Chapter 12, Problem 24CLT

To determine

TO-DETERMINE whether vectors $u=\langle -1,6,-3\rangle$ and $v=\langle 3,-18,9\rangle$ are orthogonal, parallel or neither.

Answer to Problem 24CLT

Vectors $u=\langle -1,6,-3\rangle$ and $v=\langle 3,-18,9\rangle$ are parallel but not orthogonal.

Explanation of Solution

Given information:

Consider the vectors $u=\langle -1,6,-3\rangle$ and $v=\langle 3,-18,9\rangle$ .

Concept used:

The dot product of two nonzero vectors is zero, then the angle between the vectors is 90° and the vectors are orthogonal.

Two nonzero vectors u and v are parallel when there is some scalar c such that $u=cv$ .

Calculations:

In order to find out the vectors are orthogonal or not, compute the dot product:

  $\begin{array}{c}u\cdot v= \langle -1,6,-3\rangle \cdot \langle 3,-18,9\rangle \\ =\left(-1\times 3\right)+\left(6\times \left(-18\right)\right)+\left(-3\times 9\right)\\ =-3-108-27\\ =-138\end{array}$

Since, $u\cdot v\ne 0$ , therefore, the vectors are not orthogonal.

In order to find out the vectors are parallel or not, let there be some scalar c such that $u=cv$ , then,

  $\begin{array}{c}\langle -1,6,-3\rangle =c\langle 3,-18,9\rangle \\ =\langle 3c,-18c,9c\rangle \end{array}$

Now compareboth sides,

  $-1=3c\Rightarrow c=-\frac{1}{3}$

  $6=-18c\Rightarrow c=-\frac{6}{18}=-\frac{1}{3}$

  $-3=9c\Rightarrow c=-\frac{3}{9}=-\frac{1}{3}$

Since there exist a scalar $c=-\frac{1}{3}$ , $u=cv$ , therefore, the vectors are parallel.