Video: Calculating the Scalar Triple Product and Vector Triple Product of Three Vectors in 3D

𝐔, 𝐕, and 𝐖 are three vectors, where 𝐔 = 〈1, 0, 2βŒͺ, 𝐕 = βŒ©βˆ’1, 0, 3βŒͺ, and 𝐖 = 〈2, 0, βˆ’2βŒͺ. Calculate 𝐔 β‹… (𝐕 Γ— 𝐖) and 𝐔 Γ— (𝐕 Γ— 𝐖).

07:30

Video Transcript

𝐔, 𝐕, and 𝐖 are three vectors where 𝐔 has components one zero two 𝐕 has components negative one zero three and 𝐖 has components two zero negative two. Calculate 𝐔 dot 𝐕 cross 𝐖 and 𝐔 cross 𝐕 cross 𝐖.

This question involves the dot product and the cross product of vectors. We have to find the dot product of the vector 𝐔 with the vector 𝐕 cross 𝐖 and the cross product of the vector 𝐔 with the vector 𝐕 cross 𝐖. The common thread here is 𝐕 cross 𝐖, the cross product of the vectors 𝐕 and 𝐖, which is itself a vector.

So let’s first find 𝐕 cross 𝐖. We can write this cross product as the determinant of a three-by-three matrix where the first row of the matrix contains the unit vectors in the π‘₯-, 𝑦-, and 𝑧-directions; that is, the vectors 𝑖, 𝑗, and π‘˜. The second row contains the components of the first vector in the cross product, in this case 𝐕 with components negative one, zero, and three. And the third row contains the components of the second vector in the cross product, in our case 𝐖 with components two, zero, and negative two.

Now we evaluate this determinant. We can expand this determinant along the top row. So we get 𝑖 times the determinant of the matrix you get by deleting the row and column containing 𝑖 minus 𝑗 times the determinant of the matrix you get by deleting the row and column containing 𝑗 plus π‘˜ times the determinant of the matrix you get by deleting the row and column containing π‘˜.

We can evaluate the two-by-two determinants in the normal way. And so simplifying, we get zero 𝑖 plus four 𝑗 plus zero π‘˜, which we can write in component form as zero, four, zero. Now we found the components of 𝐕 cross 𝐖. Let’s clear some space and use them to find 𝐔 dot 𝐕 cross 𝐖 and 𝐔 cross 𝐕 cross 𝐖.

Okay, what is 𝐔 dot 𝐕 cross 𝐖? Well, we’re given the components of 𝐔 in the question. 𝐔 has components one, zero, two. And of course, we’ve just worked out the components of 𝐕 cross 𝐖. They are zero, four, and zero. And finding the dot product of two vectors in component form is very straightforward. It’s the product of the π‘₯-components plus the product of the 𝑦-components plus the product of the 𝑧-components.

And as whenever a component of 𝐔 is nonzero, the corresponding component of 𝐕 cross 𝐖 is zero. This sum is very easy. It’s zero plus zero plus zero, which is zero. Before we move on to finding the cross product of 𝐔 and 𝐕 cross 𝐖, let’s take a moment to consider the geometric interpretation of the dot product of 𝐔 and 𝐕 cross 𝐖 being zero.

The dot product of two vectors 𝐴 and 𝐡 is zero when 𝐴 and 𝐡 are perpendicular. So 𝐔 dot 𝐕 cross 𝐖 being zero tells us that 𝐔 and 𝐕 cross 𝐖 are perpendicular. The vector 𝐕 cross 𝐖 is perpendicular to both vector 𝐕 and vector 𝐖. In fact, you might know that at least when 𝐕 cross 𝐖 is nonzero, 𝐕 cross 𝐖 is the normal to the plane spanned by the vectors 𝐕 and 𝐖.

As 𝐔 is perpendicular to the normal to the plane spanned by 𝐕 and 𝐖, 𝐔 must lie in this plane. And in fact, we can see from the question that the 𝑦- components of 𝐔, 𝐕, and 𝐖 are all zero. And hence 𝐔, 𝐕, and 𝐖 all lie in the π‘₯𝑧-plane. We could have reasoned that 𝐔 dot 𝐕 cross 𝐖 is zero without doing any calculation just by using the fact that the three vectors 𝐔, 𝐕, and 𝐖 lie in the same plane.

We’ll come back to this point at the end of the video, but first let’s compute 𝐔 cross 𝐕 cross 𝐖. This is just the cross product of two vectors. It’s just that one of the vectors happens to be the cross product of two other vectors. If it’s made you happier, you could just call 𝐕 cross 𝐖 𝐴 and then we’d just be looking for the cross product of 𝐔 and 𝐴.

We compute this using determinants as before with the unit vectors 𝑖, 𝑗, and π‘˜ in the first row, the components of 𝐔 in the second row, and the components of 𝐕 cross 𝐖 or 𝐴 in the third row. Again, we expand the determinant along the first row and evaluate each two-by-two determinant in the normal way.

And writing in components, we see that the cross product of 𝐔 and 𝐕 cross 𝐖 has components negative eight, zero, four. So now we’ve found both the dot product of 𝐔 and 𝐕 cross 𝐖 and the cross product of 𝐔 and 𝐕 cross 𝐖. Now earlier, I promised some more geometric interpretations of these things. 𝐔 dot 𝐕 cross 𝐖 is called the scalar triple product of 𝐔, 𝐕, and 𝐖. And 𝐔 cross 𝐕 cross 𝐖 is the vector triple product of 𝐔, 𝐕, and 𝐖.

They are both triple products because they involve three vectors 𝐔, 𝐕, and 𝐖. But 𝐔 dot 𝐕 cross 𝐖 returns a scalar, in other words a number, in our case zero, whereas 𝐔 cross 𝐕 cross 𝐖 returns a vector, in our case the vector with components negative eight, zero, four. The scalar triple product has a nice geometric interpretation.

Much as the magnitude of the vector 𝐕 cross 𝐖 is the area of the parallelogram with sides 𝐕 and 𝐖, the absolute value of the scalar triple product of 𝐔, 𝐕, and 𝐖 is the volume of the 3D parallelepiped with edges 𝐔, 𝐕, and 𝐖. And much as the cross product of two vectors being zero or having magnitude zero tells us that those two vectors are parallel or lie on the same line.

The scalar triple product of three vectors tells us that the three vectors lie in the same plane. The vector triple product is slightly harder to interpret geometrically. But as it is perpendicular to 𝐕 cross 𝐖, you can convince yourself with a bit of work that it must lie in the same plane as 𝐕 and 𝐖. And with a bit more work, you can write it explicitly as a linear combination of the vectors 𝐕 and 𝐖.

One final thing to note here is that the cross product of 𝐔 and 𝐕 cross 𝐖 is not the same as the cross product of 𝐔 cross 𝐕 and 𝐖. Where the parentheses are really matters here. The technical term to describe this phenomenon is that the cross product of vectors is not associative. This makes it’s difference not only from normal real number multiplication but also from matrix multiplication, which is associative even though it isn’t commutative.

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