# Show that $u · (u × v) = 0$

I'm unsure of how to show the following without using an example which I presume isn't the correct to show this.

The question is

Let $$u = (u_1, u_2, u_3)$$ and $$v = (v_1, v_2, v_3)$$ where $$u$$ & $$v$$ are vectors in $$\Bbb{R}^3$$.

Show that $$u · (u × v) = 0$$.

I understand how to do the dot and cross product of vectors, however, I'm not sure how to show that this equals $$0$$.

• Step 1: write down the definition of $u\times v$ Step 2: write down inner product of $u$ and $u\times v$. This is something that you can do yourself. – Kavi Rama Murthy Oct 8 '19 at 7:28

Hint. $$x\times y$$ is perpendicular (orthogonal) to both $$x$$ and $$y$$. If $$x$$ and $$y$$ are perpendicular (orthogonal), then $$x\cdot y=0$$. This is true of any vectors $$x,y\in\mathbb R^3$$.

First, calculate $$x=u\times v$$. That is, calculate $$x_1,x_2,x_3$$ such that $$(x_1,x_2,x_3)=u\times v$$.

Second, calculate $$u\cdot x$$, that is, $$u\cdot x = u_1x_1+u_2x_2+u_3x_3$$ and prove it is $$0$$.

The triple product of three vectors in $$\Bbb R^3$$ can be written as a determinant: $${\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\det {\begin{bmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\\\end{bmatrix}}={\rm {det}}\left(\mathbf {a} ,\mathbf {b} ,\mathbf {c} \right).}$$ This implies that $$\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c}) = 0$$ if any two of the three vectors are equal. This is also obvious from the geometric interpretation as the (signed) volume of a parallelepiped.

The triple product does also not change if the operands are shifted in circular order: $${\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\mathbf {b} \cdot (\mathbf {c} \times \mathbf {a} )=\mathbf {c} \cdot (\mathbf {a} \times \mathbf {b} )}$$ so if you know that the cross product of a vector $$\mathbf {u}$$ with itself is the zero vector then $$\mathbf {u} \cdot (\mathbf {u} \times \mathbf {v} ) = \mathbf {v} \cdot (\mathbf {u} \times \mathbf {u} ) = \mathbf {v} \cdot \mathbf {0} = 0 \, .$$

Generally, the triple product of three vectors in $$\Bbb R^3$$ is zero if and only if the vectors are linearly dependent (as can be seen from the determinant formula).

Here, u×v will give vector which have direction perpendicular to both u and v ,so after this cross multiplication resultant vector's dot product with any of u or v will give zero.