Question

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) Two nonzero vectors $\mathbf{u}$and $\mathbf{v}$in ${\mathrm{ℝ}}^3$such that $\mathbf{u}\times \mathbf{v}=\mathbf{v}\times \mathbf{u}$.

(b) Three vectors $\mathbf{u},\mathbf{v}$, and $\mathbf{w}$in ${\mathrm{ℝ}}^3$such that $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}\ne$$\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$.

(c) Three vectors $\mathbf{u},\mathbf{v}$, and $\mathbf{w}$in ${\mathrm{ℝ}}^3$such that $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}=$$\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$.

Short answer

Ans:

Part (a). The non-zero vectors $\mathbf{u}=⟨1,2,0⟩$and $\mathbf{v}=⟨2,4,0⟩$the equation $\mathbf{u}\times \mathbf{v}=\mathbf{v}\times \mathbf{u}$

Part (b). Vectors $\mathbf{u}=⟨2,1,-3⟩,\mathbf{v}=⟨4,0,1⟩$and$\mathbf{w}=⟨-2,6,5⟩$the result $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}\ne \mathbf{u}\times (\mathbf{v}\times \mathbf{w})$.

Part (c). Vectors $\mathbf{u}=⟨1,2,0⟩,\mathbf{v}=⟨2,4,0⟩$and $\mathbf{w}=⟨4,8,0⟩$the result $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}=\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$.

Step-by-step solution

Step 1. Given Information:

The non-zero vectors $\mathbf{u}$and $\mathbf{v}$.

Step 2. Solving part (a).

The non-zero vectors $\mathbf{u}=⟨1,2,0⟩$and $\mathbf{v}=⟨2,4,0⟩$.

The cross-product $\mathbf{u}\times \mathbf{v}$gives:

role="math" localid="1650771821570" $$\begin{gathered} \mathbf{u}\times \mathbf{v}=det\left[\begin{array}{l}\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 4& 0\\ 1& 2& 0\end{array}\end{array}\right] \\ =0\mathbf{i}-0\mathbf{j}-0\mathbf{k} \end{gathered}$$

The cross-product $\mathbf{v}\times \mathbf{u}$gives:

role="math" localid="1650771829212" $$\begin{gathered} \mathbf{v}\times \mathbf{u}=det\left[\begin{array}{l}\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 4& 0\\ 1& 2& 0\end{array}\end{array}\right] \\ =0\mathbf{i}-0\mathbf{j}-0\mathbf{k} \end{gathered}$$

Therefore, the cross-products $\mathbf{u}\times \mathbf{v}$and $\mathbf{v}\times \mathbf{u}$are equal.

Thus, for the non-zero vectors $\mathbf{u}=⟨1,2,0⟩$and $\mathbf{v}=⟨2,4,0⟩$the equation $\mathbf{u}\times \mathbf{v}=\mathbf{v}\times \mathbf{u}$ holds.

Step 3. Solving part (b). 

The non-zero vectors $\mathbf{u},\mathbf{v}$and $\mathbf{w}$.

The vectors $\mathbf{u}=⟨2,1,-3⟩,\mathbf{v}=⟨4,0,1⟩$and $\mathbf{w}=⟨-2,6,5⟩$.

The vectors $\mathbf{u}=⟨2,1,-3⟩$, and $\mathbf{v}=⟨4,0,1⟩$.

$\mathbf{u}\times \mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right|$

Now, calculate, $\mathbf{u}\times \mathbf{v}$.

$$\begin{gathered} \mathbf{u}\times \mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right| \\ =\mathbf{i}(1-0)-\mathbf{j}(2+12)+\mathbf{k}(0-4) \\ =\mathbf{i}-14\mathbf{j}-4\mathbf{k} \end{gathered}$$

Hence, the value of $\mathbf{u}\times \mathbf{v}$is $\mathbf{i}-14\mathbf{j}-4\mathbf{k}$

Step 4. Continue:

Now, calculate $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$

$$\begin{gathered} (\mathbf{u}\times \mathbf{v})\times \mathbf{w}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -14& -4\\ -2& 6& 5\end{array}\right| \\ =\mathbf{i}(-70+24)-\mathbf{j}(5-8)+\mathbf{k}(6-28) \\ =-46\mathbf{i}+3\mathbf{j}-22\mathbf{k} \end{gathered}$$

Hence, the value of $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$is $-46\mathbf{i}+3\mathbf{j}-22\mathbf{k}$.

