Question Video: Operations on Vectors Mathematics

If 𝚨 = βŸ¨βˆ’2, 2, βˆ’1⟩, 𝚩 = βŸ¨βˆ’4, βˆ’4, βˆ’5⟩, and 𝐂 = βŸ¨βˆ’4, 2, 4⟩, find (𝚨 Γ— 𝚩) β‹… (𝚨 Γ— 𝐂).

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Video Transcript

If vector 𝚨 equals negative two, two, negative one; vector 𝚩 equals negative four, negative four, negative five; and vector 𝐂 equals negative four, two, four, find 𝚨 cross 𝚩 dotted with 𝚨 cross 𝐂.

Alright, to calculate this answer, we’ll need to take two cross products, 𝚨 cross 𝚩 and 𝚨 cross 𝐂, and then a dot product between the two resulting vectors. As we get started then, let’s recall how to calculate a three-dimensional cross product. If we have two three-dimensional vectors, we’ve called them here 𝚨 and 𝚩, then their cross product equals the determinant of this three-by-three matrix. The top row of the matrix has the three orthogonal unit vectors 𝐒, 𝐣, and 𝐀. After that comes the corresponding components of vector 𝚨 and then the corresponding components of vector 𝚩. And note that this order in which the last two rows are filled matters. 𝚨 cross 𝚩 is different from 𝚩 cross 𝚨.

But anyway, this relationship gives us a template for calculating 𝚨 cross 𝚩 and then 𝚨 cross 𝐂. 𝚨 cross 𝚩 is given by this matrix, where we still need to fill in the components of 𝚨 and 𝚩. Our problem statement tells us that the components of 𝚨 are negative two, two, and negative one and those of vector 𝚩 are negative four, negative four, negative five. We’re now ready to calculate this cross product, starting with the 𝐒-component. It’s equal to the determinant of this two-by-two matrix. Two times negative five is negative 10. And from that we subtract negative one times negative four or positive four.

Now, we’ll move on to the 𝐣-component, which is equal to negative the determinant of this matrix. Negative two times negative five is positive 10 and from that we subtract negative one times negative four or four, and then lastly the 𝐀-component equal to the determinant of this matrix. Negative two times negative four is positive eight minus two times negative four or negative eight. Combining these components this way gives us 𝚨 cross 𝚩, which we see simplifies to negative 14𝐒 minus six 𝐣 plus 16𝐀.

So then, we’ve calculated our first cross product 𝚨 cross 𝚩. Now let’s move on to 𝚨 cross 𝐂. This matrix looks identical to the matrix for 𝚨 cross 𝚩 except for our last row. Note that here we have the components of vector 𝐂, negative four, positive two, and positive four. Once again, we’ll calculate this cross product starting with the 𝐒-component. The determinant of this two-by-two matrix is two times four or eight minus negative one times positive two or negative two. Our 𝐣-component is given as negative the determinant of this matrix, where that determinant is negative two times four or negative eight minus negative one times negative four or positive four.

Finally, the 𝐀-component with this determinant. Negative two times two is negative four minus two times negative four or negative eight. Combined this way, these components add up to the cross product 𝚨 cross 𝐂. They simplify to 10𝐒 plus 12𝐣 plus four 𝐀. We’re now ready to take our last mathematical step, computing the dot product of these two cross products. Recalling that in general the dot product of two three-dimensional vectors equals the product of their respective components added together, we can say that the dot product of 𝚨 cross 𝚩 and 𝚨 cross 𝐂 equals negative 14 times 10 plus negative six times 12 plus 16 times four.

Notice that we’ve multiplied together the corresponding components of our two vectors. And when we add these products together, we find they’re equal to negative 148. And that’s our answer. The dot product of 𝚨 cross 𝚩 and 𝚨 cross 𝐂 is negative 148.

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