Question

Let $\mathbf{u},\mathbf{v}$, and $\mathbf{w}$be three mutually perpendicular vectors in ${\mathrm{ℝ}}^3$.

(a) Prove that $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=\mathbf{0}$.

(b) Show that $|\mathbf{u}·(\mathbf{v}\times \mathbf{w}\left)\right|=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert$.

Short answer

Ans:

part (a).

$$\begin{gathered} \mathbf{u}\times (\mathbf{v}\times \mathbf{w})=\left(\Vert \mathbf{u}\Vert \Vert \mathbf{w}\Vert \cos 90°\right)\mathbf{v}-\left(\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos 90°\right)\mathbf{w} \\ =\left(0\right)\mathbf{v}-\left(0\right)\mathbf{w} \\ =0 \end{gathered}$$

part (b)

.$$\begin{gathered} \mathbf{u}\mathbf{·}\mathbf{(}\mathbf{v}\mathbf{\times }\mathbf{w}\mathbf{)}\mathbf{=}\mathbf{\Vert }\mathbf{u}\mathbf{\Vert }\mathbf{\Vert }\mathbf{v}\mathbf{\times }\mathbf{w}\mathbf{\Vert }\mathbf{cos}\mathit{\theta } \\ \mathbf{=}\mathbf{\Vert }\mathbf{u}\mathbf{\Vert }\mathbf{\Vert }\mathbf{v}\mathbf{\times }\mathbf{w}\mathbf{\Vert }\mathbf{cos}\mathbf{0}\mathbf{°} \\ \mathbf{=}\mathbf{\Vert }\mathbf{u}\mathbf{\Vert }\mathbf{\Vert }\mathbf{v}\mathbf{\times }\mathbf{w}\mathbf{\Vert } \\ \mathbf{=}\mathbf{\Vert }\mathbf{u}\mathbf{\Vert }\mathbf{\Vert }\mathbf{v}\mathbf{\Vert }\mathbf{\Vert }\mathbf{w}\mathbf{\Vert }\mathbf{sin}\mathbf{90}\mathbf{°} \\ \mathbf{=}\mathbf{\Vert }\mathbf{u}\mathbf{\Vert }\mathbf{\Vert }\mathbf{v}\mathbf{\Vert }\mathbf{\Vert }\mathbf{w}\mathbf{\Vert } \end{gathered}$$

Step-by-step solution

Step 1. Given information:

There are three mutually perpendicular vectors in ${\mathrm{ℝ}}^3$.

• $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=\mathbf{0}$
  
• $|\mathbf{u}·(\mathbf{v}\times \mathbf{w}\left)\right|=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert$
  

Step 2. Solving part (a):

The equation $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}$gives:

$$\begin{gathered} \mathbf{u}\times (\mathbf{v}\times \mathbf{w})=\left(\Vert \mathbf{u}\Vert \Vert \mathbf{w}\Vert \cos 90°\right)\mathbf{v}-\left(\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos 90°\right)\mathbf{w}\text{(Vectors are perpendicular)} \\ =\left(0\right)\mathbf{v}-\left(0\right)\mathbf{w}\text{(Simplify)} \\ =0 \end{gathered}$$

Therefore, if three vectors are mutually perpendicular then $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=0$holds.

Step 3. Solving part (b).

The vectors $\mathbf{u}$and $\mathbf{v}\times \mathbf{w}$are parallel to each other.

The value of $\mathbf{u}·(\mathbf{v}\times \mathbf{w})$is:

role="math" localid="1650748918202" $$\begin{gathered} \mathbf{u}·(\mathbf{v}\times \mathbf{w})=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\times \mathbf{w}\Vert \cos \theta \\ \left.=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\times \mathbf{w}\Vert \cos 0°\text{(Angle between}\mathbf{u}\text{and}\mathbf{v}\times \mathbf{w}\right) \\ =\Vert \mathbf{u}\Vert \Vert \mathbf{v}\times \mathbf{w}\Vert \\ =\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert \sin 90°\text{(Because vectors}\mathbf{v}\mathbf{}\mathbf{and}\mathbf{}\mathbf{w}\text{are perpendicular)} \\ =\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert \end{gathered}$$

Therefore, the result $\mathbf{u}·(\mathbf{v}\times \mathbf{w})=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \Vert \mathbf{w}\Vert$ is proved.