Chapter 9, Problem 40RE

(a)

To determine

To find:The cross product of $a$ and $b$ .

Explanation of Solution

Given: $a=i+j-k$ and $b=i-j+k$

The formula for the cross product vector of two vector is,

$u\times v=\left|\begin{array}{ccc}i& j& k\\ u_x& u_y& u_z\\ v_x& v_y& v_z\end{array}\right|$

The cross product of the given vector is calculated as,

$\begin{array}{c}a\times b=\left|\begin{array}{ccc}i& j& k\\ 1& 1& -1\\ 1& -1& 1\end{array}\right|\\ =\left(0\right)i-\left(1+1\right)j+\left(-1-1\right)k\\ =-2j-2k\end{array}$

Therefore, the cross product is $-2j-2k$ .

(b)

To determine

To find :A unit vector $u$ perpendicular to $a$ and $b$ .

Explanation of Solution

The unit vector perpendicular to $a$ and $b$ is calculated as,

$\begin{array}{c}u=\frac{a\times b}{\left|a\times b\right|}\\ =\frac{-2j-2k}{\sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2}}\\ =\frac{-2j-2k}{2\sqrt{2}}\end{array}$

Therefore, the required unit vector $u$ is $\frac{-2j-2k}{2\sqrt{2}}$ .