0
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.