Multiplying Vectors — Question 7

Question 7

A vector triple product needs parentheses

Use the dimensionless right-handed unit vectors $\vec a=\hat\imath$, $\vec b=\hat\jmath$ and $\vec c=\hat\jmath$. Recall $\hat\imath\times\hat\jmath=\hat k$, $\hat\jmath\times\hat k=\hat\imath$, and reversal of order changes the sign.

1. Compute $(\vec a\times\vec b)\times\vec c$ and $\vec a\times(\vec b\times\vec c)$.

2. Explain why writing $\vec a\times\vec b\times\vec c$ without parentheses is ambiguous.

3. For a separate vector $\vec d=2\hat k$, compute $\vec a\cdot(\vec b\times\vec d)$ and interpret its absolute value geometrically.

Question 7 - Solution

A vector triple product needs parentheses

(a) Perform the inner multiplication first.

$$\boxed{(\hat i\times\hat j)\times\hat j=\hat k\times\hat j=-\hat i},\qquad\hat i\times(\hat j\times\hat j)=\hat i\times\vec0=\vec0$$ For the left grouping, $(\hat\imath\times\hat\jmath)\times\hat\jmath=\hat k\times\hat\jmath=-\hat\imath$. For the right grouping, $\hat\imath\times(\hat\jmath\times\hat\jmath)=\hat\imath\times\vec0=\vec0$. The results differ.

(b) Do not assume the rules of scalar multiplication.

$$(\vec a\times\vec b)\times\vec c\ne\vec a\times(\vec b\times\vec c)\quad\text{in general}$$ The cross product is not associative. Parentheses specify different vector operations, not just a preferred calculation order. Its distributive property over sums does not imply associativity over other cross products.

(c) A dot after a cross gives a scalar.

$$\vec b\times\vec d=2\hat i,\quad\vec a\cdot(\vec b\times\vec d)=2,\quad V=|2|=2$$ $\vec b\times\vec d=\hat\jmath\times2\hat k=2\hat\imath$, so $\vec a\cdot(\vec b\times\vec d)=2$. The absolute scalar triple product is the parallelepiped volume in these dimensionless coordinates. If the three edge vectors instead carried metres, the numerical result would be $2\ \mathrm{m^3}$.