Chapter 12, Problem 23CLT

To determine

To find whether vectors $u=\langle 4,-2,10\rangle$ and $v=\langle -2,6,2\rangle$ are orthogonal, parallel or neither.

Answer to Problem 23CLT

Vectors $u=\langle 4,-2,10\rangle$ and $v=\langle -2,6,2\rangle$ are orthogonal but not parallel.

Explanation of Solution

Given information:

Consider the vectors $u=\langle 4,-2,10\rangle$ and $v=\langle -2,6,2\rangle$ .

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:

  $\begin{array}{c}u\cdot v= \langle 4,-2,10\rangle \cdot \langle -2,6,2\rangle \\ =\left(4\times \left(-2\right)\right)+\left(-2\times \left(6\right)\right)+\left(10\times 2\right)\\ =-8-12+20\\ =0\end{array}$

Since, $u\cdot v=0$ , therefore, the vectors areorthogonal.

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

  $\begin{array}{c}\langle 4,-2,10\rangle =c\langle -2,6,2\rangle \\ =\langle -2c,6c,2c\rangle \end{array}$

Now compare both sides,

  $4=-2c\Rightarrow c=-2$

  $-2=6c\Rightarrow c=-\frac{1}{3}$

  $10=2c\Rightarrow c=5$

This is not possible.

Since there doesn’t exist a scalar $c$ , such that, $u=cv$ , therefore, the vectors are not parallel.