Consider vectors $\mathbf{v}=⟨4,0,1⟩$and $\mathbf{w}=⟨-2,6,5⟩$

$\mathbf{v}\times \mathbf{w}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4& 0& 1\\ -2& 6& 5\end{array}\right|$

Now, calculate, $\mathbf{v}\times \mathbf{w}$.

$$\begin{gathered} \mathbf{v}\times \mathbf{w}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4& 0& 1\\ -2& 6& 5\end{array}\right| \\ =-6\mathbf{i}-22\mathbf{j}+24\mathbf{k} \end{gathered}$$

Hence, the value of$\mathbf{v}\times \mathbf{w}$is $-6\mathbf{i}-22\mathbf{j}+24\mathbf{k}$.

Step 5. Continue:

Now, calculate $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$

$$\begin{gathered} \mathbf{u}\times (\mathbf{v}\times \mathbf{w})=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 1& -3\\ -6& -22& 24\end{array}\right| \\ =\mathbf{i}(24-66)-\mathbf{j}(48-18)+\mathbf{k}(-44+6) \\ =-42\mathbf{i}-30\mathbf{j}-38\mathbf{k} \end{gathered}$$

Hence, the value of $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$is $-42\mathbf{i}-30\mathbf{j}-38\mathbf{k}$.

Therefore, for the vectors$\mathbf{u}=⟨2,1,-3⟩,\mathbf{v}=⟨4,0,1⟩$and $\mathbf{w}=⟨-2,6,5⟩$the result $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}\ne \mathbf{u}\times (\mathbf{v}\times \mathbf{w})$ holds.

Step 6. Solving part (b). 

The vectors $\mathbf{u},\mathbf{v}$and $\mathbf{w}$.

Consider the vectors $\mathbf{u}=⟨1,2,0⟩,\mathbf{v}=⟨2,4,0⟩$and $\mathbf{w}=⟨4,8,0⟩$.

Consider the vectors $\mathbf{u}=⟨1,2,0⟩$and $\mathbf{v}=⟨2,4,0⟩$.

$$\begin{gathered} \mathbf{u}\times \mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& 0\\ 2& 4& 0\end{array}\right| \\ =0\mathbf{i}+0\mathbf{j}+0\mathbf{k} \end{gathered}$$

Now, calculate $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$

$$\begin{gathered} (\mathbf{u}\times \mathbf{v})\times \mathbf{w}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 0& 0\\ -2& 6& 5\end{array}\right| \\ =0\mathbf{i}+0\mathbf{j}+0\mathbf{k} \end{gathered}$$

Hence, the value of $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$is$0\mathbf{i}+0\mathbf{j}+0\mathbf{k}$.

Step 7. Continue:

Consider vectors $\mathbf{v}=⟨2,4,0⟩$and $\mathbf{w}=⟨4,8,0⟩$

$$\begin{gathered} \mathbf{v}\times \mathbf{w}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 4& 0\\ 4& 8& 0\end{array}\right| \\ =0\mathbf{i}+0\mathbf{j}+0\mathbf{k} \end{gathered}$$

Hence, the value of $\mathbf{v}\times \mathbf{w}$is $0\mathbf{i}+0\mathbf{j}+0\mathbf{k}$.

Now, calculate $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$

$$\begin{gathered} \mathbf{u}\times (\mathbf{v}\times \mathbf{w})=\left|\begin{array}{lll}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& 0\\ 0& 0& 0\end{array}\right| \\ =0\mathbf{i}+0\mathbf{j}+0\mathbf{k} \end{gathered}$$

Hence, the value of $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$is $0\mathbf{i}+0\mathbf{j}+0\mathbf{k}$.

Therefore, for the vectors $\mathbf{u}=⟨1,2,0⟩,\mathbf{v}=⟨2,4,0⟩$and $\mathbf{w}=⟨4,8,0⟩$the result $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}=\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$ holds.