Chapter 9.5, Problem 11E

a.

To determine

To find: The perpendicular vector to the given vectors.

Answer to Problem 11E

The perpendicular vector to the given vectors is $\langle 14,7,0\rangle$ .

Explanation of Solution

Given information:

The vectoris $a=\frac{1}{2}i-j+\frac{2}{3}k,b=6i-12j-6k$ .

Calculation:

The cross product is given as,

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

Substitute the values in above equation.

  $\begin{array}{c}a\times b=\left|\begin{array}{ccc}i& j& k\\ \frac{1}{2}& -1& \frac{2}{3}\\ 6& -12& -6\end{array}\right|\\ =14i+7j+0k\\ =\langle 14,7,0\rangle \end{array}$

b.

To determine

To find: The unit perpendicular vector to the given vectors.

Answer to Problem 11E

The unit perpendicular vector to the given vectors is $\frac{2i+j}{\sqrt{5}}$ .

Explanation of Solution

Given information:

The vectoris $a=\frac{1}{2}i-j+\frac{2}{3}k,b=6i-12j-6k$ .

Calculation:

The unit perpendicular vector is given as,

  $\begin{array}{c}p=\frac{a\times b}{\left|a\times b\right|}\\ =\frac{14i+7j+0k}{\sqrt{{\left(14\right)}^2+{\left(7\right)}^2+{\left(0\right)}^2}}\\ =\frac{2i+j}{\sqrt{5}}\end{array}$