The portal has been deactivated. Please contact your portal admin.

Question Video: Finding the Value of an Expression of Vectors Using the Scalar Triple Product Mathematics

Find 𝐒 β‹… (𝐣 Γ— 𝐀) + 𝐣 β‹… (𝐀 Γ— 𝐒) + 𝐀 β‹… (𝐒 Γ— 𝐣).

02:33

Video Transcript

Find the sum of the scalar triple product 𝐒 dot 𝐣 cross 𝐀 plus 𝐣 dot 𝐀 cross 𝐒 plus 𝐀 dot 𝐒 cross 𝐣.

We’re given the sum of three scalar triple products of the same three unit vectors 𝐒, 𝐣, and 𝐀, where these are the unit vectors in the π‘₯-, 𝑦-, and 𝑧-directions. These three scalar triple products are in different orders. The first is a scalar product of 𝐒 with 𝐣 cross 𝐀. The second is a scalar product of 𝐣 with 𝐀 cross 𝐒. And the third is the scalar product of 𝐀 with 𝐒 cross 𝐣. We do know, however, that because these three scalar triple products have the same three vectors, the absolute values of the scalar triple products will be the same.

We also know that only if the three vectors are in the same cyclic order will they have the same sign. In our case, the cyclic order of the first triple product is 𝐒 to 𝐣 to 𝐀. And now if we look at the second triple product, we have 𝐣 to 𝐀 to 𝐒, which is in the same cycling direction. And the third triple product goes from 𝐀 to 𝐒 to 𝐣, which again is in the same cyclic direction.

And because these are all in the positive direction, if we find one of the scalar triple products, we simply multiply our result by three, since they all have the same magnitude. Now using our cyclic diagram on the right, from the cyclic properties, this tells us that 𝐣 cross 𝐀 is equal to 𝐒 and similarly 𝐀 cross 𝐒 is equal to 𝐣 and that 𝐒 cross 𝐣 is equal to 𝐀. This then tells us that the scalar product of 𝐒 with the cross product of 𝐣 and 𝐀 is simply the scalar product of 𝐒 with itself. And we know that this is simply the modulus or the magnitude of 𝐒 squared. And since 𝐒 is a unit vector, that’s equal to one.

And recall that all of our scalar triple products have the same magnitude. And since all three are in the same direction, they’re all equal. So we can say that the sum of our three scalar triple products is three times one of the scalar triple products. And since we found that the scalar triple product of 𝐒 with 𝐣 cross 𝐀 is equal to one, we have three times one, which is equal to three. The sum of our three scalar triple product is therefore three.

It’s worth noting that we could also have used the determinant method to work this out. We again use the fact that their magnitudes are all equal to one and that they’re all in the same cyclic direction. So by finding one of our determinants, multiply this by three. And again our answer is three.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.