Chapter 11.2, Problem 45E

To determine

TO-DETERMINE whether vectors $u=\langle -12,6,15\rangle$ and $v=\langle 8,-4,-10\rangle$ are orthogonal, parallel or neither.

Answer to Problem 45E

Vectors $u=\langle -12,6,15\rangle$ and $v=\langle 8,-4,-10\rangle$ are parallel.

Explanation of Solution

Given information:

Consider the vectors $u=\langle -12,6,15\rangle$ and $v=\langle 8,-4,-10\rangle$ .

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:

  $\begin{array}{c}u\cdot v=\langle -12,6,15\rangle \cdot \langle 8,-4,-10\rangle \\ =\left(-12\times 8\right)+\left(6\times \left(-4\right)\right)+\left(15\times \left(-10\right)\right)\\ =-96-24-150\\ =-270\end{array}$

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,

  $\begin{array}{c}\langle -12,6,15\rangle =c\langle 8,-4,-10\rangle \\ =\langle 8c,-4c,-10c\rangle \end{array}$

Compare the coefficients and get,

  $-12=8c\Rightarrow c=\frac{-12}{8}=-\frac{3}{2}$

  $6=-4c\Rightarrow c=\frac{6}{-4}=-\frac{3}{2}$

And,

  $15=-10c\Rightarrow c=-\frac{15}{10}=-\frac{3}{2}$

Since, $u=cv$ ,where $c=-\frac{3}{2}$ , therefore the vectors are parallel.