative two. And of course the determinant of this two-by-two matrix is negative five times negative two minus three times two. We then subtract two times the determinant of the two-by-two matrix with elements negative four, three, one, negative two.

Finally, we’re going to add three times the determinant of the remaining two-by-two matrix. We’re multiplying that three by negative four times two minus negative five times one. This all simplifies to four minus 10 plus negative nine. And that’s equal to negative 15. And so the scalar triple product of 𝐀, 𝐁, and 𝐂 is negative 15.

The second part of this question wants us to find the scalar triple product of the matrices 𝐁, 𝐀, and 𝐂. And so we could repeat the process from earlier, finding the determinant of negative four, negative five, three, one, two, three, one, two, negative two. But actually, there are some properties of the scalar triple product that can save us some time. For instance, let’s imagine what happens if we switch two vectors in our scalar triple product. Swapping two horizontal rows in a three-by-three determinant changes its sign. So, rather than working out this full determinant, we can see that it contains the same values as the previous determinant. It’s just that two rows are swapped. This means we can change its sign. And so the scalar triple product of 𝐁, 𝐀, and 𝐂 is positive 15.

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