So I came across this question: Given vector $\textbf{u} = i+j, \textbf{v} = j+k, \textbf{w} = i+k$. Find the triple scalar product $u(\textbf{v}\times \textbf{w})$.

So I tried to check my notes to see if I can solve it myself. And according to my notes, the triple scalar product of vectors $\textbf{u} = i+j+k, \textbf{v} = i+j+k \text{ and } \textbf{w} = i+j+k$ is the determinant of the $3\times3$ matrix formed by the components of the vector.

If for my case I only have $2$ for each vector, does that mean the matrix is $2\times3$? I am kinda confused.

  • 2
    $\begingroup$ Maybe you can assume that the component which is left is zero. $\endgroup$ – Eric Toporek Mar 24 at 8:27
  • $\begingroup$ So you mean 0, i and j for u, etc? $\endgroup$ – AMU Mar 24 at 9:14
  • $\begingroup$ That would result in 0 $\endgroup$ – AMU Mar 24 at 9:28
  • $\begingroup$ It's not 0. The question was already answered under the same argument. Remember that $i$ is related to the $x-axis$, $j$ to the $y$ and $k$ to $z$. $\endgroup$ – Eric Toporek Mar 24 at 9:35
  • $\begingroup$ Ah, I misunderstood what he meant. $\endgroup$ – AMU Mar 24 at 11:15

Those vectors should be interpreted as \begin{align*}u &= 1\cdot\vec{i} + 1\cdot \vec{j}+ 0\cdot\vec{k}\\ v &= 0\cdot\vec{i} + 1\cdot \vec{j} + 1\cdot\vec{k}\\ w &= 1\cdot\vec{i} + 0\cdot \vec{j}+ 1\cdot\vec{k}\end{align*} so that the coefficient matrix we'll be taking the determinant of is $\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}$.

The other components are always there, even if their coefficients are zero. These are vectors in $\mathbb{R}^3$, so we use coefficients with respect to the standard basis $\vec{i},\vec{j},\vec{k}$, and that means all three coefficients to express any particular vector.

  • $\begingroup$ So i(jk-0)-j(0-ik)+0(0-ij), then 2ijk? $\endgroup$ – AMU Mar 24 at 11:16
  • $\begingroup$ No. The scalar triple product is just a number. There is no generic product that we would ever write as "ijk". $\endgroup$ – jmerry Mar 24 at 11:27
  • $\begingroup$ So the answer should be a real number? $\endgroup$ – AMU Mar 25 at 11:25

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