In Exercises 30–35 compute the indicated quantities when $u=(-3,1,-4),v=(2,0,5),\mathrm{and}w=(1,3,13)$

role="math" localid="1649436488889" $(u\times v)·w\mathrm{and}u·(v\times w)$

The indicated quantities when $u=(-3,1,-4),v=(2,0,5),\mathrm{and}w=(1,3,13)$

We have to find the value of$(u\times v)·w\mathrm{and}u·(v\times w)$

To just take the absolute value of the determinant of the 3 × 3 matrix formed from the components of u, v and w as the rows.

The required cross product of $(u\times v)·w$

$(u\times v)·w=\det \left[\begin{array}{ccc}-3& 1& -4\\ 2& 0& 5\\ 1& 3& 13\end{array}\right]$

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

To just take the absolute value of the determinant of the 3 × 3 matrix formed from the components of u, v and w as the rows.

The required cross product of $u·(v\times w)$

$u·(v\times w)=\det \left[\begin{array}{ccc}-3& 1& -4\\ 2& 0& 5\\ 1& 3& 13\end{array}\right]$

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