Video Transcript
Find the sum of the scalar triple
product π’ dot π£ cross π€ plus π£ dot π€ cross π’ plus π€ dot π’ cross π£.
Weβre given the sum of three scalar
triple products of the same three unit vectors π’, π£, and π€, where these are the
unit vectors in the π₯-, π¦-, and π§-directions. These three scalar triple products
are in different orders. The first is a scalar product of π’
with π£ cross π€. The second is a scalar product of
π£ with π€ cross π’. And the third is the scalar product
of π€ with π’ cross π£. 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 π’ to π£ to π€. And now if we look at the second
triple product, we have π£ to π€ to π’, which is in the same cycling direction. And the third triple product goes
from π€ to π’ to π£, 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 π£ cross π€ is equal to π’ and
similarly π€ cross π’ is equal to π£ and that π’ cross π£ is equal to π€. This then tells us that the scalar
product of π’ with the cross product of π£ and π€ is simply the scalar product of π’
with itself. And we know that this is simply the
modulus or the magnitude of π’ squared. And since π’ is a unit vector,
thatβs 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, theyβre 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 π’ with π£ cross π€ 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.
Itβs 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 theyβre all in the same cyclic
direction. So by finding one of our
determinants, multiply this by three. And again our answer is three.
