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Found in: Page 556

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# To find a unit vector $$\left( a \right).$$

The unit vector $${\rm{a}} = \frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}$$ and $${\rm{a}} = - \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{{\rm{l}}}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}$$.

See the step by step solution

## Step 1: Concept of three dimensional vector

Formula:

Consider three dimensional vector $${\bf{a}}$$ as follows.

$${\rm{a}} = {a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}$$

## Step 2: Calculation for a unit vector as$${\rm{i}} + {\rm{j}}$$

As a unit vector is orthogonal to $${\rm{i}} + {\rm{j}}$$ that is, $$a \cdot (i + j) = 0$$

Substitute $${a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}$$ for $${\rm{a}}$$.

Rearrange the equation.

$${a_1} = - {a_2}$$

## Step 3: Calculation for a unit vector as$${\rm{i}} + k$$

As a unit vector is orthogonal to $${\rm{i}} + {\rm{k}}$$ that is, $${\rm{a}} \cdot ({\rm{i}} + {\rm{k}}) = 0$$

Substitute $${a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}$$ for $${\rm{a}}$$.

Rearrange the equation.

$${a_1} = - {a_3}$$

## Step 4: Calculation of the magnitude of$$a$$

Compare the above equations of$${a_1}$$.

\$$${a_1} = - {a_2} = - {a_3}$$

Find magnitude of equation$$a$$,

$$|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2}$$

Substitute $$1$$ for $$|{\rm{a}}|, - {a_1}$$ for $${a_2}$$ and $$- {a_1}$$ for $${a_3}$$.

\begin{aligned}{l}1 &= \sqrt {a_1^2 + {{\left( { - {a_1}} \right)}^2} + {{\left( { - {a_1}} \right)}^2}} \\1 &= \sqrt {a_1^2 + a_1^2 + a_1^2} \\1 &= \sqrt {3a_1^2} \\1 &= 3a_1^2\end{aligned}

Rearrange the equation.

\begin{aligned}{l}{a_1} &= \sqrt {\frac{1}{3}} \\{a_1} &= \pm \frac{1}{{\sqrt 3 }}\end{aligned}

In equation$${a_1}$$,$$\pm \frac{1}{{\sqrt 3 }}$$for $${a_1}$$. $${a_2} = \mp \frac{1}{{\sqrt 3 }}$$

In equation$${a_1}$$ , substitute $$\pm \frac{1}{{\sqrt 3 }}$$ for $${a_1}$$.

$${a_3} = \mp \frac{1}{{\sqrt 3 }}$$

## Step 5: Substitution of the values of$${a_1},{a_2}$$

In equation of$$a$$, substitute $$\frac{1}{{\sqrt 3 }}$$ for $${a_1}, - \frac{1}{{\sqrt 3 }}$$ for $${a_2}$$ and $$- \frac{1}{{\sqrt 3 }}{a_2}$$. $$a = \frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}$$

Thus, the value of unit vector $$a$$ is $$\frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}$$.

In equation $$a$$,substitute $$- \frac{1}{{\sqrt 3 }}$$ for $${a_1},\frac{1}{{\sqrt 3 }}$$ for $${a_2}$$ and $$\frac{1}{{\sqrt 3 }}{a_2}$$. $$a = - \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}$$

Thus, the value of unit vector $$a$$ is $$- \frac{1}{{\sqrt 3 }}i + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}$$.