Question

Let u, v, and w be vectors in ${\mathrm{ℝ}}^3$. Prove that $u·(v\times w)=(u\times v)·w$.

(This is part (b) of Theorem 10.37.)

Short answer

Hence, we prove that$u·(v\times w)=(u\times v)·w$

Step-by-step solution

Step 1. Given Information

Let u, v and w be vectors in ${\mathrm{ℝ}}^3$. Prove that $u·(v\times w)=(u\times v)·w$.

Step 2. Let u=(−3,1,−4), v=(2,0,5), and w=(1,3,13)

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

$$\begin{gathered} u·(v\times w)=\det \left[\begin{array}{ccc}-3& 1& -4\\ 2& 0& 5\\ 1& 3& 13\end{array}\right] \\ u·(v\times w)=-3\left|\begin{array}{cc}0& 5\\ 3& 13\end{array}\right|-1\left|\begin{array}{cc}2& 5\\ 1& 13\end{array}\right|-4\left|\begin{array}{cc}2& 0\\ 1& 3\end{array}\right| \\ u·(v\times w)=-3(0\times 13-5\times 3)-1(2\times 13-5\times 1)-4(2\times 3-0\times 1) \\ u·(v\times w)=-3(0-15)-1(26-5)-4(6-0) \\ u·(v\times w)=45-21-24 \\ u·(v\times w)=0 \end{gathered}$$

Step 3. Now finding the value of (u×v)·w

$$\begin{gathered} (u\times v)·w=\det \left[\begin{array}{ccc}-3& 1& -4\\ 2& 0& 5\\ 1& 3& 13\end{array}\right] \\ (u\times v)·w=-3\left|\begin{array}{cc}0& 5\\ 3& 13\end{array}\right|-1\left|\begin{array}{cc}2& 5\\ 1& 13\end{array}\right|-4\left|\begin{array}{cc}2& 0\\ 1& 3\end{array}\right| \\ (u\times v)·w=-3(0\times 13-5\times 3)-1(2\times 13-5\times 1)-4(2\times 3-0\times 1) \\ (u\times v)·w=-3(0-15)-1(26-5)-4(6-0) \\ (u\times v)·w=45-21-24 \\ (u\times v)·w=0 \end{gathered}$$

Hence, prove that$u·(v\times w)=(u\times v)·w$