We have vectors $x, y, z$ where $z = x \times y$.
What is $x \cdot z$?
From my intuition, the cross product is perpendicular to both vectors, so dot product should be 0?
vector times cross product
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linear-algebra
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1You are right, and that's in fact one of the basic properties of the vector [scalar triple product](https://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product). – 2017-01-22
3 Answers
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Correct. If $x$ is not collinear to $y$, $z$ defines the direction of a normal vector to the plane that both $x$ and $y$ lie in. So $z$ is normal to both. In the case where $x$ and $y$ are collinear (one is a scalar multiple of the other), then the cross product between them is the null vector, so the assertion is trivially true.
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As defined $z$ will be perpendicular to both $x$ and $y$ from the definition of cross product and hence $x.z = 0$ since $z$ is perpendicular to both $x$ and $y$ and as dot product of any two perpendicular vectors is zero due to the involvement of $cos(90^{\circ})$ term.
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The geometric reasoning you mentioned is correct, and it probably the best way to understand what's going on.
Another option is as follows: We know that $x \cdot z = x \cdot (x \times y)$. In general, the value of the triple product $u \cdot (v \times w)$ is given by the determinant that has $u$, $v$, $w$ as its rows. But, in our case, since $x$ occurs twice in our triple product, the determinant will have two identical rows, so its value will be zero. In short $$ x \cdot z = x \cdot (x \times y) = \det \left[\begin{matrix} \leftarrow & x & \rightarrow \\ \leftarrow & x & \rightarrow \\ \leftarrow & y & \rightarrow \\ \end{matrix} \right] = 0 $$