Question

Let u, v and w be vectors in ${\mathrm{ℝ}}^3$. Prove:

role="math" localid="1649919341939" $u\times (v+w)=u\times v+u\times w\mathrm{and}(u+v)\times w=u\times w+v\times w$

(This is Theorem 10.29.)

Short answer

Hence, we prove that$u\times (v+w)=u\times v+u\times w\mathrm{and}(u+v)\times w=u\times w+v\times w$

Step-by-step solution

Step 1. Given Information

Let u, v and w be vectors in ${\mathrm{ℝ}}^3$. Prove:

$u\times (v+w)=u\times v+u\times w\mathrm{and}(u+v)\times w=u\times w+v\times w$

Step 2. We have to prove u×(v+w)=u×v+u×w and (u+v)×w=u×w+v×w

Let $u=(1,0,-8),v=(0,1,6)\mathrm{and}w=(-1,9,3)$

Firstly proving role="math" localid="1649920209467" $u\times (v+w)=u\times v+u\times w$

Now finding the value of $u\times (v+w)$

role="math" localid="1649921394781" $$\begin{gathered} u\times (v+w)=(1,0,-8)\times \left\{(0,1,6)+(-1,9,3)\right\} \\ u\times (v+w)=(1,0,-8)\times (0+(-1),1+9,6+3) \\ u\times (v+w)=(1,0,-8)\times (-1,10,9) \\ u\times (v+w)=\left[\begin{array}{ccc}i& j& k\\ 1& 0& -8\\ -1& 10& 9\end{array}\right] \\ u\times (v+w)=i\left|\begin{array}{cc}0& -8\\ 10& 9\end{array}\right|-j\left|\begin{array}{cc}1& -8\\ -1& 9\end{array}\right|+k\left|\begin{array}{cc}1& 0\\ -1& 10\end{array}\right| \\ u\times (v+w)=i(9-(-80\left)\right)-j(9-8)+k(10-0) \\ u\times (v+w)=89i-j+10k \end{gathered}$$

Step 3. Now finding the value of u×v+u×w

$$\begin{gathered} u\times v+u\times w=\left[\begin{array}{ccc}i& j& k\\ 1& 0& -8\\ 0& 1& 6\end{array}\right]+\left[\begin{array}{ccc}i& j& k\\ 1& 0& -8\\ -1& 9& 3\end{array}\right] \\ u\times v+u\times w=\left[i\left|\begin{array}{cc}0& -8\\ 1& 6\end{array}\right|-j\left|\begin{array}{cc}1& -8\\ 0& 6\end{array}\right|+k\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|\right]+\left[i\left|\begin{array}{cc}0& -8\\ 9& 3\end{array}\right|-j\left|\begin{array}{cc}1& -8\\ -1& 3\end{array}\right|+k\left|\begin{array}{cc}1& 0\\ -1& 9\end{array}\right|\right] \\ u\times v+u\times w=\left\{i(6-(-8\left)\right)-j(6-0)+k(1-0)\right\}+\left\{i(0-(-72\left)\right)-j(3-8)+k(9-0)\right\} \\ u\times v+u\times w=14i-6j+k+72i+5j+9k \\ u\times v+u\times w=89i-j+10k \end{gathered}$$

Now we prove that$u\times (v+w)=u\times v+u\times w$

Step 4. Now proving (u+v)×w=u×w+v×w

Firstly finding the value of $(u+v)\times w$

role="math" localid="1649922089509" $$\begin{gathered} (u+v)\times w=\left\{(1,0,-8)+(0,1,6)\right\}\times (-1,9,3) \\ (u+v)\times w=(1+0,0+1,-8+6)\times (-1,9,3) \\ (u+v)\times w=(1,1,-2)\times (-1,9,3) \\ (u+v)\times w=\left[\begin{array}{ccc}i& j& k\\ 1& 1& -2\\ -1& 9& 3\end{array}\right] \\ (u+v)\times w=i\left|\begin{array}{cc}1& -2\\ 9& 3\end{array}\right|-j\left|\begin{array}{cc}1& -2\\ -1& 3\end{array}\right|+k\left|\begin{array}{cc}1& 1\\ -1& 9\end{array}\right| \\ (u+v)\times w=i(3-(-18\left)\right)-j(3-2)+k(9-(-1\left)\right) \\ (u+v)\times w=21i-j+10k \end{gathered}$$

Step 3. Now finding the value of u×w+v×w

$$\begin{gathered} u\times w+v\times w=\left[\begin{array}{ccc}i& j& k\\ 1& 0& -8\\ -1& 9& 3\end{array}\right]+\left[\begin{array}{ccc}i& j& k\\ 0& 1& 6\\ -1& 9& 3\end{array}\right] \\ u\times w+v\times w=\left[i\left|\begin{array}{cc}0& -8\\ 9& 3\end{array}\right|-j\left|\begin{array}{cc}1& -8\\ -1& 3\end{array}\right|+k\left|\begin{array}{cc}1& 0\\ -1& 9\end{array}\right|\right]+\left[i\left|\begin{array}{cc}1& 6\\ 9& 3\end{array}\right|-j\left|\begin{array}{cc}0& 6\\ -1& 3\end{array}\right|+k\left|\begin{array}{cc}0& 1\\ -1& 9\end{array}\right|\right] \\ u\times w+v\times w=\left\{i(0-(-72\left)\right)-j(3-8)+k(9-0)\right\}+\left\{i(3-54)-j(0-(-6\left)\right)+k(0-(-1\left)\right)\right\} \\ u\times w+v\times w=72i+5j+9k+51i-6j+k \\ u\times w+v\times w=21i-j+10k \end{gathered}$$

Now we prove that$(u+v)\times w=u\times w+v\times w$