# Triple scalar product

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.

• Maybe you can assume that the component which is left is zero. – Eric Toporek Mar 24 at 8:27
• So you mean 0, i and j for u, etc? – AMU Mar 24 at 9:14
• That would result in 0 – AMU Mar 24 at 9:28
• 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 $z$. – Eric Toporek Mar 24 at 9:35
• Ah, I misunderstood what he meant. – 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.