Chapter 9, Problem 39RE

(a)

To determine

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

Explanation of Solution

Given: $a=i-j$ and $b=2j-k$

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,

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

Therefore, the cross product is $i+j+2k$ .

(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,

$\begin{array}{c}u=\frac{a\times b}{\left|a\times b\right|}\\ =\frac{i+j+2k}{\sqrt{{\left(1\right)}^2+1^2+{\left(2\right)}^2}}\\ =\frac{i+j+2k}{\sqrt{6}}\end{array}$

Therefore, the required unit vector $u$ is $\frac{i+j+2k}{\sqrt{6}}$ .